πŸ“–Topic Explanations

🌐 Overview
Hello students! Welcome to Solution of a Homogeneous Differential Equation! Get ready to unlock a powerful tool that describes how things change and evolve in the world around us.

Have you ever wondered how scientists predict the trajectory of a rocket, the spread of a disease, or the cooling of a hot object? Often, the answer lies in understanding differential equations. These are not just any equations; they are mathematical statements that involve not just variables, but also their derivatives, representing rates of change. They are, in essence, the language of dynamic systems, allowing us to model and predict how quantities change over time or space.

Within the vast and fascinating universe of differential equations, we encounter various types, each with its own unique characteristics and solution methods. Today, we're focusing on a specific, yet incredibly significant category: homogeneous differential equations. Don't let the term "homogeneous" intimidate you; in this context, it refers to a particular structural property within the equation that makes it solvable using a clever and elegant substitution technique. It's like finding a secret key that unlocks a specific type of mathematical puzzle!

Mastering the solution of homogeneous differential equations is a cornerstone of your differential equations chapter. For your JEE Main & Advanced exams, questions involving these types are frequent and require a solid understanding of both identification and systematic solution techniques. Similarly, for your board exams, a clear, step-by-step approach to solving these equations is often tested and can fetch you full marks. It's a fundamental skill that significantly builds your analytical prowess and problem-solving toolkit.

In this section, we will embark on an exciting journey to:
* Identify what makes a differential equation "homogeneous" and how to recognize it.
* Understand the ingenious method of substitution that simplifies these equations.
* Systematically solve homogeneous differential equations to find their general and particular solutions.

This topic will not only enhance your problem-solving skills but also provide a deeper appreciation for the mathematical elegance used to model real-world phenomena. So, buckle up and prepare to master this crucial concept – it's an essential step in your journey through advanced mathematics!
πŸ“š Fundamentals
Hello, aspiring mathematicians! Welcome to this deep dive into understanding and solving a very important type of differential equation: the Homogeneous Differential Equation. Don't let the name intimidate you; by the end of this session, you'll find them quite approachable!

We're going to start from scratch, building our intuition step-by-step. So, grab your virtual pen and paper, and let's unravel this together!

### What's a Homogeneous Function? (The Prerequisite!)

Before we tackle homogeneous differential equations, we first need to understand what a "homogeneous function" is. Think of it as a special kind of function where all its terms behave similarly under scaling.

Imagine you have a function of two variables, say `f(x, y)`. This function is called a homogeneous function of degree 'n' if, when you replace `x` with `tx` and `y` with `ty` (where 't' is any non-zero constant), the 't' factor comes out as `t^n`.

Mathematically, it looks like this:

f(tx, ty) = tn f(x, y)



Here, 'n' is the degree of homogeneity.

Let's look at some examples to build intuition:


  1. Example 1: Consider `f(x, y) = x^2 + y^2`.

    Let's replace `x` with `tx` and `y` with `ty`:

    `f(tx, ty) = (tx)^2 + (ty)^2 = t^2x^2 + t^2y^2 = t^2(x^2 + y^2)`

    Notice that `t^2` came out, and `(x^2 + y^2)` is our original `f(x, y)`.

    So, `f(tx, ty) = t^2 f(x, y)`.

    This means `f(x, y) = x^2 + y^2` is a homogeneous function of degree 2.


  2. Example 2: Consider `g(x, y) = x^3 + xy^2`.

    Replace `x` with `tx` and `y` with `ty`:

    `g(tx, ty) = (tx)^3 + (tx)(ty)^2 = t^3x^3 + t * x * t^2 * y^2 = t^3x^3 + t^3xy^2 = t^3(x^3 + xy^2)`

    So, `g(tx, ty) = t^3 g(x, y)`.

    This is a homogeneous function of degree 3.


  3. Example 3: Consider `h(x, y) = sin(y/x)`.

    Replace `x` with `tx` and `y` with `ty`:

    `h(tx, ty) = sin((ty)/(tx)) = sin(y/x)`

    Here, `t` effectively cancelled out. We can write this as `t^0 * sin(y/x)`.

    So, `h(tx, ty) = t^0 h(x, y)`.

    This is a homogeneous function of degree 0. Functions of the form `f(y/x)` or `f(x/y)` are always homogeneous of degree zero.



Quick Check: Non-Homogeneous Function


Consider `k(x, y) = x^2 + y`.
`k(tx, ty) = (tx)^2 + (ty) = t^2x^2 + ty`.
Can you factor out a single `t^n` completely from this expression? No! You have `t^2` in one term and `t` in another. So, `k(x, y)` is not a homogeneous function.

The key idea here is that every term in a homogeneous function has the same "overall power" when you consider `x` and `y` together. For example, in `x^2 + y^2`, both `x^2` and `y^2` have power 2. In `x^3 + xy^2`, `x^3` has power 3, and `xy^2` has `1+2=3` (sum of powers).

### What is a Homogeneous Differential Equation?

Now that we understand homogeneous functions, let's connect it to differential equations.

A differential equation of the form `dy/dx = F(x, y)` is called a homogeneous differential equation if the function `F(x, y)` is a homogeneous function of degree zero.

Wait, why degree zero? This is crucial! If `F(x, y)` is homogeneous of degree zero, it means `F(tx, ty) = t^0 F(x, y) = F(x, y)`.
This special property allows us to rewrite `F(x, y)` in a very useful form: as a function purely of the ratio `y/x`.
That is, if `F(x,y)` is homogeneous of degree zero, then it can always be expressed as `F(x, y) = g(y/x)` for some function `g`.

Let's see why:
Since `F(x, y)` is homogeneous of degree zero, we have `F(tx, ty) = F(x, y)`.
Let's choose `t = 1/x` (assuming `x β‰  0`).
Then `F(x, y) = F( (1/x) * x, (1/x) * y ) = F(1, y/x)`.
Let `g(y/x)` be `F(1, y/x)`. So, `dy/dx = g(y/x)`. This transformation is the core idea!

Alternative Form:
Sometimes, you might encounter homogeneous differential equations in the form:

M(x, y) dx + N(x, y) dy = 0


This equation is homogeneous if both `M(x, y)` and `N(x, y)` are homogeneous functions of the same degree.
If you rearrange this to `dy/dx = -M(x, y) / N(x, y)`, then the ratio `-M(x, y) / N(x, y)` will be a homogeneous function of degree zero.
(For example, if M and N are both degree 'k', then `(t^k M(x,y)) / (t^k N(x,y))` will have `t^k` cancelling out, leaving a degree zero homogeneous function).

Example of a Homogeneous Differential Equation:
Consider `dy/dx = (x^2 + y^2) / (xy)`.
Let `F(x, y) = (x^2 + y^2) / (xy)`.
Let's check for homogeneity:
`F(tx, ty) = ((tx)^2 + (ty)^2) / ((tx)(ty))`
`= (t^2x^2 + t^2y^2) / (t^2xy)`
`= t^2(x^2 + y^2) / (t^2xy)`
`= (x^2 + y^2) / (xy) = F(x, y)`
Since `F(tx, ty) = t^0 F(x, y)`, `F(x, y)` is a homogeneous function of degree zero.
Therefore, `dy/dx = (x^2 + y^2) / (xy)` is a homogeneous differential equation.

### The "Magic" Substitution: `y = vx`

Alright, so we've identified our homogeneous differential equation as `dy/dx = g(y/x)`. How do we solve it? This is where a clever substitution comes in handy!

The trick is to introduce a new variable, `v`, such that `v = y/x`.
This implies `y = vx`.

Now, we need to find `dy/dx` in terms of `v`, `x`, and `dv/dx`.
We differentiate `y = vx` with respect to `x` using the product rule:
`dy/dx = d/dx (vx)`
`dy/dx = v * (d/dx x) + x * (d/dx v)`

dy/dx = v + x (dv/dx)



This is a fundamental step you'll use every time!

Now, substitute `y = vx` and `dy/dx = v + x (dv/dx)` into our original homogeneous differential equation `dy/dx = g(y/x)`:
`v + x (dv/dx) = g(v)`

Look at what happened! We've successfully transformed our equation into one where the variables `v` and `x` can be separated! This is the most exciting part because we already know how to solve variable separable differential equations.

`x (dv/dx) = g(v) - v`
`dv / (g(v) - v) = dx / x`

Now, you can integrate both sides to find the solution!

### Step-by-Step Algorithm for Solving Homogeneous Differential Equations

Let's summarize the process into a clear set of steps:

1. Identify Homogeneity: Check if the given differential equation is homogeneous. If it's in the form `dy/dx = F(x, y)`, verify that `F(x, y)` is a homogeneous function of degree zero (i.e., `F(tx, ty) = F(x, y)`). If it's `M(x, y) dx + N(x, y) dy = 0`, check if `M` and `N` are homogeneous functions of the same degree.
2. Make the Substitution: Substitute `y = vx` and `dy/dx = v + x (dv/dx)` into the differential equation.
3. Separate Variables: After substitution, rearrange the equation to separate the variables `v` and `x`. You should get something like `(some function of v) dv = (some function of x) dx`.
4. Integrate: Integrate both sides of the separated equation. Don't forget the constant of integration, `C`.
5. Substitute Back: Replace `v` with `y/x` in the final solution to express it in terms of the original variables `x` and `y`.

### Let's Work Through Some Examples!

#### Example 1: Solve the differential equation `dy/dx = (x + y) / x`

Step 1: Identify Homogeneity
Let `F(x, y) = (x + y) / x`.
`F(tx, ty) = (tx + ty) / (tx) = t(x + y) / (tx) = (x + y) / x = F(x, y)`.
Yes, `F(x, y)` is a homogeneous function of degree zero. So, this is a homogeneous differential equation.

Step 2: Make the Substitution
Substitute `y = vx` and `dy/dx = v + x (dv/dx)`:
`v + x (dv/dx) = (x + vx) / x`
`v + x (dv/dx) = x(1 + v) / x`
`v + x (dv/dx) = 1 + v`

Step 3: Separate Variables
`x (dv/dx) = 1 + v - v`
`x (dv/dx) = 1`
`dv = dx / x`

Step 4: Integrate
Integrate both sides:
`∫ dv = ∫ (1/x) dx`
`v = ln|x| + C`

Step 5: Substitute Back
Replace `v` with `y/x`:
`y/x = ln|x| + C`
`y = x (ln|x| + C)`
This is the general solution to the given homogeneous differential equation.

#### Example 2: Solve the differential equation `(x^2 + y^2) dx = 2xy dy`

Step 1: Identify Homogeneity
First, let's rewrite it in the `dy/dx = F(x, y)` form:
`dy/dx = (x^2 + y^2) / (2xy)`
Let `F(x, y) = (x^2 + y^2) / (2xy)`.
Check for homogeneity:
`F(tx, ty) = ((tx)^2 + (ty)^2) / (2(tx)(ty))`
`= (t^2x^2 + t^2y^2) / (2t^2xy)`
`= t^2(x^2 + y^2) / (2t^2xy)`
`= (x^2 + y^2) / (2xy) = F(x, y)`
Yes, it's homogeneous of degree zero.

Step 2: Make the Substitution
Substitute `y = vx` and `dy/dx = v + x (dv/dx)`:
`v + x (dv/dx) = (x^2 + (vx)^2) / (2x(vx))`
`v + x (dv/dx) = (x^2 + v^2x^2) / (2vx^2)`
`v + x (dv/dx) = x^2(1 + v^2) / (2vx^2)`
`v + x (dv/dx) = (1 + v^2) / (2v)`

Step 3: Separate Variables
`x (dv/dx) = (1 + v^2) / (2v) - v`
`x (dv/dx) = (1 + v^2 - 2v^2) / (2v)`
`x (dv/dx) = (1 - v^2) / (2v)`
Now, separate `v` and `x`:
`2v / (1 - v^2) dv = 1/x dx`

Step 4: Integrate
Integrate both sides:
`∫ (2v / (1 - v^2)) dv = ∫ (1/x) dx`

For the left side, let `u = 1 - v^2`. Then `du = -2v dv`. So, `2v dv = -du`.
`∫ (-1/u) du = ln|x| + C`
`-ln|u| = ln|x| + C`
`-ln|1 - v^2| = ln|x| + C`

Step 5: Substitute Back
Replace `v` with `y/x`:
`-ln|1 - (y/x)^2| = ln|x| + C`
`-ln|(x^2 - y^2) / x^2| = ln|x| + C`
`ln|(x^2 - y^2) / x^2| = -ln|x| - C`
`ln|(x^2 - y^2) / x^2| = ln|x^(-1)| - C` (using `a ln b = ln b^a`)
`ln|(x^2 - y^2) / x^2| = ln(1/|x|) - C`

Let's rename `-C` as `ln|A|` (where `A` is an arbitrary constant, `A = e^(-C)`).
`ln|(x^2 - y^2) / x^2| = ln(1/|x|) + ln|A|`
`ln|(x^2 - y^2) / x^2| = ln(|A|/|x|)`
Taking exponential on both sides:
`(x^2 - y^2) / x^2 = Β± A/x`
Let `K = Β± A`.
`(x^2 - y^2) / x^2 = K/x`
`x^2 - y^2 = Kx`
`x^2 - y^2 = Cx` (where `C` is the arbitrary constant)
This is the general solution.

### CBSE vs. JEE Focus

* CBSE (and other board exams): You need to know the entire method thoroughly. Step-by-step derivation and clear presentation of each stage (identifying, substituting, separating, integrating, back-substituting) are crucial for full marks. Questions will typically be straightforward applications of this method.
* JEE (Main & Advanced): While the core method is the same, JEE problems rarely ask you to simply solve a homogeneous DE. Instead, this method becomes a tool within a larger problem. You might need to recognize a homogeneous DE quickly and solve it efficiently to proceed to the next part of the question. Sometimes, a problem might *not look* homogeneous initially but can be reduced to one with a clever manipulation. Speed and accuracy in integration are paramount. Also, implicit solutions are perfectly acceptable unless otherwise specified.

I hope this detailed explanation of homogeneous differential equations and their solution method makes perfect sense! Practice is key to mastering these concepts, so pick up some problems and start applying what you've learned. Good luck!
πŸ”¬ Deep Dive

Hello, future engineers! Welcome to this deep dive into one of the most fundamental and frequently tested methods for solving differential equations: the Solution of a Homogeneous Differential Equation. In this section, we will systematically unravel the concept, the 'why' behind the method, and master the 'how' with detailed examples, equipping you for both CBSE/ICSE boards and the challenging JEE examinations.



Before we jump into solving, let's establish a strong conceptual foundation, starting with what a 'homogeneous function' actually means.



1. Understanding Homogeneous Functions: The Foundation



A differential equation is called homogeneous if it can be expressed in a specific form, which relies on the concept of a homogeneous function. So, what is a homogeneous function?



A function f(x, y) is said to be a homogeneous function of degree 'n' if, for any non-zero constant t, we have:


$$f(tx, ty) = t^n f(x, y)$$



This definition might seem a bit abstract, so let's break it down with an analogy. Imagine you have a recipe for a cake. If you double all the ingredients (x becomes 2x, y becomes 2y), the amount of cake you get (the output of the function) also scales in a predictable way (t^n times the original). Here, 'n' tells you how much the output scales.



Examples:



  1. Let $f(x, y) = x^2 + xy$.

    Then $f(tx, ty) = (tx)^2 + (tx)(ty) = t^2x^2 + t^2xy = t^2(x^2 + xy) = t^2 f(x, y)$.

    Here, $n=2$. So, $f(x, y) = x^2 + xy$ is a homogeneous function of degree 2.

  2. Let $f(x, y) = frac{x+y}{x-y}$.

    Then $f(tx, ty) = frac{tx+ty}{tx-ty} = frac{t(x+y)}{t(x-y)} = frac{x+y}{x-y} = t^0 f(x, y)$.

    Here, $n=0$. So, $f(x, y) = frac{x+y}{x-y}$ is a homogeneous function of degree 0.

  3. Let $f(x, y) = x sinleft(frac{y}{x}
    ight) + y cosleft(frac{x}{y}
    ight)$.

    Then $f(tx, ty) = tx sinleft(frac{ty}{tx}
    ight) + ty cosleft(frac{tx}{ty}
    ight) = tx sinleft(frac{y}{x}
    ight) + ty cosleft(frac{x}{y}
    ight) = t left[x sinleft(frac{y}{x}
    ight) + y cosleft(frac{x}{y}
    ight)
    ight] = t^1 f(x, y)$.

    Here, $n=1$. So, $f(x, y)$ is a homogeneous function of degree 1.



Key Insight: A homogeneous function of degree zero can always be expressed as a function of the ratio y/x (or x/y). For example, $f(x, y) = frac{x+y}{x-y} = frac{1+y/x}{1-y/x}$. This insight is crucial for solving homogeneous differential equations.



2. Defining a Homogeneous Differential Equation



A first-order, first-degree differential equation is called homogeneous if it can be written in one of the following forms:



  1. $frac{dy}{dx} = Fleft(frac{y}{x}
    ight)$

  2. $frac{dy}{dx} = frac{g(x, y)}{h(x, y)}$, where $g(x, y)$ and $h(x, y)$ are homogeneous functions of the same degree.



The second form can always be converted to the first. If $g(x, y)$ and $h(x, y)$ are homogeneous functions of degree 'n', then:


$frac{dy}{dx} = frac{g(x, y)}{h(x, y)} = frac{x^n g(1, y/x)}{x^n h(1, y/x)} = frac{g(1, y/x)}{h(1, y/x)} = Fleft(frac{y}{x}
ight)$.


This transformation is key, as it shows why the substitution we're about to learn works so effectively.



3. The Method of Solution: Substitution (Derivation)



The core idea behind solving a homogeneous differential equation lies in a clever substitution that transforms it into a variable separable differential equation, which we already know how to solve.



Consider a homogeneous differential equation of the form:


$$frac{dy}{dx} = Fleft(frac{y}{x}
ight) quad ext{... (1)}$$



Let's introduce a new variable, $v$, such that:


$$v = frac{y}{x} quad Rightarrow quad y = vx quad ext{... (2)}$$



Now, we need to find an expression for $frac{dy}{dx}$ in terms of $v$, $x$, and $frac{dv}{dx}$. We differentiate equation (2) with respect to $x$ using the product rule:


$$frac{dy}{dx} = v cdot frac{d}{dx}(x) + x cdot frac{d}{dx}(v)$$

$$frac{dy}{dx} = v cdot 1 + x frac{dv}{dx}$$

$$frac{dy}{dx} = v + x frac{dv}{dx} quad ext{... (3)}$$



Now, substitute equations (2) and (3) back into the original homogeneous differential equation (1):


$$v + x frac{dv}{dx} = F(v)$$



Our goal is to separate the variables $v$ and $x$. Let's rearrange the terms:


$$x frac{dv}{dx} = F(v) - v$$



Assuming $F(v) - v
eq 0$, we can separate the variables:


$$frac{dv}{F(v) - v} = frac{dx}{x}$$



This is a variable separable form! We can now integrate both sides to find the solution:


$$int frac{dv}{F(v) - v} = int frac{dx}{x}$$



After performing the integration, we will get an equation involving $v$ and $x$. The final step is to substitute back $v = frac{y}{x}$ to express the solution in terms of the original variables $x$ and $y$.



JEE Advanced Tip: Sometimes, it might be more convenient to make the substitution $x = vy$, especially if the equation is easier to write as $frac{dx}{dy} = Gleft(frac{x}{y}
ight)$. In this case, you differentiate $x=vy$ with respect to $y$, getting $frac{dx}{dy} = v + y frac{dv}{dy}$.


4. Step-by-Step Procedure for Solving Homogeneous DEs



Let's consolidate the method into a clear, actionable set of steps:




  1. Identify Homogeneity: Check if the given differential equation is homogeneous. This usually involves:

    • Expressing it in the form $frac{dy}{dx} = frac{M(x, y)}{N(x, y)}$.

    • Checking if $M(x, y)$ and $N(x, y)$ are homogeneous functions of the same degree.

    • Alternatively, try to rewrite it directly as $frac{dy}{dx} = Fleft(frac{y}{x}
      ight)$.




  2. Make the Substitution: Substitute $y = vx$ (or $x = vy$ if preferred).


  3. Differentiate: Calculate $frac{dy}{dx} = v + x frac{dv}{dx}$ (or $frac{dx}{dy} = v + y frac{dv}{dy}$).


  4. Substitute into DE: Replace $y$ with $vx$ and $frac{dy}{dx}$ with $v + x frac{dv}{dx}$ in the original differential equation. This will eliminate $y$ and $frac{dy}{dx}$, leaving an equation in terms of $v$, $x$, and $frac{dv}{dx}$.


  5. Separate Variables: Rearrange the transformed equation to separate the variables $v$ and $x$. It will typically take the form:

    $$frac{dv}{F(v) - v} = frac{dx}{x} quad ext{or} quad frac{dv}{G(v) - v} = frac{dy}{y}$$




  6. Integrate: Integrate both sides of the separated equation. Don't forget the constant of integration, $C$.


  7. Back-Substitute: Replace $v$ with $frac{y}{x}$ (or $frac{x}{y}$) in the general solution to obtain the final answer in terms of $x$ and $y$.



CBSE/ICSE vs. JEE Focus: For board exams, simply following these steps meticulously is usually sufficient. For JEE, you often need to be proficient in complex integration techniques and sometimes solve Initial Value Problems (IVPs), which means using a given condition to find the specific value of the constant C.

5. Illustrative Examples



Example 1: Basic Application


Solve the differential equation: $(x^2 + xy) dy = (x^2 + y^2) dx$



Solution:


Step 1: Identify Homogeneity.


Rewrite the equation as $frac{dy}{dx} = frac{x^2 + y^2}{x^2 + xy}$.


Let $M(x, y) = x^2 + y^2$ and $N(x, y) = x^2 + xy$.


For $M(x, y)$: $M(tx, ty) = (tx)^2 + (ty)^2 = t^2x^2 + t^2y^2 = t^2(x^2 + y^2) = t^2 M(x, y)$. (Degree 2)


For $N(x, y)$: $N(tx, ty) = (tx)^2 + (tx)(ty) = t^2x^2 + t^2xy = t^2(x^2 + xy) = t^2 N(x, y)$. (Degree 2)


Since both are homogeneous functions of the same degree (2), the differential equation is homogeneous.


Also, we can write $frac{dy}{dx} = frac{x^2(1 + (y/x)^2)}{x^2(1 + y/x)} = frac{1 + (y/x)^2}{1 + y/x} = Fleft(frac{y}{x}
ight)$.



Step 2: Make the Substitution.


Let $y = vx$.



Step 3: Differentiate.


$frac{dy}{dx} = v + x frac{dv}{dx}$.



Step 4: Substitute into DE.


Substitute $y=vx$ and $frac{dy}{dx} = v + x frac{dv}{dx}$ into the equation:


$v + x frac{dv}{dx} = frac{x^2 + (vx)^2}{x^2 + x(vx)}$


$v + x frac{dv}{dx} = frac{x^2(1 + v^2)}{x^2(1 + v)}$


$v + x frac{dv}{dx} = frac{1 + v^2}{1 + v}$



Step 5: Separate Variables.


$x frac{dv}{dx} = frac{1 + v^2}{1 + v} - v$


$x frac{dv}{dx} = frac{1 + v^2 - v(1 + v)}{1 + v}$


$x frac{dv}{dx} = frac{1 + v^2 - v - v^2}{1 + v}$


$x frac{dv}{dx} = frac{1 - v}{1 + v}$


Separating variables:


$frac{1 + v}{1 - v} dv = frac{dx}{x}$



Step 6: Integrate.


$int frac{1 + v}{1 - v} dv = int frac{dx}{x}$


For the left integral, we can rewrite the numerator: $frac{1 + v}{1 - v} = frac{-(1 - v) + 2}{1 - v} = -1 + frac{2}{1 - v}$.


So, $int left(-1 + frac{2}{1 - v}
ight) dv = int frac{dx}{x}$


$-v + 2 frac{ln|1 - v|}{-1} = ln|x| + C$


$-v - 2 ln|1 - v| = ln|x| + C$


$C_1 - v = ln|x| + 2 ln|1 - v|$ (where $C_1 = -C$)


$C_1 - v = ln|x(1 - v)^2|$



Step 7: Back-Substitute.


Replace $v$ with $frac{y}{x}$:


$C_1 - frac{y}{x} = lnleft|xleft(1 - frac{y}{x}
ight)^2
ight|$


$C_1 - frac{y}{x} = lnleft|xleft(frac{x - y}{x}
ight)^2
ight|$


$C_1 - frac{y}{x} = lnleft|x frac{(x - y)^2}{x^2}
ight|$


$C_1 - frac{y}{x} = lnleft|frac{(x - y)^2}{x}
ight|$


This is the general solution.



Example 2: An Initial Value Problem (IVP)


Solve: $(x^2 - y^2) dy = 2xy dx$, given $y(1) = 1$.



Solution:


Step 1: Identify Homogeneity.


Rewrite as $frac{dy}{dx} = frac{2xy}{x^2 - y^2}$.


Numerator $2xy$ is degree 2. Denominator $x^2 - y^2$ is degree 2. Homogeneous.


Alternatively: $frac{dy}{dx} = frac{2(y/x)}{1 - (y/x)^2}$. This is $F(y/x)$ form.



Step 2 & 3: Substitution & Differentiation.


Let $y = vx implies frac{dy}{dx} = v + x frac{dv}{dx}$.



Step 4: Substitute into DE.


$v + x frac{dv}{dx} = frac{2x(vx)}{x^2 - (vx)^2}$


$v + x frac{dv}{dx} = frac{2vx^2}{x^2(1 - v^2)}$


$v + x frac{dv}{dx} = frac{2v}{1 - v^2}$



Step 5: Separate Variables.


$x frac{dv}{dx} = frac{2v}{1 - v^2} - v$


$x frac{dv}{dx} = frac{2v - v(1 - v^2)}{1 - v^2}$


$x frac{dv}{dx} = frac{2v - v + v^3}{1 - v^2}$


$x frac{dv}{dx} = frac{v + v^3}{1 - v^2} = frac{v(1 + v^2)}{1 - v^2}$


Separating variables:


$frac{1 - v^2}{v(1 + v^2)} dv = frac{dx}{x}$



Step 6: Integrate.


$int frac{1 - v^2}{v(1 + v^2)} dv = int frac{dx}{x}$


Use partial fractions for the left integral: $frac{1 - v^2}{v(1 + v^2)} = frac{A}{v} + frac{Bv + C}{1 + v^2}$


$1 - v^2 = A(1 + v^2) + (Bv + C)v$


$1 - v^2 = A + Av^2 + Bv^2 + Cv$


Comparing coefficients:



  • Constant term: $1 = A implies A = 1$

  • Coefficient of $v$: $0 = C implies C = 0$

  • Coefficient of $v^2$: $-1 = A + B implies -1 = 1 + B implies B = -2$


So, $frac{1 - v^2}{v(1 + v^2)} = frac{1}{v} - frac{2v}{1 + v^2}$.


Now integrate:


$int left(frac{1}{v} - frac{2v}{1 + v^2}
ight) dv = int frac{dx}{x}$


$ln|v| - ln|1 + v^2| = ln|x| + ln|C|$ (Using $ln|C|$ for the constant for convenience)


$lnleft|frac{v}{1 + v^2}
ight| = ln|Cx|$


$frac{v}{1 + v^2} = Cx$



Step 7: Back-Substitute.


Replace $v$ with $frac{y}{x}$:


$frac{y/x}{1 + (y/x)^2} = Cx$


$frac{y/x}{(x^2 + y^2)/x^2} = Cx$


$frac{y}{x} cdot frac{x^2}{x^2 + y^2} = Cx$


$frac{xy}{x^2 + y^2} = Cx$


$frac{y}{x^2 + y^2} = C$ (This is the general solution)



Step 8: Apply Initial Condition (for IVP).


Given $y(1) = 1$, so when $x=1, y=1$.


Substitute these values into the general solution:


$frac{1}{1^2 + 1^2} = C$


$frac{1}{1 + 1} = C implies C = frac{1}{2}$



Therefore, the particular solution is:


$frac{y}{x^2 + y^2} = frac{1}{2}$


$2y = x^2 + y^2$


$x^2 + y^2 - 2y = 0$


$x^2 + (y - 1)^2 = 1$ (This is the equation of a circle centered at $(0,1)$ with radius 1).



This deep dive has provided you with a thorough understanding of homogeneous differential equations, from their foundational definition to the detailed steps of their solution, including advanced considerations for JEE and practical examples. Master this method, as it forms a cornerstone for more complex differential equation types!

🎯 Shortcuts

Solving homogeneous differential equations involves a specific sequence of steps. Remembering these steps accurately and efficiently is crucial for exam success. Here are some mnemonics and shortcuts to help you master this topic:



The "HVSIR" Mnemonic for Homogeneous Equations


This mnemonic helps you remember the five essential steps in order:



  • Homogeneous Check

  • V-Substitution

  • Separate Variables

  • Integrate

  • Replace V



Let's break down each part with its corresponding shortcut:





  1. H - Homogeneous Check



    • Mnemonic/Shortcut: "Same Degree Power" (SDP)

    • Explanation: A differential equation $dy/dx = f(x,y)$ is homogeneous if $f(x,y)$ can be expressed as $g(y/x)$ (or $g(x/y)$). Practically, if you express $dy/dx = P(x,y)/Q(x,y)$, check if all terms in $P(x,y)$ have the same degree and all terms in $Q(x,y)$ have the same degree, and these degrees are equal. For example, $(x^2+y^2)/(xy)$ is homogeneous (degree 2 in numerator, degree 2 in denominator).

    • JEE Tip: Quickly test by replacing $x o tx$ and $y o ty$. If $f(tx,ty) = t^0 f(x,y) = f(x,y)$, it's homogeneous.




  2. V - V-Substitution



    • Mnemonic/Shortcut: "Very X-citing derivative"

    • Explanation: Always substitute $y = vx$. The crucial part is remembering its derivative: $dy/dx = v + x frac{dv}{dx}$ (using the product rule). This is a non-negotiable step.

    • Alternative: If the equation is easier to write as $dx/dy = g(x,y)$, then use $x = vy$ and its derivative $dx/dy = v + y frac{dv}{dy}$. (Think "Flip to V-Y").




  3. S - Separate Variables



    • Mnemonic/Shortcut: "Get V with dv, Get X with dx"

    • Explanation: After substituting $y=vx$ and $dy/dx$, the equation will transform into a form where $v$ and $x$ can be separated. The goal is to get all terms involving $v$ with $dv$ on one side and all terms involving $x$ with $dx$ on the other side. This typically looks like $frac{dv}{f(v)} = frac{dx}{g(x)}$.




  4. I - Integrate



    • Mnemonic/Shortcut: "Integrate Both Sides - Don't Forget C" (IBS-DFC)

    • Explanation: Once the variables are separated, integrate both sides of the equation. Remember to add the constant of integration, $C$, to one side (usually the right side). This is a common point for losing marks in both CBSE and JEE.

    • CBSE & JEE Tip: Sometimes, writing the constant as $log C$ simplifies the final expression, especially if all other terms are logarithmic.




  5. R - Replace V



    • Mnemonic/Shortcut: "V-turn back to Y/X"

    • Explanation: The final solution must be in terms of the original variables $x$ and $y$. After integration, replace $v$ back with $y/x$ (or $x/y$ if you used $x=vy$). This gives the general solution of the homogeneous differential equation.





Common Pitfalls to Avoid (Shortcuts for Error Prevention):



  • Forgetting Chain Rule: Always remember $dy/dx = v + x(dv/dx)$, not just $dv/dx$. This is the most frequent mistake.

  • Incorrect Separation: Double-check that all terms with $v$ are on the $dv$ side and all terms with $x$ are on the $dx$ side before integrating. A misplaced $x$ or $v$ can ruin the integration.

  • Missing Constant: Never forget to add the constant of integration '$C$'.

  • Unsimplified Answer: Always simplify the final expression and ensure it is in terms of $x$ and $y$.



By consistently applying the "HVSIR" mnemonic and being mindful of these common pitfalls, you can efficiently and accurately solve homogeneous differential equations in your exams!

πŸ’‘ Quick Tips

Quick Tips: Solving Homogeneous Differential Equations


Homogeneous differential equations are a fundamental type in differential equations, often appearing in both board exams and competitive tests. Master these quick tips to solve them efficiently:




  • Identify Homogeneity:

    • A first-order differential equation `dy/dx = f(x, y)` is homogeneous if `f(x, y)` can be expressed as a function of `y/x` or `x/y`.

    • Practical Check (JEE Tip): Look at each term in the numerator and denominator if the equation is in the form `dy/dx = (P(x,y))/(Q(x,y))`. If the total degree of x and y is the same for every term in both the numerator and denominator, it's homogeneous. For example, `(x^2 + y^2) / (xy)` is homogeneous because all terms (`x^2`, `y^2`, `xy`) have a total degree of 2.



  • Standard Substitution:

    • For equations of the form `dy/dx = f(y/x)`, always use the substitution `y = vx`.

    • Differentiate this with respect to `x` using the product rule: `dy/dx = v + x(dv/dx)`.

    • JEE Pro Tip: If the equation is more naturally expressed as `dx/dy = g(x/y)` (meaning `x` is easier to isolate in terms of `y`), then use `x = vy`, which implies `dx/dy = v + y(dv/dy)`. Choose the substitution that simplifies the algebra the most.



  • Transformation to Variable Separable Form:

    • After substituting `y = vx` and `dy/dx = v + x(dv/dx)` (or `x=vy` and `dx/dy = v+y(dv/dy)`), the given homogeneous equation will always transform into a variable separable form, involving `v` and `x` (or `v` and `y`).

    • Your primary goal in this step is to rearrange the equation such that all `v` terms and `dv` are on one side, and all `x` terms and `dx` are on the other side (e.g., `dv / g(v) = dx / x`).



  • Integration Step:

    • Integrate both sides of the separated equation. This often involves standard integration formulas.

    • Common Pitfall: Don't forget the constant of integration, `C`, immediately after integrating. It's often convenient to write it as `ln|C|` if both sides of the equation result in `ln` terms, as this can simplify the final solution.

    • For JEE, be prepared for complex integrals that may require techniques like partial fractions, specific trigonometric substitutions, or integration by parts.



  • Back-Substitution (Crucial):

    • After finding the solution in terms of `v` and `x` (or `v` and `y`), always substitute back `v = y/x` (or `v = x/y`) to express the final solution in terms of the original variables `x` and `y`.

    • Failure to perform this back-substitution will result in an incomplete or incorrect answer, especially in subjective board exams.



  • CBSE vs JEE Emphasis:

    • CBSE: Focuses more on the clear application of the method and basic integration. Homogeneous equations are usually explicitly identifiable.

    • JEE: May require you to rearrange the equation to identify its homogeneous nature. The integration steps can be more demanding, requiring a strong command over various integration techniques. Also, pay close attention to initial value problems to find the particular solution.




Mastering these steps ensures you can efficiently solve homogeneous differential equations in your exams!

🧠 Intuitive Understanding

Welcome to the intuitive understanding of homogeneous differential equations! This section aims to demystify why a particular method works, rather than just presenting it as a formula.



What is a Homogeneous Differential Equation?


Imagine a function where if you scale both x and y by the same factor, the 'essence' of the function remains the same, just scaled. A homogeneous differential equation has this property, though it's easier to think of it in terms of 'degree':




  • Intuition 1 (Degree): In a homogeneous differential equation like dy/dx = F(x, y) or M(x, y)dx + N(x, y)dy = 0, if you look at each term in M(x, y) and N(x, y), the sum of the powers of x and y in each term is identical. For example, in (xΒ² + yΒ²)dx - xy dy = 0, the degree of xΒ² is 2, yΒ² is 2, and xy is 1+1=2. All terms have a degree of 2.


  • Intuition 2 (Ratio Dependency): More importantly, a homogeneous differential equation dy/dx = F(x, y) can always be rewritten such that dy/dx depends *only* on the ratio y/x (or x/y).

    For example, if dy/dx = (xΒ² + yΒ²)/(xy), dividing numerator and denominator by xΒ² gives dy/dx = (1 + (y/x)Β²)/(y/x). Notice how the entire right-hand side is now a function of y/x.


This second intuition is crucial: it means that the slope of the solution curve (dy/dx) is constant along any line passing through the origin (i.e., where y/x is constant). All points on the line y = mx will have the same slope.



Why the Substitution `y = vx` (or `x = vy`)?


Since the slope dy/dx in a homogeneous equation depends only on the ratio y/x, it's natural to make this ratio our new variable.
Let v = y/x, which implies y = vx.


Now, let's see how this substitution simplifies the equation:




  1. Transformation of dy/dx: Differentiating y = vx with respect to x using the product rule, we get:
    dy/dx = v + x(dv/dx)


  2. Transformation of F(x, y): The right-hand side of the differential equation, F(x, y), which we established depends only on y/x, will transform into a function of v alone.

    So, F(x, y) = G(y/x) becomes G(v).


  3. The Magic - Variable Separable Form: Equating the transformed expressions, we get:
    v + x(dv/dx) = G(v)

    Rearranging this:
    x(dv/dx) = G(v) - v
    dv / (G(v) - v) = dx / x

    Voila! The equation is now in a variable separable form, meaning all v terms are on one side with dv, and all x terms are on the other side with dx. This form is much easier to integrate.


The substitution y = vx (or x = vy, if dx/dy is easier to work with) is a powerful trick because it converts a seemingly complex homogeneous differential equation into a simpler form that can be solved by direct integration.



Key Takeaways for Exams:



  • CBSE & JEE: Identifying and solving homogeneous differential equations is a standard and frequently tested topic in both board exams and JEE. Mastering this substitution is essential.

  • Recognition: The first step is always to recognize a homogeneous DE, either by checking the degree of terms or by seeing if dy/dx can be expressed as a function of y/x.

  • Substitution: The substitution y = vx is the cornerstone of the solution method. Remember to also substitute for dy/dx.

  • Transformation: This substitution *guarantees* the equation will become variable separable, making it solvable by direct integration.


Understanding *why* the y = vx substitution works, by recognizing the inherent ratio-dependency of homogeneous equations, makes the solution process intuitive and memorable.

🌍 Real World Applications

Differential equations are fundamental tools for modeling phenomena across various scientific and engineering disciplines. While specific scenarios might simplify to homogeneous differential equations, their core utility lies in describing systems where the rate of change of one variable with respect to another depends on the ratio of the variables themselves. This often occurs when scale invariance or proportional relationships are key aspects of the system.



Here are some real-world applications where solutions of homogeneous differential equations find utility:




  • Orthogonal Trajectories: This is a classic mathematical application. Given a family of curves, an orthogonal trajectory is a curve that intersects every member of the family at a right angle. The differential equation for the orthogonal trajectories of a given family can often turn out to be a homogeneous differential equation. This concept is used in various fields like fluid flow (streamlines and equipotential lines), electrostatics (electric field lines and equipotential surfaces), and thermodynamics (isotherms and adiabatic curves).

  • Chemical Kinetics: In certain chemical reactions, particularly those involving competitive or consecutive steps, the rate of change of reactant or product concentrations might depend on the ratio of their current concentrations. For example, in some disproportionation reactions or competitive adsorption processes, the rate law might simplify to a homogeneous differential equation, allowing for the determination of concentration profiles over time.

  • Fluid Dynamics and Heat Transfer: In simplified models of fluid flow or heat transfer, especially when dealing with geometrically similar systems or boundary layer phenomena, homogeneous differential equations can arise. For instance, in studying specific types of boundary layer flows where the velocity profile's shape is crucial, or heat conduction in specific geometries where ratios of dimensions are significant, these equations provide analytical solutions.

  • Population Dynamics: In ecological models, particularly those describing competition between two species or predator-prey relationships, a simplified model might lead to a homogeneous differential equation if the growth rate of one population depends on its ratio to another, or to resources. This provides insights into how the ratio of populations evolves over time.

  • Economics: In economic models, particularly those involving growth and distribution, homogeneous differential equations can describe situations where the rate of change of economic variables (like capital, labor, or output) depends on their ratios. For example, in some models of economic growth, the capital-labor ratio might evolve according to such an equation.



Illustrative Example: Orthogonal Trajectories


Consider a family of circles centered at the origin, given by $x^2 + y^2 = c^2$. To find their orthogonal trajectories, we first differentiate implicitly with respect to $x$ to find the differential equation of the family:


$2x + 2y frac{dy}{dx} = 0 implies frac{dy}{dx} = -frac{x}{y}$.


For the orthogonal trajectories, the slope must be the negative reciprocal. So, the differential equation for the orthogonal trajectories is:


$frac{dy}{dx} = - left( frac{1}{-x/y}
ight) = frac{y}{x}$.


This is a homogeneous differential equation of the form $frac{dy}{dx} = fleft(frac{y}{x}
ight)$.


Solving this homogeneous equation:



  1. Let $y = vx implies frac{dy}{dx} = v + x frac{dv}{dx}$.

  2. Substitute into the equation: $v + x frac{dv}{dx} = v$.

  3. This simplifies to $x frac{dv}{dx} = 0$. Since $x
    eq 0$ (as the original family is circles not points), we must have $frac{dv}{dx} = 0$.

  4. Integrating, $v = k$, where $k$ is a constant.

  5. Substitute back $v = y/x$: $y/x = k implies y = kx$.


The orthogonal trajectories are thus the family of straight lines passing through the origin. This intuitively makes sense, as lines through the origin intersect circles centered at the origin at 90 degrees.



JEE Main Perspective: While direct "real-world application" questions are rare in JEE Main for differential equations, understanding the underlying principles and being able to solve homogeneous differential equations is crucial. Problems involving orthogonal trajectories are common and directly test the ability to form and solve such equations. Focus on the solution methodology rather than memorizing applications for the exam.

πŸ”„ Common Analogies

Changing Perspective: The Homogeneous Differential Equation Analogy



Understanding the solution method for homogeneous differential equations can be greatly simplified by drawing an analogy to changing one's perspective or coordinate system to simplify a complex problem.

Imagine you are trying to describe the slope (dy/dx) of a landscape that has a very particular characteristic: it looks the same no matter how much you zoom in or out, provided you're looking along a straight line from a central point. Think of a perfectly conical hill or a valley whose cross-section is always proportional.

Here's how the analogy breaks down:



  • The "Complex Landscape" (Homogeneous DE): A homogeneous differential equation often appears complicated in its original (x, y) form. All its terms involve 'x' and 'y' in a way that makes it look daunting to separate variables directly. The key characteristic is that if you replace x with tx and y with ty, the derivative dy/dx remains unchanged in form. This means the 'slope' only depends on the *ratio* y/x, not on the individual values of x and y themselves.


  • The "Hidden Feature" (Homogeneity): The fact that the equation is homogeneous is like a special clue. It tells you that the problem's underlying structure is related to the ratio of variables (y/x). It's an invitation to view the problem differently.


  • "Changing Your Gaze" (The Substitution y=vx or v=y/x): This is the crucial step. Instead of focusing on 'x' and 'y' individually, you decide to focus on their ratio, 'v'.

    • It's like saying, "Instead of describing points by their horizontal and vertical distances (x, y), I'll describe them by their horizontal distance (x) and the *slope* of the line connecting them to the origin (v = y/x)."

    • This transformation is similar to switching from Cartesian coordinates (x, y) to a form that is better suited for problems with radial symmetry or proportionality. For instance, sometimes a problem is easier in polar coordinates (r, θ) than in Cartesian. Here, 'v' acts like an angle or a direction from the origin.




  • "Simplifying the View" (Variable Separable Form): Once you make the substitution y=vx (and consequently dy/dx = v + x dv/dx), the seemingly complex homogeneous differential equation magically transforms into a much simpler form where the variables 'v' and 'x' can be separated.

    • Before the substitution, it was like trying to solve a puzzle where all the pieces were jumbled together without any clear pattern.

    • After the substitution, it's like the puzzle pieces neatly fall into two distinct piles – one for 'v' and one for 'x' – making it straightforward to solve each part independently.




  • "Solving the Parts and Recombining" (Integration and Back-Substitution): Once in variable separable form, you integrate both sides. This is like solving the 'v' puzzle and the 'x' puzzle independently. Finally, you substitute back 'v = y/x' to express the solution in terms of the original variables 'x' and 'y'. This is like recombining the solved puzzle parts to get the complete picture in its original context.



Key Takeaway: The essence of solving a homogeneous differential equation lies in recognizing its inherent proportionality and then applying a specific transformation (y=vx) that shifts your mathematical perspective, simplifying a coupled problem into a separable one. This strategy of transforming a problem into a simpler, solvable form is a powerful concept in mathematics.
πŸ“‹ Prerequisites
To effectively solve homogeneous differential equations, a solid foundation in certain fundamental concepts is essential. These prerequisites ensure that students can understand the underlying structure of these equations and apply the necessary techniques to find their solutions.

Key Prerequisites for Homogeneous Differential Equations



Before delving into the solution methods for homogeneous differential equations, ensure you are proficient in the following concepts:



  • 1. Basic Concepts of Differential Equations:


    • Understand the definition of a differential equation, its order, and degree. This forms the very basis of classification and problem-solving.


    • Familiarity with various forms of differential equations is helpful.




  • 2. Homogeneous Functions:


    • This is a critical prerequisite, as the definition of a homogeneous differential equation directly relies on the concept of a homogeneous function.


    • You must know how to identify a homogeneous function. A function $f(x, y)$ is homogeneous of degree $n$ if $f(lambda x, lambda y) = lambda^n f(x, y)$ for some non-zero constant $lambda$.


    • Understanding this concept is key to recognizing a homogeneous differential equation, which is typically of the form $frac{dy}{dx} = f(x, y)$ where $f(x, y)$ is a homogeneous function of degree zero, or $M(x, y)dx + N(x, y)dy = 0$ where $M$ and $N$ are homogeneous functions of the same degree.




  • 3. Variables Separable Method:


    • Solving homogeneous differential equations involves a substitution that transforms them into the variables separable form. Therefore, a thorough understanding and proficiency in solving differential equations by separation of variables are absolutely crucial.


    • This is often the first method taught for first-order first-degree differential equations, and its mastery is non-negotiable for homogeneous equations.




  • 4. Methods of Integration:


    • After applying the substitution and separating variables, the final step in solving homogeneous differential equations always involves integration.


    • You must be adept at various integration techniques, including:


      • Standard Integration Formulas: Basic integrals like $int x^n dx$, $int frac{1}{x} dx$, $int e^x dx$, $int sin x dx$, etc.


      • Integration by Substitution: This is frequently used after variable separation.


      • Integration by Partial Fractions: Occasionally required when the integrand is a rational function.


      • Integration of Trigonometric Functions: Often needed if trigonometric substitutions are involved or if the original equation contains trigonometric terms.




    • JEE Specific: JEE Main often includes integrals that require clever substitutions or a combination of techniques, so a strong command over advanced integration is highly beneficial.




  • 5. Algebraic Manipulation and Simplification:


    • Proficiency in basic algebra, including substitution, factorization, and simplification of expressions, is vital for carrying out the necessary transformations and reaching the final solution.





Mastering these prerequisites will provide a strong foundation, making the process of understanding and solving homogeneous differential equations much smoother and more efficient.
⚠️ Common Exam Traps

Common Exam Traps: Solution of Homogeneous Differential Equations



Solving homogeneous differential equations requires careful attention at every step. Students often fall into specific traps during exams, leading to loss of marks. Being aware of these common pitfalls can significantly improve accuracy.





  • Misidentification of Homogeneous Equations:

    • A common mistake is applying the homogeneous method to equations that are not homogeneous. Remember, an equation of the form dy/dx = f(x, y) is homogeneous if f(Ξ»x, Ξ»y) = f(x, y) or if all terms in the numerator and denominator have the same degree.

    • JEE Tip: In JEE, options might include solutions derived from incorrect method applications. Always verify homogeneity first.




  • Incorrect Substitution and Differentiation:

    • The standard substitution is y = vx, leading to dy/dx = v + x(dv/dx). Errors frequently occur here:

      • Forgetting the product rule and simply writing dy/dx = dv/dx or dy/dx = v.

      • Forgetting to substitute y = vx into the original function f(x, y), particularly when dealing with y/x terms.



    • Similarly, if dx/dy is used (by substituting x = vy), ensure dx/dy = v + y(dv/dy) is correctly applied.




  • Algebraic Manipulation and Variable Separation Errors:

    • After substitution, the equation becomes separable. Common errors include:

      • Incorrectly isolating terms involving v with dv and x with dx.

      • Sign errors when moving terms across the equality.

      • Mistakes in simplifying complex rational expressions involving v.



    • CBSE Boards: Show all intermediate steps clearly for algebraic manipulation to avoid losing marks for calculation errors.




  • Integration Blunders:

    • This is a major source of errors. Students might incorrectly integrate standard forms, especially when partial fractions or specific trigonometric substitutions are required for the dv/f(v) term.

    • Forgetting the constant of integration, C. This is critical for obtaining the general solution.

    • Incorrectly choosing the form of the constant (e.g., log|C| vs. C) when it simplifies the final expression (e.g., when all other terms are logarithmic).




  • Failure to Back-Substitute:

    • After finding the solution in terms of v and x (or v and y), students sometimes forget to substitute back v = y/x (or v = x/y) to express the final answer in terms of the original variables x and y. This leads to an incomplete solution.




  • Incorrect Application of Initial Conditions:

    • If initial conditions are given, apply them to the general solution after back-substitution to find the particular value of C. Applying them prematurely or incorrectly is a common error.






By systematically checking each step – identification, substitution, separation, integration, and back-substitution – you can minimize these common errors and secure full marks in differential equation problems. Practice makes perfect!


⭐ Key Takeaways

Key Takeaways: Solution of Homogeneous Differential Equations



Solving homogeneous differential equations is a fundamental technique in differential equations. Mastering this method is crucial for both JEE Main and board examinations. Here are the essential points to remember:





  • Definition and Recognition:

    • A differential equation of the form dy/dx = f(x, y) is homogeneous if f(tx, ty) = f(x, y) for any non-zero constant t.

    • Alternatively, if dy/dx can be expressed as a function of y/x (i.e., dy/dx = g(y/x)), it is homogeneous.

    • For equations of the form M(x, y)dx + N(x, y)dy = 0, it is homogeneous if M and N are homogeneous functions of the same degree.




  • The Core Substitution:

    • The standard substitution for homogeneous equations is y = vx. This implies dy/dx = v + x(dv/dx) (using the product rule).

    • JEE Tip: While y=vx is most common, if the equation is more easily expressed as dx/dy = h(x/y), then using x = vy (implying dx/dy = v + y(dv/dy)) can simplify calculations. Choose wisely!




  • Transformation to Variable Separable Form:

    • After substitution, the differential equation transforms into a variable separable form involving v and x (or v and y if x=vy was used).

    • This is the critical step, as it allows direct integration.




  • Integration and Back-Substitution:

    • Integrate both sides of the separable equation. Don't forget the constant of integration, C.

    • Finally, substitute v = y/x (or v = x/y) back into the integrated solution to express the answer in terms of x and y.




  • Common Pitfalls & Best Practices:

    • Don't forget the constant of integration! This is a frequent mistake in both board exams and competitive tests.

    • Always simplify the expression before integrating. Algebraic manipulation skills are key here.

    • Be careful with the domain of the solution, especially when dividing by x or y (implies x β‰  0 or y β‰  0).

    • JEE Specific: Sometimes, the integral may require advanced integration techniques (e.g., partial fractions, specific trigonometric substitutions). Be proficient with these.




  • Geometric Interpretation:

    • The slope dy/dx at any point (x, y) depends only on the ratio y/x. This means all points on a line passing through the origin have the same slope.

    • This implies that the integral curves are geometrically similar and scale-invariant.





Mastering these points will ensure you can confidently tackle homogeneous differential equations in your exams!


🧩 Problem Solving Approach

Solving a homogeneous differential equation involves a systematic approach to transform it into a variable separable form. This method is fundamental for both JEE Main and CBSE Board exams.



Problem Solving Approach: Homogeneous Differential Equations



The core idea behind solving a homogeneous differential equation is to use a suitable substitution that converts it into a variable separable form, which can then be integrated easily.



Step 1: Identify if the Differential Equation is Homogeneous



  • A first-order differential equation of the form $frac{dy}{dx} = f(x, y)$ is said to be homogeneous if $f(x, y)$ is a homogeneous function of degree zero. This means $f(lambda x, lambda y) = lambda^0 f(x, y) = f(x, y)$.

  • Practical Tip: Check if each term in the numerator and denominator of $frac{dy}{dx}$ has the same total degree. For example, if $frac{dy}{dx} = frac{x^2+y^2}{xy}$, the degree of $x^2$ is 2, $y^2$ is 2, and $xy$ is $1+1=2$. All terms have degree 2, so it's homogeneous.

  • Alternatively, you can often write a homogeneous equation in the form $frac{dy}{dx} = Fleft(frac{y}{x}
    ight)$
    or $frac{dx}{dy} = Gleft(frac{x}{y}
    ight)$
    .



Step 2: Apply the Appropriate Substitution



  • If the equation is in the form $frac{dy}{dx} = Fleft(frac{y}{x}
    ight)$
    , substitute $y = vx$.

  • Differentiate $y = vx$ with respect to $x$ using the product rule: $frac{dy}{dx} = v cdot 1 + x frac{dv}{dx}$.

  • If the equation is in the form $frac{dx}{dy} = Gleft(frac{x}{y}
    ight)$
    , substitute $x = vy$.

  • Differentiate $x = vy$ with respect to $y$: $frac{dx}{dy} = v cdot 1 + y frac{dv}{dy}$.

    Warning (JEE Focus): Sometimes, substituting $x=vy$ might simplify the integration step more than $y=vx$, especially if $frac{dx}{dy}$ appears more naturally or if integration of $dv/F(v)-v$ becomes complicated. Always be open to both substitutions.



Step 3: Transform into Variable Separable Form



  • Substitute $y = vx$ and $frac{dy}{dx} = v + x frac{dv}{dx}$ into the original differential equation.

  • The equation will now be in terms of $v$ and $x$ only, and can be rearranged to separate variables:
    $v + x frac{dv}{dx} = F(v)$
    $x frac{dv}{dx} = F(v) - v$
    $frac{dv}{F(v) - v} = frac{dx}{x}$ (provided $F(v) - v
    eq 0$
    )



Step 4: Integrate Both Sides



  • Integrate both sides of the separated equation:
    $int frac{dv}{F(v) - v} = int frac{dx}{x}$

  • Remember to add an arbitrary constant of integration, $C$, on one side.

  • JEE Tip: The integration step can often involve standard integrals like $int frac{1}{x} dx = ln|x|$, or methods like partial fractions, or specific trigonometric/logarithmic forms.



Step 5: Back-Substitute to Obtain the General Solution



  • After integration, replace $v$ with $frac{y}{x}$ (or $frac{x}{y}$ if $x=vy$ was used).

  • Simplify the expression to get the general solution in terms of $x$ and $y$.

  • If an initial condition $y(x_0) = y_0$ is given (for particular solution), substitute $x_0$ and $y_0$ into the general solution to find the value of $C$.



Example Walkthrough:


Solve the differential equation: $(x^2 + y^2)dx - 2xy dy = 0$



  1. Identify:
    Rearrange to $frac{dy}{dx} = frac{x^2 + y^2}{2xy}$.
    Numerator terms ($x^2, y^2$) have degree 2. Denominator term ($2xy$) has degree $1+1=2$. It's homogeneous.

  2. Substitute: Let $y = vx$. Then $frac{dy}{dx} = v + xfrac{dv}{dx}$.

  3. Transform:
    $v + xfrac{dv}{dx} = frac{x^2 + (vx)^2}{2x(vx)} = frac{x^2(1+v^2)}{2x^2v} = frac{1+v^2}{2v}$
    $xfrac{dv}{dx} = frac{1+v^2}{2v} - v = frac{1+v^2 - 2v^2}{2v} = frac{1-v^2}{2v}$
    $frac{2v}{1-v^2} dv = frac{1}{x} dx$ (Separable form)

  4. Integrate:
    $int frac{2v}{1-v^2} dv = int frac{1}{x} dx$
    Let $u = 1-v^2 Rightarrow du = -2v dv$.
    So, $int frac{-du}{u} = int frac{1}{x} dx$
    $-ln|u| = ln|x| + ln|C'|$ (using $ln|C'|$ for constant for simplification)
    $-ln|1-v^2| = ln|C'x|$
    $ln|(1-v^2)^{-1}| = ln|C'x|$
    $frac{1}{1-v^2} = C'x$ or $1-v^2 = frac{1}{C'x} = Cx$ (where $C = 1/C'$, or simply a new constant).

  5. Back-Substitute: Replace $v = frac{y}{x}$.
    $1 - left(frac{y}{x}
    ight)^2 = Cx$

    $1 - frac{y^2}{x^2} = Cx$
    $frac{x^2 - y^2}{x^2} = Cx$
    $x^2 - y^2 = Cx^3$ (This is the general solution)


Mastering this methodical approach will significantly improve your speed and accuracy in solving homogeneous differential equations in exams.

πŸ“ CBSE Focus Areas

For students preparing for the CBSE board examinations, mastering the solution of homogeneous differential equations is crucial. This topic frequently appears in the long-answer section, carrying significant marks. The focus is on a systematic approach and clear presentation of steps.



CBSE Focus Areas for Homogeneous Differential Equations:



  • Identification of Homogeneity:

    • You must be able to recognize a homogeneous differential equation of the first order and first degree. This usually involves checking if dy/dx = f(x, y) where f(x, y) is a homogeneous function of degree zero.

    • Alternatively, express it as M(x, y) dx + N(x, y) dy = 0, where M and N are homogeneous functions of the same degree.

    • Common Mistake: Not rearranging the equation correctly before attempting to identify homogeneity.



  • Standard Substitution:

    • The key to solving a homogeneous differential equation is the substitution y = vx, which implies dy/dx = v + x (dv/dx).

    • In some cases, especially when dy/dx is difficult to express, the substitution x = vy (and thus dx/dy = v + y (dv/dy)) might be more convenient. However, y = vx is more commonly tested.

    • CBSE Expectation: Clear mention of the substitution used and its differentiation.



  • Separation of Variables:

    • After substitution, the equation must transform into a separable form involving variables v and x (or v and y).

    • Careful algebraic manipulation is required to group all terms involving v with dv and all terms involving x with dx.



  • Integration Techniques:

    • The next step involves integrating both sides of the separated equation. This often requires knowledge of basic integration formulas, substitution methods, or sometimes partial fractions.

    • Tip: Practice various forms of integration that commonly arise from homogeneous equations.



  • Resubstitution and General Solution:

    • After integration and finding the constant C, remember to substitute back v = y/x to express the solution in terms of x and y. This gives the general solution.

    • Simplify the expression to its simplest form where possible.



  • Particular Solution (Initial Value Problems):

    • CBSE frequently asks for the particular solution, where an initial condition (e.g., y(xβ‚€) = yβ‚€) is given.

    • After finding the general solution, substitute the given values of x and y into it to determine the value of the constant C.

    • Then, write the general solution with the calculated value of C. This step carries distinct marks.





CBSE vs. JEE Main Perspective:


While the fundamental method for solving homogeneous differential equations remains the same, the complexity differs:



























Aspect CBSE Board Exams JEE Main
Equation Complexity Relatively straightforward algebraic terms, direct application. Can involve more complex functions (trigonometric, exponential, logarithmic) leading to more intricate algebraic manipulation and integration.
Integration Standard integration methods, often requiring basic partial fractions or simple substitution. May require advanced integration techniques, clever substitutions, or properties of definite integrals (if part of a broader problem).
Problem Context Direct solution, finding general/particular solution. Often part of a multi-concept problem, requiring integration with other topics (e.g., area under curve, limits, properties of functions).

For CBSE, focus on neatness, step-by-step clarity, and accuracy in calculations, especially during integration and finding the constant C.

πŸŽ“ JEE Focus Areas

Homogeneous differential equations are a crucial topic for JEE Main, bridging the concepts of differential equations with various integration techniques. A strong understanding of this method is essential for scoring well.



Understanding Homogeneous Differential Equations


A first-order, first-degree differential equation is said to be homogeneous if it can be expressed in the form:



  • $frac{dy}{dx} = fleft(frac{y}{x}
    ight)$

  • OR $frac{dx}{dy} = gleft(frac{x}{y}
    ight)$


Alternatively, a differential equation $M(x,y)dx + N(x,y)dy = 0$ is homogeneous if both $M(x,y)$ and $N(x,y)$ are homogeneous functions of the same degree. A function $f(x,y)$ is homogeneous of degree $n$ if $f(tx,ty) = t^n f(x,y)$.



JEE Focus Areas: Method of Solution


The standard method to solve a homogeneous differential equation involves a clever substitution that converts it into a variable separable form. Here are the steps and key considerations for JEE:





  1. Identification:

    • Look for equations where the sum of powers of $x$ and $y$ in each term is the same. For example, $x^2 + y^2$, $xy$, $y^2$, $x^2$ are all homogeneous terms of degree 2.

    • If $M(x,y)dx + N(x,y)dy = 0$, check if $M(x,y)$ and $N(x,y)$ are homogeneous functions of the same degree.




  2. Strategic Substitution:

    • If the equation is in the form $frac{dy}{dx} = fleft(frac{y}{x}
      ight)$, substitute $mathbf{y = vx}$.


      • This implies $frac{dy}{dx} = v + x frac{dv}{dx}$ (using the product rule).


    • If the equation is in the form $frac{dx}{dy} = gleft(frac{x}{y}
      ight)$, substitute $mathbf{x = vy}$.


      • This implies $frac{dx}{dy} = v + y frac{dv}{dy}$.


    • JEE Tip: Choose the substitution that minimizes algebraic complexity. If it's easier to express as $dx/dy$, use $x=vy$. Otherwise, $y=vx$ is generally preferred.




  3. Separation of Variables:

    • After substitution, the equation will transform into a variable separable form, usually involving $v$ and $x$ (or $v$ and $y$).

    • Rearrange the terms to get all $v$ terms with $dv$ and all $x$ terms with $dx$ (or $y$ terms with $dy$). For example, $frac{dv}{f(v) - v} = frac{dx}{x}$.




  4. Integration:

    • Integrate both sides of the separated equation. This step is often the most challenging part in JEE problems.

    • Be proficient with various integration techniques: partial fractions, trigonometric substitutions, standard integral formulas (e.g., $int frac{dx}{x^2+a^2}$, $int frac{dx}{x^2-a^2}$), and integration by substitution.

    • Common Pitfall: Forgetting the constant of integration 'C' or making algebraic errors during integration. Also, remember to use absolute values when integrating $frac{1}{x}$ or $frac{1}{v}$ to get $ln|x|$ or $ln|v|$.




  5. Resubstitution:

    • After integration, replace $v$ with $frac{y}{x}$ (or $frac{x}{y}$) to obtain the general solution in terms of $x$ and $y$.




  6. Particular Solution (if required):

    • If initial conditions (e.g., $y(x_0) = y_0$) are given, substitute these values into the general solution to find the specific value of the constant 'C'. This gives the particular solution.





CBSE vs. JEE Perspective


While CBSE also covers homogeneous differential equations, JEE questions often present a higher level of complexity:



  • Integration Difficulty: JEE problems frequently involve more intricate integrals requiring advanced techniques like partial fractions or specific standard forms that demand good recall.

  • Algebraic Manipulation: The initial algebraic setup and subsequent rearrangement after substitution can be more involved in JEE.

  • Initial Value Problems: JEE often focuses on finding particular solutions given initial conditions, testing your ability to correctly use these to determine the constant of integration.

  • Problem Recognition: Quickly identifying the homogeneous form amongst other types of differential equations (e.g., linear, variable separable) is key in JEE.



Mastering homogeneous differential equations requires not just understanding the method but also strong algebraic skills and a comprehensive grasp of integration techniques. Practice a variety of problems to build speed and accuracy.

🌐 Overview
A first-order homogeneous differential equation has RHS as a homogeneous function of x and y of the same degree: dy/dx = F(y/x). Use substitution y = v x (or x = v y) to reduce it to a separable equation in v and x, then integrate and apply initial conditions.
πŸ“š Fundamentals
β€’ Homogeneous ODE: dy/dx = F(y/x).
β€’ With y = v x: v + x dv/dx = F(v) β‡’ x dv/dx = F(v) βˆ’ v β‡’ ∫ dv/(F(v) βˆ’ v) = ∫ dx/x.
β€’ Solve for v, then y = v x; include constant solutions if F(v) βˆ’ v = 0.
πŸ”¬ Deep Dive
Scale invariance and similarity solutions; connection to direction fields in slope coordinates; uniqueness near singular points (brief).
🎯 Shortcuts
β€œHomog? Try y = v x; then split and integrate.”
πŸ’‘ Quick Tips
β€’ Sometimes x = v y is easier if RHS depends on x/y.
β€’ Watch domains where division by x (or y) is invalid.
β€’ Consider implicit final answers when inversion is messy.
🧠 Intuitive Understanding
If the ODE depends only on the ratio y/x, scale invariance suggests setting y proportional to x (y = v x) to simplify; then solve for how v changes with x.
🌍 Real World Applications
β€’ Scale-invariant models in physics/economics where only ratios matter.
β€’ Direction fields with radial symmetry after re-scaling.
πŸ”„ Common Analogies
β€’ Reducing a two-variable dependence to a single parameter (slope v = y/x) that captures the essential ratio behavior.
πŸ“‹ Prerequisites
Definition of homogeneity for functions; separable equations; basic integration; initial value problems.
⚠️ Common Exam Traps
β€’ Forgetting dy/dx = v + x dv/dx after substitution.
β€’ Treating non-homogeneous RHS as homogeneous.
β€’ Missing constant-slope branches from F(v) = v.
⭐ Key Takeaways
β€’ Test for homogeneity by checking dependence on y/x.
β€’ Substitution y = v x reduces problem to separable form.
β€’ Don’t miss constant-slope (v = const) solutions from F(v) = v.
🧩 Problem Solving Approach
1) Confirm homogeneity; set y = v x.
2) Derive v + x dv/dx = F(v); separate variables.
3) Integrate and back-substitute y = v x.
4) Apply initial conditions; consider special v roots.
πŸ“ CBSE Focus Areas
Identify homogeneous equations and execute y = v x substitution cleanly; integrate standard forms.
πŸŽ“ JEE Focus Areas
Edge cases with constant v; disguised homogeneity after algebraic manipulation; partial fractions in v-integral.

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🎯IIT-JEE Main Problems (6)

Problem 255
Easy 4 Marks
If <code>y(x)</code> is the solution of the differential equation <code>x<sup>2</sup> dy + (xy + y<sup>2</sup>) dx = 0</code> such that <code>y(1) = 1</code>, then <code>y(e)</code> is equal to:
Show Solution
1. Rewrite the differential equation as <code>dy/dx = -(xy + y<sup>2</sup>)/x<sup>2</sup> = -(y/x + (y/x)<sup>2</sup>)</code>. This is a homogeneous differential equation. 2. Substitute <code>y = vx</code>, so <code>dy/dx = v + x(dv/dx)</code>. 3. Substitute into the equation: <code>v + x(dv/dx) = -(v + v<sup>2</sup>)</code>. 4. Separate variables: <code>x(dv/dx) = -2v - v<sup>2</sup> = -v(2+v)</code>, leading to <code>dv / (v(2+v)) = -dx / x</code>. 5. Integrate both sides using partial fractions: <code>(1/2) ∫(1/v - 1/(2+v)) dv = -∫(1/x) dx</code>. 6. This yields <code>(1/2) (ln|v| - ln|2+v|) = -ln|x| + C</code>, which simplifies to <code>ln|v / (2+v)| = -2ln|x| + K</code>. 7. Exponentiate to get <code>v / (2+v) = A/x<sup>2</sup></code>. 8. Substitute back <code>v = y/x</code>: <code>(y/x) / (2 + y/x) = A/x<sup>2</sup></code>, which simplifies to <code>y / (2x + y) = A/x<sup>2</sup></code> or <code>y x<sup>2</sup> = A (2x + y)</code>. 9. Use the initial condition <code>y(1) = 1</code>: <code>1 * 1<sup>2</sup> = A (2*1 + 1) => A = 1/3</code>. 10. The particular solution is <code>y x<sup>2</sup> = (1/3) (2x + y)</code>. 11. To find <code>y(e)</code>, substitute <code>x=e</code>: <code>y(e) * e<sup>2</sup> = (1/3) (2e + y(e))</code>. 12. Solve for <code>y(e)</code>: <code>3e<sup>2</sup> y(e) = 2e + y(e) => y(e) (3e<sup>2</sup> - 1) = 2e => y(e) = 2e / (3e<sup>2</sup> - 1)</code>.
Final Answer: <code>2e / (3e<sup>2</sup> - 1)</code>
Problem 255
Easy 4 Marks
If <code>y(x)</code> is the solution of the differential equation <code>dy/dx = (x+y)/x</code> such that <code>y(1)=1</code>, then the value of <code>y(e)</code> is:
Show Solution
1. Rewrite the differential equation as <code>dy/dx = 1 + y/x</code>. This is a homogeneous differential equation. 2. Substitute <code>y = vx</code>, so <code>dy/dx = v + x(dv/dx)</code>. 3. Substitute into the equation: <code>v + x(dv/dx) = 1 + v</code>. 4. Simplify: <code>x(dv/dx) = 1</code>. 5. Separate variables: <code>dv = dx/x</code>. 6. Integrate both sides: <code>∫dv = ∫(1/x)dx</code>, which gives <code>v = ln|x| + C</code>. 7. Substitute back <code>v = y/x</code>: <code>y/x = ln|x| + C</code>, so <code>y = x(ln|x| + C)</code>. 8. Use initial condition <code>y(1)=1</code>: <code>1 = 1(ln|1| + C) => 1 = 0 + C => C = 1</code>. 9. The particular solution is <code>y = x(ln|x| + 1)</code>. 10. To find <code>y(e)</code>, substitute <code>x=e</code>: <code>y(e) = e(ln|e| + 1) = e(1 + 1) = 2e</code>.
Final Answer: <code>2e</code>
Problem 255
Easy 4 Marks
The general solution of the differential equation <code>(x<sup>2</sup> + y<sup>2</sup>)dx - 2xy dy = 0</code> is:
Show Solution
1. Rewrite the differential equation as <code>dy/dx = (x<sup>2</sup> + y<sup>2</sup>) / (2xy) = (1 + (y/x)<sup>2</sup>) / (2(y/x))</code>. This is a homogeneous differential equation. 2. Substitute <code>y = vx</code>, so <code>dy/dx = v + x(dv/dx)</code>. 3. Substitute into the equation: <code>v + x(dv/dx) = (1 + v<sup>2</sup>) / (2v)</code>. 4. Separate variables: <code>x(dv/dx) = (1 + v<sup>2</sup>) / (2v) - v = (1 - v<sup>2</sup>) / (2v)</code>, leading to <code>2v / (1 - v<sup>2</sup>) dv = dx / x</code>. 5. Integrate both sides. For the left side, use substitution <code>u = 1 - v<sup>2</sup></code>, so <code>du = -2v dv</code>: <code>∫(-1/u)du = ∫(1/x)dx</code>. 6. This yields <code>-ln|1 - v<sup>2</sup>| = ln|x| + ln|C'|</code>, which simplifies to <code>ln|1 / (1 - v<sup>2</sup>)| = ln|C'x|</code>. 7. Exponentiate to get <code>1 / (1 - v<sup>2</sup>) = C'x</code>. 8. Substitute back <code>v = y/x</code>: <code>1 / (1 - (y/x)<sup>2</sup>) = C'x</code>. 9. Simplify: <code>x<sup>2</sup> / (x<sup>2</sup> - y<sup>2</sup>) = C'x</code>. 10. Further simplification (assuming <code>x β‰  0</code>) gives <code>x / (x<sup>2</sup> - y<sup>2</sup>) = C'</code>, or <code>x<sup>2</sup> - y<sup>2</sup> = x/C'</code>. Let <code>K = 1/C'</code>. 11. The general solution is <code>x<sup>2</sup> - y<sup>2</sup> = Kx</code>.
Final Answer: <code>x<sup>2</sup> - y<sup>2</sup> = Kx</code> (or equivalent forms like <code>x = C(x<sup>2</sup> - y<sup>2</sup>)</code>)
Problem 255
Easy 4 Marks
The solution of the differential equation <code>(x<sup>2</sup> - y<sup>2</sup>)dx + 2xy dy = 0</code> which passes through the point <code>(1, 1)</code> is:
Show Solution
1. Rewrite the differential equation as <code>2xy dy = -(x<sup>2</sup> - y<sup>2</sup>)dx</code>, so <code>dy/dx = (y<sup>2</sup> - x<sup>2</sup>) / (2xy) = ((y/x)<sup>2</sup> - 1) / (2(y/x))</code>. This is a homogeneous differential equation. 2. Substitute <code>y = vx</code>, so <code>dy/dx = v + x(dv/dx)</code>. 3. Substitute into the equation: <code>v + x(dv/dx) = (v<sup>2</sup> - 1) / (2v)</code>. 4. Separate variables: <code>x(dv/dx) = (v<sup>2</sup> - 1) / (2v) - v = (-v<sup>2</sup> - 1) / (2v) = -(v<sup>2</sup> + 1) / (2v)</code>, leading to <code>2v / (v<sup>2</sup> + 1) dv = -dx / x</code>. 5. Integrate both sides: <code>∫(2v / (v<sup>2</sup> + 1)) dv = -∫(1/x) dx</code>, which gives <code>ln|v<sup>2</sup> + 1| = -ln|x| + ln|C|</code>. 6. Simplify to <code>ln|v<sup>2</sup> + 1| = ln|C/x|</code>, so <code>v<sup>2</sup> + 1 = C/x</code>. 7. Substitute back <code>v = y/x</code>: <code>(y/x)<sup>2</sup> + 1 = C/x</code>, which simplifies to <code>(y<sup>2</sup> + x<sup>2</sup>) / x<sup>2</sup> = C/x</code> or <code>y<sup>2</sup> + x<sup>2</sup> = Cx</code>. 8. Use the initial condition <code>(1,1)</code>: <code>1<sup>2</sup> + 1<sup>2</sup> = C*1 => 2 = C</code>. 9. The particular solution is <code>x<sup>2</sup> + y<sup>2</sup> = 2x</code>.
Final Answer: <code>x<sup>2</sup> + y<sup>2</sup> = 2x</code>
Problem 255
Easy 4 Marks
The differential equation <code>x dy/dx = y + x sin(y/x)</code> with <code>y(1) = Ο€/2</code> has a solution <code>y(x)</code> such that <code>y(e)</code> is equal to:
Show Solution
1. Rewrite the differential equation as <code>dy/dx = y/x + sin(y/x)</code>. This is a homogeneous differential equation. 2. Substitute <code>y = vx</code>, so <code>dy/dx = v + x(dv/dx)</code>. 3. Substitute into the equation: <code>v + x(dv/dx) = v + sin(v)</code>. 4. Simplify: <code>x(dv/dx) = sin(v)</code>. 5. Separate variables: <code>dv / sin(v) = dx / x</code>, which is <code>csc(v) dv = dx / x</code>. 6. Integrate both sides: <code>∫csc(v) dv = ∫(1/x) dx</code>, which gives <code>ln|tan(v/2)| = ln|x| + ln|C'|</code>. 7. Simplify to <code>tan(v/2) = C'x</code>. 8. Substitute back <code>v = y/x</code>: <code>tan(y/(2x)) = C'x</code>. 9. Use initial condition <code>y(1) = Ο€/2</code>: <code>tan((Ο€/2)/(2*1)) = C'*1 => tan(Ο€/4) = C' => C' = 1</code>. 10. The particular solution is <code>tan(y/(2x)) = x</code>. 11. To find <code>y(e)</code>, substitute <code>x=e</code>: <code>tan(y(e)/(2e)) = e</code>. 12. Solve for <code>y(e)</code>: <code>y(e)/(2e) = arctan(e) => y(e) = 2e * arctan(e)</code>.
Final Answer: <code>2e * arctan(e)</code>
Problem 255
Easy 4 Marks
If <code>y(x)</code> is the solution of the differential equation <code>x dy/dx = y(ln y - ln x + 1)</code> such that <code>y(1) = e</code>, then <code>y(e)</code> is equal to:
Show Solution
1. Rewrite the differential equation as <code>dy/dx = (y/x)(ln(y/x) + 1)</code>. This is a homogeneous differential equation. 2. Substitute <code>y = vx</code>, so <code>dy/dx = v + x(dv/dx)</code>. 3. Substitute into the equation: <code>v + x(dv/dx) = v(ln(v) + 1)</code>. 4. Simplify: <code>v + x(dv/dx) = v ln(v) + v => x(dv/dx) = v ln(v)</code>. 5. Separate variables: <code>dv / (v ln(v)) = dx / x</code>. 6. Integrate both sides. For the left side, use substitution <code>u = ln(v)</code>, so <code>du = (1/v) dv</code>: <code>∫(1/u)du = ∫(1/x)dx</code>. 7. This yields <code>ln|ln(v)| = ln|x| + ln|C|</code>, which simplifies to <code>ln|ln(v)| = ln|Cx|</code>. 8. Exponentiate to get <code>ln(v) = Cx</code> (assuming <code>ln(v) > 0</code>), then <code>v = e^(Cx)</code>. 9. Substitute back <code>v = y/x</code>: <code>y/x = e^(Cx)</code>, so <code>y = x e^(Cx)</code>. 10. Use initial condition <code>y(1) = e</code>: <code>e = 1 * e^(C*1) => e = e^C => C = 1</code>. 11. The particular solution is <code>y = x e^x</code>. 12. To find <code>y(e)</code>, substitute <code>x=e</code>: <code>y(e) = e * e^e = e^(e+1)</code>.
Final Answer: <code>e^(e+1)</code>

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⚠️Common Mistakes to Avoid (60)

Minor Other

❌ Incomplete Back-Substitution of `v = y/x`

Students often correctly apply the substitution `y = vx` (or `x = vy`), differentiate, separate variables, and integrate the transformed equation. However, a common minor error is to forget to substitute `v` back as `y/x` (or `x/y`) in the final integrated expression, or to leave the solution in a form that is not fully expressed in terms of the original variables `x` and `y`.
πŸ’­ Why This Happens:
This mistake typically arises from:
  • Haste and time pressure during exams, leading to overlooking the final critical step.
  • A mindset that the problem is 'solved' once the integration is completed, without realizing the solution must be in terms of the original dependent and independent variables.
  • Lack of attention to the complete form required for the final answer.
βœ… Correct Approach:
After successfully separating variables and integrating, always remember to substitute `v = y/x` (or `x = vy` if `x` is the dependent variable) back into your integrated equation. This ensures the solution is expressed in terms of `y` and `x`. Simplify the resulting equation to present `y` explicitly as a function of `x` if possible and required (for CBSE), or in an implicit form (often acceptable in JEE Advanced if not explicitly asked otherwise).
πŸ“ Examples:
❌ Wrong:
Consider the homogeneous equation `dy/dx = (x+y)/x`. After substitution `y=vx` and simplification, we get `dv = dx/x`. Integrating both sides yields `v = ln|x| + C`.
A common minor mistake is to present this as the final answer, i.e., v = ln|x| + C, without replacing v with y/x.
βœ… Correct:
Continuing from the wrong example, once `v = ln|x| + C` is obtained, the correct final step is to substitute `v = y/x` back:
y/x = ln|x| + C
Then, to fully express `y` in terms of `x` (as is often expected):
y = x(ln|x| + C)
This is the complete and correct solution in terms of the original variables.
πŸ’‘ Prevention Tips:
  • Develop a mental checklist for solving homogeneous equations: Identify homogeneity → Substitute `y=vx` → Separate variables → Integrate → Back-substitute `v=y/x` → Simplify.
  • Always pause at the end to check if your answer is expressed in terms of the original variables (`y` and `x`).
  • Practice writing out full solutions, including all algebraic manipulations, to reinforce the complete process.
JEE_Advanced
Minor Conceptual

❌ Incorrect Transformation of <code>dy/dx</code> after Substitution in Homogeneous Equations

Students frequently apply the substitution y=vx (or x=vy) in homogeneous differential equations but fail to correctly transform dy/dx (or dx/dy). Instead of using the product rule, they might incorrectly replace dy/dx with just dv/dx or v, leading to an entirely wrong variable separable form.
πŸ’­ Why This Happens:
This error stems from a lack of understanding that v is a function of x (i.e., v = y/x), not a constant. Students often forget to apply the product rule of differentiation when transforming dy/dx = d/dx(vx). This indicates a conceptual gap in the application of calculus fundamentals combined with the specific substitution technique.
βœ… Correct Approach:
When substituting y=vx, it's crucial to remember that both v and x are variables. Therefore, differentiate y=vx with respect to x using the product rule:
dy/dx = d/dx(vx) = v(d/dx(x)) + x(d/dx(v)) = v(1) + x(dv/dx) = v + x(dv/dx).
Similarly, if x=vy, then dx/dy = v + y(dv/dy). Always substitute this full expression into the differential equation.
πŸ“ Examples:
❌ Wrong:
Consider the homogeneous equation dy/dx = (x^2 + y^2) / (xy).
After substituting y=vx, we get dy/dx = (x^2 + (vx)^2) / (x(vx)) = (x^2 + v^2x^2) / (vx^2) = (1 + v^2)/v.
A common mistake is to then incorrectly write:
dv/dx = (1 + v^2)/v
This omits the 'v' term from the left-hand side, leading to an incorrect differential equation.
βœ… Correct:
For the same equation dy/dx = (x^2 + y^2) / (xy):
1. Substitute y=vx, which simplifies the right-hand side to (1 + v^2)/v.
2. Crucially, replace dy/dx with its correct transformation: v + x(dv/dx).
3. The equation becomes: v + x(dv/dx) = (1 + v^2)/v
4. Now, separate variables: x(dv/dx) = (1 + v^2)/v - v = (1 + v^2 - v^2)/v = 1/v
5. This correctly leads to the variable separable form: v dv = dx/x.
πŸ’‘ Prevention Tips:
  • Understand the Derivation: Don't just memorize the formula dy/dx = v + x(dv/dx); understand its derivation using the product rule.
  • Practice the Substitution Step: Explicitly write out the differentiation step for y=vx until it becomes second nature.
  • Double-Check: Before separating variables, always verify that dy/dx has been replaced by the complete expression v + x(dv/dx) (or dx/dy = v + y(dv/dy)).
JEE_Main
Minor Calculation

❌ Algebraic Errors in Isolating <code>dv/dx</code> Term

After performing the substitution (e.g., y=vx) and simplifying the differential equation to the form v + x(dv/dx) = f(v), students frequently make algebraic errors when isolating the x(dv/dx) term. This typically involves mistakes in subtracting v from f(v), leading to an incorrect expression for x(dv/dx). These errors can be due to improper fraction subtraction, sign errors, or incorrect distribution, which subsequently makes variable separation and integration incorrect.

πŸ’­ Why This Happens:
  • Haste and Lack of Meticulousness: Rushing through algebraic steps, especially with fractions and negative signs.
  • Weak Foundation in Fraction Arithmetic: Forgetting the rules for adding/subtracting algebraic fractions, particularly when one term is a simple variable or integer (e.g., treating v as v/1).
  • Sign Errors: Mistakes in distributing negative signs when expanding terms.
βœ… Correct Approach:

Always approach the algebraic simplification of f(v) - v with extreme care. Treat v as v/1 and find a common denominator before combining terms. Double-check all distributions and sign changes. The goal is to obtain an accurate expression for x(dv/dx) which can then be easily separated into dv/g(v) = dx/x for integration.

πŸ“ Examples:
❌ Wrong:

Consider dy/dx = (y-x)/(y+x). After substitution y=vx, we get:

v + x(dv/dx) = (vx-x)/(vx+x)
v + x(dv/dx) = (v-1)/(v+1)
x(dv/dx) = (v-1)/(v+1) - v

Incorrect Calculation:

x(dv/dx) = (v-1 - v)/(v+1)  (Mistake: 'v' is incorrectly subtracted only from the numerator)
x(dv/dx) = -1/(v+1)
βœ… Correct:

Continuing from x(dv/dx) = (v-1)/(v+1) - v:

Correct Calculation:

x(dv/dx) = (v-1)/(v+1) - v/1
x(dv/dx) = (v-1 - v*(v+1))/(v+1)  (Correct: 'v' is multiplied by the denominator before subtracting)
x(dv/dx) = (v-1 - v2 - v)/(v+1)
x(dv/dx) = (-1 - v2)/(v+1)
x(dv/dx) = -(1+v2)/(v+1)
πŸ’‘ Prevention Tips:
  • Practice Fraction Simplification: Regularly solve problems involving algebraic fraction addition/subtraction.
  • Write Down Every Step: For complex algebraic manipulations, avoid mental calculations. Write each step explicitly to reduce errors.
  • Verify Signs and Distribution: After each step, quickly re-check if signs are correctly managed and terms are distributed properly.
  • JEE Focus: In JEE, these small calculation errors can lead to entirely wrong answers, costing valuable marks. Prioritize accuracy over speed in such critical steps.
JEE_Main
Minor Formula

❌ Incorrect Application of the Derivative Formula in Homogeneous Substitution

Students often correctly identify a differential equation as homogeneous and decide to use the substitution y = vx (or x = vy). However, a common minor mistake is misapplying the product rule when differentiating this substitution to find dy/dx or dx/dy. They might incorrectly write dy/dx = dv/dx, dy/dx = v, or miss one of the terms, leading to an incorrect transformed equation.
πŸ’­ Why This Happens:
This error primarily stems from a rushed approach to differentiation or a lack of thorough understanding of the product rule in calculus. Students might forget that v is a function of x (when y = vx), treating it as a constant, or misapply the product rule itself. This is a foundational calculus error that impacts the solution of differential equations.
βœ… Correct Approach:
For a homogeneous differential equation of the form dy/dx = f(y/x), the correct substitution is y = vx. Differentiating both sides with respect to x using the product rule (d(uv)/dx = u(dv/dx) + v(du/dx)) yields:
dy/dx = v + x(dv/dx).
Similarly, for dx/dy = g(x/y), the substitution x = vy implies:
dx/dy = v + y(dv/dy).
Using these complete and correct derivative forms is crucial for the subsequent separation of variables.
πŸ“ Examples:
❌ Wrong:
If y = vx, some students incorrectly write:
  • dy/dx = dv/dx
  • dy/dx = v
βœ… Correct:
Given y = vx, the correct differentiation using the product rule is:
dy/dx = (d/dx(v)) * x + v * (d/dx(x)) = dv/dx * x + v * 1 = v + x(dv/dx).
πŸ’‘ Prevention Tips:
  • Always write out the differentiation step: Explicitly apply the product rule when finding dy/dx or dx/dy after substitution.
  • Review the product rule: Revisit and strengthen your understanding of basic differentiation rules, especially the product rule. Remember that v is a function of x (or y).
  • Cross-check your transformation: After substituting dy/dx (or dx/dy) and y/x (or x/y) into the original equation, ensure the resulting equation only contains v and x (or y) variables, ready for separation.
JEE_Main
Minor Unit Conversion

❌ Ignoring Unit Consistency in Applied Homogeneous DE Problems

Students correctly solve a homogeneous differential equation mathematically but fail to ensure unit consistency for physical quantities represented by variables when evaluating the solution at specific points or using initial conditions. They might substitute values directly without converting them to a uniform system of units, especially when variables represent quantities like length, time, or concentration, which are given in different units.
πŸ’­ Why This Happens:
  • Homogeneous differential equations are often taught in a purely mathematical context where units are not explicitly discussed, leading students to assume all variables are dimensionless.
  • Lack of careful attention to detail in problem statements that implicitly define units for physical variables.
  • A habit of solving purely mathematical problems without considering the physical interpretations or unit requirements when the problem has an applied context.
βœ… Correct Approach:
  • Identify Physical Variables: Always check if the variables (e.g., x, y) in the differential equation represent physical quantities (e.g., distance, time, concentration).
  • Ensure Unit Uniformity: Before applying initial conditions or evaluating the solution at specific points, ensure all given numerical values are converted to a single, consistent system of units (e.g., meters for length, seconds for time).
  • Consistent Substitution: If x represents length in meters when the constant of integration is determined, then any subsequent x value (e.g., 100 cm) must be converted to meters (1 m) before substitution into the solved equation.
πŸ“ Examples:
❌ Wrong:

Problem: A homogeneous differential equation's solution is found to be y = x * (C + ln|x|), where x is a length. Given an initial condition y(1 meter) = 2. Find y when x = 50 cm.

Wrong Approach:

  1. From y(1 meter) = 2: 2 = 1 * (C + ln|1|) => C = 2. (Correctly determined C assuming x in meters)
  2. Student directly substitutes x = 50 (from 50 cm) into the solution: y = 50 * (2 + ln|50|). This mixes units as C was derived using x in meters, but x=50 is treated as meters, leading to an incorrect numerical value.
βœ… Correct:

Using the same problem:

  1. Determine Constant C: Using the initial condition y(1 meter) = 2 in y = x * (C + ln|x|):
    2 = 1 * (C + ln|1|)
    => C = 2.
    The solution is y = x * (2 + ln|x|), where x is assumed to be in meters for consistency with how C was derived.
  2. Convert x to Consistent Units: To find y when x = 50 cm, convert x to meters:
    x = 50 cm = 0.5 meters.
  3. Substitute Consistently: Substitute the consistent value into the solution:
    y = 0.5 * (2 + ln|0.5|). This yields the correct numerical result.
πŸ’‘ Prevention Tips:
  • Read Carefully: Always read problem statements thoroughly to identify if variables represent physical quantities and if any units are specified or implied.
  • Standardize Units: Before performing any numerical calculations (especially for initial conditions or final evaluation), explicitly convert all given values to a single, consistent system of units (e.g., SI units or the units used to establish constants).
  • Dimensional Analysis (JEE Advanced): For more complex problems, briefly consider the dimensions of terms in the equation to catch inconsistencies, though this is less common for pure homogeneous DEs in JEE Main.
JEE_Main
Minor Sign Error

❌ <span style='color: #FF5733;'>Incorrect Sign Handling during Variable Separation and Integration</span>

Students frequently make sign errors when rearranging terms to separate variables or during the integration step, particularly with expressions involving a negative coefficient or a reciprocal with a negative exponent. This minor algebraic oversight can lead to a completely different and incorrect solution for the homogeneous differential equation.
πŸ’­ Why This Happens:
  • Rushing Algebra: Students often rush through the algebraic manipulations to separate variables (dv and dx terms), leading to errors in transposing terms or factoring out negative signs.
  • Integration Oversight: A common mistake is integrating ∫ -1/x dx as ln|x| instead of the correct -ln|x|, or similarly for other functions like ∫ 1/(a-x) dx = -ln|a-x|.
  • Lack of Double-Checking: Failing to re-verify the signs after each step of rearrangement or integration.
βœ… Correct Approach:
Always be meticulous with signs throughout the entire process. After substituting y=vx (or x=vy) and simplifying, carefully rearrange terms to separate v and x variables. Pay extra attention to negative signs that arise from subtraction or division. During integration, recall standard integral formulas precisely, especially those involving signs (e.g., ∫ f'(x)/f(x) dx = ln|f(x)| and ∫ 1/(a-x) dx = -ln|a-x|).
πŸ“ Examples:
❌ Wrong:
Consider: dy/dx = (y^2 - x^2) / (2xy)
After substitution y=vx and simplification, we get:
x(dv/dx) = -(1+v^2)/(2v)
Separating variables:
2v/(1+v^2) dv = -dx/x
Wrong Integration Step: A student might write ∫ 2v/(1+v^2) dv = ∫ dx/x (missing the negative sign on the right side).
This leads to: ln(1+v^2) = ln|x| + C
βœ… Correct:
Correct Integration Step for the above equation:
∫ 2v/(1+v^2) dv = ∫ -dx/x
Performing the integration correctly:
ln(1+v^2) = -ln|x| + C
Applying logarithm properties:
ln(1+v^2) + ln|x| = C
ln|x(1+v^2)| = C
x(1+v^2) = e^C = K (where K is a constant)
Substituting back v=y/x:
x(1 + y^2/x^2) = K
x((x^2+y^2)/x^2) = K
(x^2+y^2)/x = K or x^2+y^2 = Kx
πŸ’‘ Prevention Tips:
  • Step-by-Step Writing: Avoid doing multiple algebraic steps mentally. Write down each rearrangement clearly.
  • Highlight Negative Signs: Mentally or physically highlight negative signs that appear during separation of variables to ensure they are carried correctly.
  • Recall Integral Formulas: Before integrating, pause and explicitly recall the correct integral formula, especially for forms like ∫ 1/(a-bx) dx.
  • Verification: After integrating, do a quick mental differentiation of your answer to see if you get the integrand back (including the sign).
  • JEE Specific: In JEE Main, even a minor sign error can lead to selecting an incorrect option or getting zero marks. Practice is key to internalizing these steps.
JEE_Main
Minor Approximation

❌ Premature or Incorrect Approximation in Algebraic Simplification

Students often attempt to simplify expressions involving the substituted variable `v = y/x` (or `x/y`) by using approximation techniques (e.g., binomial approximation, `(1+x)^n β‰ˆ 1+nx`, or `sin(x) β‰ˆ x` for small `x`). This is incorrect for solving homogeneous differential equations, as the method requires exact algebraic manipulation and integration, not approximation, unless explicitly stated otherwise in the problem.
πŸ’­ Why This Happens:
  • Confusion with Other Topics: Students may misapply approximation techniques learned in limits, series expansions, or physics problems where such approximations are valid under specific conditions.
  • Desire for Simplification: An eagerness to simplify complex-looking expressions or make the subsequent integration step easier can lead to premature approximation.
  • Misunderstanding of `v`: Not recognizing `v = y/x` as a general variable throughout the solution process, but rather treating it as an infinitesimally small quantity that can be approximated, without any such condition given in the problem.
βœ… Correct Approach:
The standard method for solving homogeneous differential equations involves an exact substitution (`y=vx` or `x=vy`) and then separating variables. Every algebraic manipulation, rearrangement, and integration step must be performed precisely without any approximations. Only if the problem specifically asks for an approximate solution under certain limiting conditions should approximations be considered, which is rare for this topic in JEE Main.
πŸ“ Examples:
❌ Wrong:
Consider solving `dy/dx = (x^2 + y^2)/(x^2 + xy)`.
Substitute `y=vx`, so `dy/dx = v + x dv/dx`.
`v + x dv/dx = (x^2 + (vx)^2) / (x^2 + x(vx))`
`= (x^2(1+v^2)) / (x^2(1+v))`
`= (1+v^2) / (1+v)`.
A student might incorrectly approximate: If `v` is considered 'small', then `(1+v^2)/(1+v) β‰ˆ (1+v^2)(1-v) β‰ˆ 1 - v + v^2 - v^3`. They would then proceed with this approximated expression, leading to an incorrect solution.
βœ… Correct:
Continuing from the previous step for `dy/dx = (x^2 + y^2)/(x^2 + xy)`:
`v + x dv/dx = (1+v^2) / (1+v)`.
The correct approach involves exact algebraic manipulation:
`x dv/dx = (1+v^2) / (1+v) - v`
`x dv/dx = (1+v^2 - v(1+v)) / (1+v)`
`x dv/dx = (1+v^2 - v - v^2) / (1+v)`
`x dv/dx = (1-v) / (1+v)`.
Now, separate variables and integrate:
`(1+v)/(1-v) dv = dx/x`.
This exact integral would lead to the precise solution of the differential equation. No approximation is used or implied.
πŸ’‘ Prevention Tips:
  • Strictly adhere to Exact Algebra: Always perform precise algebraic substitutions and simplifications. Do not introduce approximations unless explicitly instructed by the problem or dealing with specific limits/boundary conditions.
  • Understand `v`'s Nature: Remember that `v = y/x` is a variable, not necessarily a 'small' quantity. Its value changes as `x` and `y` change.
  • Review Problem Context: For JEE Main, problems involving homogeneous differential equations almost always require exact solutions. Approximation is generally not part of the standard solution methodology for this topic.
  • Double-Check Steps: Carefully verify each step of algebraic manipulation to ensure no term has been approximated without justification.
JEE_Main
Minor Other

❌ <span style='color: #FF0000;'>Confusing or Incorrect Derivative Substitution (y=vx vs x=vy)</span>

Students often correctly identify a differential equation as homogeneous but then mechanically apply the substitution y = vx and dy/dx = v + x(dv/dx), even when the equation is more naturally expressed in the form dx/dy = f(x/y), for which the appropriate substitution is x = vy and dx/dy = v + y(dv/dy). This oversight, while not always leading to an incorrect answer, often results in significantly more complex algebraic manipulation and integration steps, increasing the likelihood of errors and wasting valuable exam time.
πŸ’­ Why This Happens:
  • Over-reliance on a single method: Students are typically introduced to homogeneous equations with the y=vx substitution first and tend to stick to it.
  • Lack of analytical thinking: Not taking a moment to analyze which form (dy/dx or dx/dy) simplifies the expression and subsequent integration.
  • Time pressure: In a rush, students might jump to the first familiar substitution rather than optimizing.
βœ… Correct Approach:
After confirming a differential equation is homogeneous:
  1. Attempt to express it in both dy/dx = f(y/x) and dx/dy = g(x/y) forms.
  2. Choose the form that looks simpler or leads to a less complicated algebraic expression for f(v) or g(v) after substitution.
  3. If dy/dx = f(y/x) is simpler, use y = vx and dy/dx = v + x(dv/dx).
  4. If dx/dy = g(x/y) is simpler, use x = vy and dx/dy = v + y(dv/dy).
  5. Proceed with separation of variables and integration.
πŸ“ Examples:
❌ Wrong:
Consider the equation: (x2 + y2) dy = 2xy dx.
Student tries to solve using y=vx:
dy/dx = 2xy / (x2 + y2)
Substituting y=vx and dy/dx = v + x(dv/dx):
v + x(dv/dx) = 2x(vx) / (x2 + (vx)2) = 2v / (1+v2)
x(dv/dx) = (v - v3) / (1+v2)
Separating variables leads to: ∫ (1+v2) / (v(1-v2)) dv = ∫ (1/x) dx.
The integral on the LHS requires partial fraction decomposition, which is time-consuming.
βœ… Correct:
For the same equation: (x2 + y2) dy = 2xy dx.
Rewrite as dx/dy = (x2 + y2) / (2xy).
Substitute x=vy and dx/dy = v + y(dv/dy):
v + y(dv/dy) = ( (vy)2 + y2) / (2(vy)y) = (v2+1) / (2v)
y(dv/dy) = (1-v2) / (2v)
Separating variables leads to: ∫ 2v / (1-v2) dv = ∫ (1/y) dy.
The integral on the LHS is a direct logarithmic integral (by substituting u=1-v2, du=-2v dv), which is much simpler and faster to solve.
πŸ’‘ Prevention Tips:
  • Evaluate Both Forms: Before making any substitution, quickly check which form (dy/dx or dx/dy) simplifies the right-hand side of the equation the most.
  • Practice Diverse Problems: Ensure you practice problems where x=vy is the more efficient substitution, not just y=vx.
  • Derivative Recall: Always remember the correct derivative forms: dy/dx = v + x(dv/dx) for y=vx and dx/dy = v + y(dv/dy) for x=vy.
JEE_Main
Minor Other

❌ Incorrect Algebraic Manipulation During Variable Separation

A common minor error students make is algebraic miscalculation when separating variables after substituting y = vx (or x = vy) in a homogeneous differential equation. This can lead to an incorrectly simplified expression, making subsequent integration difficult or erroneous.
πŸ’­ Why This Happens:
This mistake primarily stems from a lack of strong fundamental algebraic skills, particularly with fractions and polynomial manipulation. Students might rush the simplification, overlook signs, fail to find a common denominator correctly, or improperly factor terms, preventing the equation from being transformed into a true separable form.
βœ… Correct Approach:
After performing the substitution (e.g., y = vx, leading to dy/dx = v + x(dv/dx)), the homogeneous equation is transformed into an equation involving v and x. The critical next step is to algebraically rearrange this equation to isolate all terms involving v with dv on one side and all terms involving x with dx on the other side. This requires careful and precise algebraic manipulation, including finding common denominators and factoring.
πŸ“ Examples:
❌ Wrong:
Consider a step where you have x(dv/dx) = v^2 + 1/v - v. A student might incorrectly simplify the RHS as:
x(dv/dx) = (v^2 + 1 - v) / v
This algebraic error prevents proper separation of variables and leads to an incorrect integration.
βœ… Correct:
Following the previous step, the correct simplification should be:
x(dv/dx) = (v^2 + 1)/v - v = (v^2 + 1 - v*v)/v = (v^2 + 1 - v^2)/v = 1/v
Then, the correct separation of variables follows:
v dv = (1/x) dx
This form is readily integrable.
πŸ’‘ Prevention Tips:
  • Strengthen Algebra Fundamentals: Dedicate time to practice algebraic simplification, especially with rational expressions and fractions. This is crucial for both JEE and CBSE exams.
  • Work Systematically: Break down complex algebraic steps into smaller, manageable parts. Avoid combining too many operations in a single step.
  • Double-Check All Signs: Errors in signs are very common. Carefully review each step for correct positive/negative signs.
  • Verify Common Denominators: When combining fractions, ensure you've used the correct common denominator and adjusted numerators accordingly.
CBSE_12th
Minor Approximation

❌ <span style='color: #FF4500;'>Incorrectly Simplifying `f(v)` after `y=vx` Substitution</span>

Students often correctly identify a homogeneous differential equation and apply the substitution (e.g., y = vx or x = vy). However, a common minor error occurs during the algebraic simplification of the resulting expression on the right-hand side. Instead of simplifying it to its exact separable form, they might make algebraic mistakes, leading to a function that is either difficult to integrate or, more subtly, is an incorrect 'approximation' of the true f(v). This leads to an incorrect separation of variables and an erroneous final solution.
πŸ’­ Why This Happens:
  • Algebraic Weakness: Lack of proficiency in simplifying rational expressions involving the substitution variable (e.g., v).
  • Rushing Steps: Skipping intermediate algebraic simplification steps.
  • Misunderstanding of Equivalence: Not fully grasping that the right-hand side f(y/x) must be meticulously reduced to a function purely of v for proper separation of variables.
βœ… Correct Approach:

1. After substituting y = vx (and dy/dx = v + x dv/dx), carefully replace all y terms with vx in the original equation.

2. Factor out common x terms from the numerator and denominator to ensure the right-hand side is purely a function of v (i.e., f(v)).

3. Perform algebraic simplifications meticulously to obtain the exact f(v) function.

4. Then, proceed with separating variables (x dv/dx = f(v) - v) and integrating.

πŸ“ Examples:
❌ Wrong:

Consider the equation: dy/dx = (x^2 + y^2) / (xy)

Student's Incorrect Simplification:

1. Substitute y=vx, dy/dx = v + x dv/dx.

2. v + x dv/dx = (x^2 + (vx)^2) / (x(vx))

3. v + x dv/dx = (x^2 + v^2 x^2) / (vx^2)

4. v + x dv/dx = x^2(1 + v^2) / (vx^2)

5. v + x dv/dx = (1 + v^2) / v

6. Incorrect step: Student mistakenly simplifies (1 + v^2) / v as 1 + v (e.g., cancelling v from v^2 only, or incorrectly distributing division by v).

7. This leads to x dv/dx = (1 + v) - v = 1, which simplifies to dv = dx/x. This is an incorrect separation and will yield an erroneous solution.

βœ… Correct:

Using the same equation: dy/dx = (x^2 + y^2) / (xy)

Correct Simplification:

1. After substitution and cancellation of x^2, we correctly arrive at: v + x dv/dx = (1 + v^2) / v

2. Isolate x dv/dx: x dv/dx = (1 + v^2) / v - v

3. Combine terms on the RHS using a common denominator:

x dv/dx = (1 + v^2 - v * v) / v

x dv/dx = (1 + v^2 - v^2) / v

x dv/dx = 1 / v

4. Now, variables are correctly separated for integration: v dv = dx / x

This leads to the correct integration and the exact solution.

πŸ’‘ Prevention Tips:
  • CBSE & JEE: Both exams demand high precision in algebraic steps. A single mistake here can lead to loss of full marks.
  • Double-check Algebraic Simplifications: Always review the algebraic reduction of f(v), especially involving rational expressions.
  • Factor Out `x` Completely: Ensure that after substitution, all instances of x (or y if x=vy is used) in the numerator and denominator cancel out, leaving a function purely of v.
  • Practice Fraction Operations: Strengthen skills in adding, subtracting, multiplying, and dividing algebraic fractions.
  • Verify Form: Before proceeding to separate variables, mentally (or physically) confirm that the RHS is indeed a well-defined function of v that can be algebraically manipulated to facilitate separation.
CBSE_12th
Minor Sign Error

❌ Sign Errors in Variable Separation after Substitution

Students frequently make sign errors when rearranging terms to separate variables after making the standard substitution (y = vx or x = vy) in homogeneous differential equations. This often occurs when subtracting v (or u) from both sides or during subsequent algebraic simplification involving negative signs, leading to an incorrect integral.
πŸ’­ Why This Happens:
  • Hasty Algebraic Manipulation: Students rush through the steps, overlooking the sign changes when transposing terms.
  • Lack of Attention to Negatives: Failing to correctly distribute negative signs, especially when simplifying complex fractions or combining terms.
  • Mental Arithmetic Errors: Attempting too many steps mentally rather than writing them down, increasing the chance of sign errors.
βœ… Correct Approach:
Always perform algebraic manipulations systematically. When moving a term to the other side of the equation, explicitly change its sign. Use parentheses to group terms, particularly when dealing with subtractions or negative multipliers. Double-check each step of simplification, focusing on how signs are affected.
πŸ“ Examples:
❌ Wrong:
Consider an intermediate step:
x (dv/dx) = (v - 1)/(v + 1) + v (Incorrectly adding v instead of subtracting after moving it from LHS to RHS)
x (dv/dx) = (v - 1 + v(v + 1))/(v + 1)
x (dv/dx) = (v - 1 + v^2 + v)/(v + 1)
x (dv/dx) = (v^2 + 2v - 1)/(v + 1) (Incorrect form for integration due to wrong sign)
βœ… Correct:
Following the substitution y = vx and differentiation dy/dx = v + x (dv/dx) for an equation like dy/dx = f(v):
v + x (dv/dx) = (v - 1)/(v + 1)
x (dv/dx) = (v - 1)/(v + 1) - v (Correctly subtracting v)
x (dv/dx) = (v - 1 - v(v + 1))/(v + 1)
x (dv/dx) = (v - 1 - v^2 - v)/(v + 1)
x (dv/dx) = (-v^2 - 1)/(v + 1)
x (dv/dx) = -(v^2 + 1)/(v + 1)
Then, separating variables: (v + 1)/(v^2 + 1) dv = - (1/x) dx (Ready for correct integration)
πŸ’‘ Prevention Tips:
  • Isolate Carefully: Always ensure x (dv/dx) (or its equivalent) is isolated correctly before simplifying.
  • Use Parentheses: Employ parentheses liberally to manage terms, especially when subtracting an entire expression or when a negative sign is involved.
  • Step-by-Step Simplification: Break down complex algebraic steps into smaller, manageable ones. Avoid combining too many operations into a single step.
  • Review Signs: After each major algebraic manipulation, quickly re-check the signs of all terms to catch errors early.
CBSE_12th
Minor Unit Conversion

❌ Incomplete Variable Transformation (x, y to v)

Students often fail to completely 'convert' the original right-hand side function f(x,y) into an expression purely in terms of the new variable v (where v = y/x or v = x/y) after applying the necessary substitution. This means leaving behind x or y terms in the simplified f(v) expression, which should ideally contain only v and constants.
πŸ’­ Why This Happens:
This mistake stems from a lack of complete understanding of the algebraic manipulation required for homogeneous functions. It is often compounded by careless algebraic simplification, particularly incorrect factorization or incomplete cancellation of x (or y) terms after substituting y=vx. Students might not fully grasp that the ultimate goal is to eliminate all instances of x and y from the RHS and represent it solely as a function of v.
βœ… Correct Approach:
After substituting y=vx (or x=vy) into the homogeneous function f(x,y), the crucial step is to meticulously manipulate the algebraic expression. This involves replacing all y terms with vx (or x with vy), and then carefully factoring out the highest possible common power of x (or y) from both the numerator and the denominator. This process *must* lead to a complete cancellation of all x (or y) terms, resulting in an expression that is purely a function of v (i.e., f(v)). If any x or y terms remain, the transformation is incomplete or incorrect.
πŸ“ Examples:
❌ Wrong:
Consider the homogeneous equation dy/dx = (xΒ² + yΒ²) / (xy).
After substituting y=vx, the RHS becomes: (xΒ² + (vx)Β²) / (x(vx)) = (xΒ² + vΒ²xΒ²) / (vxΒ²).
A common minor mistake is to proceed without explicitly factoring out xΒ² before cancelling, or to make an error during implicit cancellation. For example, a student might incorrectly write something like: (xΒ² + vΒ²xΒ²) / (vxΒ²) = 1 + vΒ²/v * x if they make a mistake in assuming partial cancellation or incorrectly distribute terms. While in this simple case, direct cancellation might seem to work, failing to systematically factor can lead to errors in more complex homogeneous equations or a misunderstanding of the transformation principle.
βœ… Correct:
For dy/dx = (xΒ² + yΒ²) / (xy).
1. Substitute y=vx into the RHS: (xΒ² + (vx)Β²) / (x(vx))
2. Simplify: (xΒ² + vΒ²xΒ²) / (vxΒ²)
3. Factor out xΒ² from the numerator: xΒ²(1 + vΒ²) / (vxΒ²)
4. Cancel the xΒ² terms completely: (1 + vΒ²) / v.
This expression is now a function of v only (i.e., f(v)), demonstrating a complete and correct variable transformation.
πŸ’‘ Prevention Tips:
  • Always Verify Homogeneity: Before substitution, ensure the equation is indeed homogeneous, meaning f(tx, ty) = f(x,y), which implies f(x,y) can be written as g(y/x).
  • Systematic Factorization: After substituting y=vx (or x=vy), always explicitly factor out the common x (or y) term from both the numerator and denominator. Do not attempt direct cancellation without clear factorization.
  • Check for `v` Only: After simplification, rigorously verify that the right-hand side contains absolutely no x or y variables, only v and constants. If any x or y terms remain, a mistake has occurred.
  • Practice Algebraic Skills: Strengthen your skills in factorization, distribution, and cancellation of algebraic terms, as these are fundamental to successfully converting variables.
CBSE_12th
Minor Formula

❌ Incorrect derivation of <code>dy/dx</code> or <code>dx/dy</code> after substitution in homogeneous equations

Students often err in deriving dy/dx from y = vx (or dx/dy from x = vy), incorrectly applying the product rule, or treating v as a constant. This is a common formula understanding mistake.
πŸ’­ Why This Happens:
This error primarily stems from hasty application of the product rule of differentiation or not fully understanding that v is a function of x (when y=vx) or y (when x=vy). It represents a fundamental calculus oversight.
βœ… Correct Approach:
The correct application of the product rule is crucial for these substitutions:
  • For the substitution y = vx, differentiate both sides with respect to x:
    dy/dx = d/dx(vx) = (dv/dx)x + v(dx/dx) = v + x(dv/dx)
  • For the substitution x = vy, differentiate both sides with respect to y:
    dx/dy = d/dy(vy) = (dv/dy)y + v(dy/dy) = v + y(dv/dy)
Always remember that v is a variable function, not a constant, in these contexts.
πŸ“ Examples:
❌ Wrong:
Consider the equation dy/dx = (x+y)/x.
A student correctly identifies it as homogeneous and makes the substitution y=vx. However, they might incorrectly write dy/dx = x(dv/dx) or dy/dx = dv/dx instead of the correct v + x(dv/dx). This leads to an entirely incorrect separable equation.
βœ… Correct:
Let's solve the homogeneous differential equation dy/dx = (x+y)/x.
1. Identify as homogeneous. Substitute y = vx.
2. Correctly derive dy/dx:
dy/dx = d/dx(vx) = v + x(dv/dx)
3. Substitute this into the original equation:
v + x(dv/dx) = (x + vx)/x
v + x(dv/dx) = x(1+v)/x
v + x(dv/dx) = 1 + v
4. Separate variables:
x(dv/dx) = 1
dv = dx/x
5. Integrate both sides:
∫dv = ∫dx/x
v = ln|x| + C
6. Substitute back v = y/x:
y/x = ln|x| + C
y = x(ln|x| + C)
This demonstrates the correct application of the substitution formula for dy/dx.
πŸ’‘ Prevention Tips:
  • Explicitly write the differentiation: Always write out the product rule step carefully to avoid omissions.
  • Understand v's nature: Remember that v is a function of x (or y), so its derivative is not zero.
  • Practice differentiation: Regular practice of differentiation, especially the product rule, will solidify this concept.
  • CBSE & JEE Tip: While a minor error in concept, such mistakes can lead to significant loss of marks. Always double-check your substitution steps before proceeding to variable separation.
CBSE_12th
Minor Calculation

❌ Algebraic Errors in Variable Separation after Homogeneous Substitution

Students frequently make minor algebraic mistakes when simplifying the differential equation after substituting y = vx (or x = vy) and its derivative. These errors occur during the rearrangement of terms to separate variables v and x (or v and y). Common issues include incorrect signs, misplacing terms in fractions, or errors in combining rational expressions before the integration step.
πŸ’­ Why This Happens:
This typically stems from a lack of careful step-by-step calculation, rushing through algebraic simplification, or a weak foundation in basic algebraic manipulation and fraction operations. It's often an oversight due to lack of practice or attention, rather than a fundamental misunderstanding of the method of solving homogeneous differential equations.
βœ… Correct Approach:
After substituting y = vx and dy/dx = v + x dv/dx (or x = vy and dx/dy = v + y dv/dy), carefully isolate the v terms on one side with dv and the x (or y) terms on the other side with dx (or dy). Always perform algebraic operations (like combining fractions, taking common factors, or cross-multiplication) systematically. Double-check each step before moving to integration. Pay meticulous attention to signs and the placement of terms in numerators and denominators.
πŸ“ Examples:
❌ Wrong:
Consider a step where you have x dv/dx = v - (v^2 + 1)/v.
Wrong separation attempt:
x dv/dx = (v^2 - v^2 - 1)/v = -1/v
∫ dv = ∫ (-1/v) dx/x
(Incorrectly grouping -1/v with dx/x, leading to wrong integration variable.)
βœ… Correct:
Continuing from x dv/dx = v - (v^2 + 1)/v:
Correct simplification and separation:
x dv/dx = (v^2 - (v^2 + 1))/v = (v^2 - v^2 - 1)/v = -1/v
Now, separate variables correctly:
∫ v dv = ∫ -1/x dx
(All v terms with dv, all x terms with dx, ready for integration.)
πŸ’‘ Prevention Tips:
  • Systematic Approach: Do not skip intermediate algebraic steps. Write down each simplification clearly.
  • Practice Algebra: Regularly practice algebraic manipulation of fractions and rational expressions.
  • Sign Vigilance: Be extremely cautious with positive and negative signs during rearrangement and simplification.
  • Self-Correction: After separating variables, quickly verify that all terms involving 'v' are with 'dv' and all terms involving 'x' (or 'y') are with 'dx' (or 'dy') and that the equation is dimensionally consistent.
  • CBSE vs. JEE: While JEE problems might involve more complex algebraic forms, the fundamental principle of careful variable separation is crucial for both exams. Even minor calculation errors can lead to significant loss of marks in CBSE board exams.
CBSE_12th
Minor Conceptual

❌ Mistaking Non-Homogeneous Differential Equations for Homogeneous Ones

Students frequently misidentify differential equations as homogeneous without properly verifying the condition. This usually happens when an equation contains a constant term or when the terms in the numerator and denominator (or M and N for Mdx+Ndy=0 form) are not all of the same degree when considered as functions of x and y. Applying the standard substitution y = vx or x = vy to a non-homogeneous equation will lead to an unsolvable or unnecessarily complicated form, consuming valuable exam time.

πŸ’­ Why This Happens:

This conceptual error arises from an incomplete understanding of what constitutes a homogeneous function and, consequently, a homogeneous differential equation. Students often look for superficial similarities in the powers of x and y rather than rigorously applying the definition. They might overlook constant terms or terms where the sum of powers of x and y differs from other terms.

βœ… Correct Approach:

A first-order differential equation dy/dx = f(x,y) is homogeneous if f(x,y) is a homogeneous function of degree zero. This means f(tx, ty) = t^0 * f(x,y) = f(x,y) for any non-zero constant t. Alternatively, check if all terms in the numerator and denominator (or M(x,y) and N(x,y)) have the same total degree. If not, the equation is not homogeneous and the substitution y=vx is not applicable.

πŸ“ Examples:
❌ Wrong:

Consider the equation: dy/dx = (x + y + 1) / (x - y)

A common mistake is to assume this is homogeneous because x and y appear in both numerator and denominator. However, the constant term +1 in the numerator makes the numerator non-homogeneous. If one attempts the substitution y=vx, it will lead to an expression that cannot be easily separated into functions of v and x alone, indicating the incorrect application of the method.

βœ… Correct:

To correctly identify, let's verify if dy/dx = (x^2 + y^2) / (xy) is homogeneous.

  1. Let f(x,y) = (x^2 + y^2) / (xy).
  2. Replace x with tx and y with ty:
    f(tx, ty) = ((tx)^2 + (ty)^2) / ((tx)(ty))
    = (t^2x^2 + t^2y^2) / (t^2xy)
    = t^2(x^2 + y^2) / (t^2xy)
    = (x^2 + y^2) / (xy) = f(x,y)

Since f(tx, ty) = f(x,y), the function is homogeneous of degree zero, and thus the differential equation is homogeneous. Alternatively, observe that all terms (x^2, y^2, xy) are of degree 2. Hence, it is homogeneous.

πŸ’‘ Prevention Tips:
  • Always verify homogeneity first: Before applying any substitution, explicitly check if f(tx, ty) = f(x,y) or if all terms have the same total degree.
  • Look for constant terms: The presence of an isolated constant term (without x or y) in the numerator or denominator immediately indicates a non-homogeneous equation.
  • Practice identification: Solve problems specifically designed to identify homogeneous and non-homogeneous equations without actually solving them to strengthen your conceptual understanding.
CBSE_12th
Minor Conceptual

❌ Forgetting to Substitute Back the Auxiliary Variable 'v'

A common minor conceptual mistake is presenting the final solution in terms of the auxiliary variable 'v' (introduced as y=vx or x=vy) rather than substituting back its original expression (y/x or x/y) to obtain the solution purely in terms of the original variables (x and y).
πŸ’­ Why This Happens:
This oversight often stems from haste during the exam or a momentary lapse in understanding that 'v' was merely a temporary tool for simplifying the differential equation. Students might correctly solve for 'v' but forget the final step of expressing the solution in the required format.
βœ… Correct Approach:
After successfully separating variables and integrating, always perform the final substitution to replace 'v' with its equivalent expression in terms of the original variables. The general solution of a differential equation must be in terms of the independent and dependent variables given in the problem.
πŸ“ Examples:
❌ Wrong:
Consider the equation dy/dx = (x+y)/x. Substituting y=vx, we get v + x dv/dx = (x+vx)/x = 1+v. This simplifies to x dv/dx = 1, or dv = dx/x. Integrating gives v = ln|x| + C. A common mistake is to stop here and present v = ln|x| + C as the solution.
βœ… Correct:
Continuing from the wrong example, after obtaining v = ln|x| + C, the correct next step is to substitute back v = y/x. This yields the final general solution as y/x = ln|x| + C, or y = x(ln|x| + C). This completely expresses 'y' as a function of 'x' and the constant of integration.
πŸ’‘ Prevention Tips:
  • Systematic Approach: Always follow a clear sequence: identify type, substitute (e.g., y=vx), separate variables, integrate, and then always substitute back v=y/x.
  • Final Check: Before concluding the problem, quickly verify that the solution only contains the original variables (x, y) and the arbitrary constant of integration.
  • Understanding 'v': Remember that 'v' is an auxiliary variable introduced for method application; it is not part of the original problem's variables.
JEE_Advanced
Minor Calculation

❌ Incorrect Differentiation During Substitution (Product Rule Error)

Students frequently make a minor calculation error when differentiating the substitution `y = vx` (or `x = vy`) with respect to `x` (or `y`), failing to correctly apply the product rule or treating `v` as a constant.
πŸ’­ Why This Happens:
This mistake primarily occurs due to rushing through steps, a momentary lapse in recalling the product rule for differentiation, or incorrectly assuming `v` is a constant when it is, in fact, a function of `x` (since `v = y/x`). Students might write `dy/dx = dv/dx` or `dy/dx = v`, omitting the product rule's second term.
βœ… Correct Approach:
When using the standard substitution `y = vx` for a homogeneous differential equation, one must differentiate both sides with respect to `x` using the product rule. This yields:
  • dy/dx = d(vx)/dx = v(dx/dx) + x(dv/dx) = v + x(dv/dx)
Similarly, if `x = vy` is used (which is sometimes more convenient), then:
  • dx/dy = d(vy)/dy = v(dy/dy) + y(dv/dy) = v + y(dv/dy)
πŸ“ Examples:
❌ Wrong:

Given: dy/dx = (x^2 + y^2) / (xy)

Incorrect substitution step:

Let y = vx. Student writes: dy/dx = dv/dx (or dy/dx = v)

This leads to: dv/dx = (x^2 + (vx)^2) / (x(vx)) which is algebraically simpler but fundamentally incorrect as the left side is wrong.

βœ… Correct:

Given: dy/dx = (x^2 + y^2) / (xy)

Correct substitution step:

Let y = vx. Then, by product rule, dy/dx = v + x(dv/dx).

Substitute this into the original equation:

v + x(dv/dx) = (x^2 + (vx)^2) / (x(vx))

v + x(dv/dx) = (x^2 + v^2x^2) / (vx^2)

v + x(dv/dx) = x^2(1 + v^2) / (vx^2)

v + x(dv/dx) = (1 + v^2) / v

This is the correct equation to proceed with separation of variables and integration.

πŸ’‘ Prevention Tips:
  • Always write out the product rule: Don't skip the step of differentiating `y = vx`. Explicitly show `d(vx)/dx = v(1) + x(dv/dx)`.
  • Conceptual Clarity: Remember that `v` is a function of `x` (or `y`), not a constant. This is crucial for applying the product rule correctly.
  • JEE Advanced Check: This initial step is foundational. A mistake here will invalidate all subsequent calculations, wasting time and marks. Double-check it rigorously.
  • Practice: Work through several examples, focusing on the differentiation step until it becomes second nature.
JEE_Advanced
Minor Formula

❌ Incorrect Differentiation of the Substitution <code>y = vx</code>

A common mistake is incorrectly differentiating the substitution y = vx with respect to x. Students frequently forget to apply the product rule, or misinterpret v as a constant, leading to an incorrect expression for dy/dx.
πŸ’­ Why This Happens:
This error primarily stems from a conceptual misunderstanding that v is also a function of x (i.e., v = y/x). Students often treat v as a constant during differentiation or make a product rule error due to rushing or lack of practice.
βœ… Correct Approach:
Always apply the product rule correctly when differentiating y = vx with respect to x. Remember that both v and x are variables in this context. The correct differentiation is:
d/dx (y) = d/dx (vx)
dy/dx = v(d/dx(x)) + x(d/dx(v))
dy/dx = v + x(dv/dx).
This is a crucial formula for solving homogeneous differential equations.
πŸ“ Examples:
❌ Wrong:
When substituting y = vx into a homogeneous differential equation, students might mistakenly write:
  • dy/dx = dv/dx (treating x as constant)
  • dy/dx = v (treating v as constant)
  • dy/dx = x(dv/dx) (forgetting the v term)
βœ… Correct:
Given the substitution y = vx, the correct derivative used in solving homogeneous differential equations is:
y = vx
Applying product rule:
dy/dx = v(1) + x(dv/dx)
dy/dx = v + x(dv/dx)
This derived form is then substituted into the original differential equation.
πŸ’‘ Prevention Tips:
  • CBSE/JEE Tip: Always write down the complete differentiation step for y = vx explicitly until it becomes second nature.
  • Practice: Solve multiple problems, focusing on this specific substitution step.
  • Conceptual Clarity: Understand that v is a dependent variable (a function of x), not a constant.
JEE_Advanced
Minor Unit Conversion

❌ Ignoring Dimensional Consistency of Constants and Transcendental Function Arguments in Physical Contexts

Students often correctly solve homogeneous differential equations mathematically (e.g., using the y=vx substitution). However, when the variables 'x' and 'y' represent physical quantities (e.g., distance, time), they might overlook the dimensional consistency requirements for constants of integration or arguments of transcendental functions (like logarithms, exponentials, trigonometric functions) that arise in the solution. This can lead to misinterpretation of the constant's units or incorrect numerical calculations if specific unit conversions are later applied.
πŸ’­ Why This Happens:
  • Focus on Pure Mathematics: Students tend to prioritize the algebraic steps, neglecting the physical dimensions of variables until the very end, or sometimes not at all.
  • Lack of Explicit Dimensional Analysis: While dimensional analysis is crucial in physics, its explicit application within the mathematical solution of differential equations is less frequently emphasized in a purely mathematical context like JEE.
  • Misconception about Constants: A belief that the constant of integration 'C' can implicitly 'absorb' any dimensional inconsistencies, which is incorrect.
βœ… Correct Approach:
When dealing with variables that have units and appear in a homogeneous differential equation:
  • Unit Consistency: Ensure all terms in the equation maintain dimensional consistency throughout the derivation.
  • Dimensionless Arguments: Arguments of transcendental functions (e.g., ln(X), eX, sin(X)) must be dimensionless. If a variable 'x' has units (e.g., meters), ln(x) is dimensionally incorrect. It should be written as ln(x/x0), where x0 is a reference unit (e.g., 1 meter), making the argument x/x0 dimensionless.
  • Consistent Constant: The constant of integration 'C' must have dimensions consistent with the term it is added to. If y/x = ln|x/x0| + C, then C must also be dimensionless.
πŸ“ Examples:
❌ Wrong:
Consider a homogeneous differential equation arising from a physical problem where y represents position (meters) and x represents time (seconds). If the solution yields y = x(C + ln|x|), a student might incorrectly assume C is a dimensionless constant and interpret ln|x| directly, leading to a dimensionally inconsistent expression or misinterpreting the units of C.
βœ… Correct:
Following from the above example (assuming a different problem yielding y/x = C + ln|x| where y and x have the same units, say meters):
For dimensional consistency, the argument of ln must be dimensionless. It should be written as ln|x/x0|, where x0 is a reference length (e.g., 1 meter). The constant C is then also dimensionless.
The correct form would be: y/x = C + ln|x/x0|.
This ensures both sides of the equation are dimensionless, and C correctly represents a dimensionless constant.
πŸ’‘ Prevention Tips:
  • Be Unit-Conscious: If the problem context hints at physical quantities, always be mindful of units throughout the solution process.
  • Check Function Arguments: Explicitly check the arguments of all transcendental functions (ln, exp, sin, etc.) to ensure they are dimensionless. Introduce reference units (e.g., x/x0) if necessary.
  • Verify Constant Dimensions: Always ensure that the constant of integration 'C' has dimensions consistent with the terms it is added to.
  • JEE vs. CBSE: While JEE problems often omit physical units for mathematical clarity, developing this dimensional analysis habit is crucial for deeper understanding and for physics/chemistry problems where such nuances are vital.
JEE_Advanced
Minor Sign Error

❌ Sign Error in Integrating `1/(a - bx)` Forms

A common minor sign error in solving homogeneous differential equations occurs during the integration step, particularly when the separated variables result in an integral of the form ∫1/(a - bx) dx or ∫1/(a - bv) dv. Students often incorrectly integrate this as ln|a - bx|, failing to account for the negative coefficient of `x` (or `v`). This omission leads to an incorrect sign in the final solution.
πŸ’­ Why This Happens:
This mistake primarily stems from a lack of careful application of the integration rule for 1/f(x), specifically when f(x) is a linear function with a negative slope. Students might rush, treating (a - bx) similarly to (a + bx), or neglect the chain rule's implication that the derivative of (a - bx) is -b. This requires multiplying the integral by -1/b, which is frequently overlooked.
βœ… Correct Approach:
When integrating an expression of the form ∫1/(a - bx) dx, it is crucial to remember that the derivative of (a - bx) with respect to x is -b. Therefore, for the integral to be ln|a - bx|, we would need -b in the numerator. Since it's not present, we must compensate by multiplying by -1/b. The correct integration rule is: 1/(a - bx) dx = (-1/b)ln|a - bx| + C. Apply this meticulously to both x and v terms after variable separation.
πŸ“ Examples:
❌ Wrong:
Consider a step after variable separation in a homogeneous differential equation:
∫ dv / (1 - v) = ∫ dx / x
Wrong Integration: ln|1 - v| = ln|x| + C
βœ… Correct:
Using the correct integration rule for 1/(1 - v) (where a=1, b=1, so -1/b = -1):
Correct Integration: -ln|1 - v| = ln|x| + C
This can be further simplified using logarithm properties, e.g., ln|1/(1 - v)| = ln|x| + C or ln|1/(1 - v)| - ln|x| = C.
πŸ’‘ Prevention Tips:
  • Double-check Coefficients: Always explicitly identify the coefficient of the variable (x or v) in the denominator when integrating 1/(a ± bx) forms.
  • Mnemonic: For 1/(a - bx), think 'minus b in the denominator, so minus 1/b outside'.
  • Practice Integration by Substitution: Mentally or explicitly perform u = a - bx, then du = -b dx, so dx = du / (-b) to reinforce the sign.
  • Review Standard Integrals: Ensure a solid grasp of basic integral formulas, especially those involving linear functions in the denominator.
JEE_Advanced
Important Calculation

❌ Algebraic Errors Post-Substitution in Homogeneous Differential Equations

Students frequently make algebraic errors immediately after applying the standard substitution (e.g., y = vx and dy/dx = v + x(dv/dx)) in homogeneous differential equations. These mistakes often involve incorrect simplification of fractions, sign errors when rearranging terms, or improper combination of terms, which prevents effective variable separation.
πŸ’­ Why This Happens:
This mistake primarily stems from a lack of meticulousness and strong algebraic manipulation skills. Haste during examinations, especially with complex expressions involving multiple 'v' and 'x' terms, contributes significantly. Not writing down each simplification step clearly also leads to overlooked errors.
βœ… Correct Approach:
After substituting y = vx and dy/dx = v + x(dv/dx), systematically simplify the equation. Ensure all 'x' terms cancel out from the numerator and denominator (after factoring). Then, isolate x(dv/dx) by taking 'v' to the other side and combining terms using a common denominator. This step-by-step approach minimizes errors and prepares the equation for variable separation.
πŸ“ Examples:
❌ Wrong:
Consider dy/dx = (x^2 + y^2) / (2xy).
Substitute y = vx and dy/dx = v + x(dv/dx):
v + x(dv/dx) = (x^2 + (vx)^2) / (2x(vx))
v + x(dv/dx) = (x^2(1 + v^2)) / (2vx^2)
v + x(dv/dx) = (1 + v^2) / (2v)
WRONG STEP: x(dv/dx) = (1 + v^2) / (2v) + v
(Incorrectly added 'v' without common denominator, might lead to (1 + v^2 + v) / (2v) instead of (1 + v^2 + 2v^2) / (2v) )
βœ… Correct:
Continuing from the correct simplification:
v + x(dv/dx) = (1 + v^2) / (2v)
x(dv/dx) = (1 + v^2) / (2v) - v
x(dv/dx) = (1 + v^2 - 2v^2) / (2v)
x(dv/dx) = (1 - v^2) / (2v)
This allows for correct variable separation: 2v / (1 - v^2) dv = dx/x.
πŸ’‘ Prevention Tips:
  • Step-by-Step Simplification: Avoid skipping intermediate algebraic steps, especially when dealing with fractions.
  • Common Denominators: Always use a common denominator correctly when adding or subtracting fractional terms.
  • Sign Awareness: Double-check signs when moving terms across the equality or distributing.
  • Practice Targeted Problems: Focus on solving problems specifically up to the point of variable separation to master algebraic manipulation.
  • Verify Separability: After simplification, quickly check if the equation is indeed separable into functions of 'v' and 'x'. If not, re-examine your algebra.
JEE_Advanced
Important Formula

❌ <strong>Incorrect Homogeneity Check and Substitution Choice</strong>

Students often fail to rigorously check if a differential equation is truly homogeneous or misapply the substitution. They might universally use y = vx, even when expressing the equation as dx/dy and using x = vy would be simpler or correct, leading to algebraic complexity and errors in JEE Advanced.
πŸ’­ Why This Happens:
  • Definition Misconception: Not thoroughly understanding that dy/dx = f(x,y) is homogeneous if f(x,y) can be written as F(y/x) (or, conversely, dx/dy = G(x/y)).
  • Rote Learning: Applying y=vx blindly without considering the alternative x=vy or the structural form of the equation.
  • Algebraic Avoidance: Reluctance to rearrange the differential equation into its potentially simpler dx/dy form when appropriate.
βœ… Correct Approach:
  1. Verify Form: Confirm the differential equation is homogeneous, expressible as either dy/dx = F(y/x) or dx/dy = G(x/y).
  2. Choose Substitution:
    • For dy/dx = F(y/x), use y = vx. Then, differentiate to get dy/dx = v + x(dv/dx).
    • For dx/dy = G(x/y), use x = vy. Then, differentiate to get dx/dy = v + y(dv/dy).
  3. The resulting equation will be variable-separable, which can then be integrated to find the solution.
πŸ“ Examples:
❌ Wrong:

Given dx/dy = (x^2 + y^2) / (xy).
Mistake: Applying y = vx. This forces complex differentiation for dx/dy, which is an algebraic pitfall and leads to a more cumbersome solution process than necessary.

βœ… Correct:

For dx/dy = (x^2 + y^2) / (xy)dx/dy = x/y + y/x.
This is clearly a function of x/y.
Correct: Substitute x = vy. Thus dx/dy = v + y(dv/dy).
Substituting into the DE: v + y(dv/dy) = v + 1/vy(dv/dy) = 1/v∫ v dv = ∫ dy/y.
Integrating yields v^2/2 = ln|y| + C. Back-substitute v = x/y to obtain the final solution: x^2/(2y^2) = ln|y| + C.

πŸ’‘ Prevention Tips:
  • Master Definition: Understand the precise conditions for a differential equation to be homogeneous.
  • Analyze Form First: Always convert the DE to either dy/dx = F(y/x) or dx/dy = G(x/y) before deciding on the substitution.
  • Choose Wisely: Use y=vx for F(y/x) and x=vy for G(x/y). This choice dramatically simplifies the subsequent steps.
  • Practice Both: Solve various problems where both types of substitutions are applicable to develop intuition and speed.
JEE_Advanced
Important Unit Conversion

❌ <span style='color: #FF0000;'>Misinterpreting Homogeneity: A 'Dimensional' Error</span>

Students frequently apply the substitution `y = vx` (or `x = vy`) without first confirming if the given differential equation is truly homogeneous. While 'unit conversion' in the physical sense isn't applicable here, this error is analogous to a 'dimensional inconsistency' in mathematics. For a differential equation to be homogeneous, all terms in the expressions `M(x,y)` and `N(x,y)` (if in the form `M dx + N dy = 0`) or in `f(x,y)` (if `dy/dx = f(x,y)`) must have the same 'degree' (sum of powers of `x` and `y`). Failing to check this 'degree consistency' is a fundamental mistake, as it leads to an incorrect classification and an unsolvable form using the homogeneous method.
πŸ’­ Why This Happens:
This mistake primarily stems from a lack of thorough understanding of the definition of a homogeneous function and the conditions required for a differential equation to be homogeneous. Students often rush to apply the standard substitution `y=vx` upon seeing `x` and `y` terms, without a rigorous 'dimensional' check of the degree of each term. In JEE Advanced, problems are often designed to test this nuanced understanding.
βœ… Correct Approach:
Always begin by rigorously verifying the homogeneity of the differential equation. For `dy/dx = f(x,y)`, check if `f(Ξ»x, Ξ»y) = f(x,y)`. For `M(x,y)dx + N(x,y)dy = 0`, check if both `M(x,y)` and `N(x,y)` are homogeneous functions of the same degree. Only after confirming homogeneity should the substitution `y=vx` (or `x=vy`) be applied. This ensures the equation transforms into a separable form in `v` and `x` (or `v` and `y`).
πŸ“ Examples:
❌ Wrong:
Consider the equation `dy/dx = (x + y + 1) / (x - y)`. A common mistake is to attempt `y=vx` directly. However, the term `+1` in the numerator has a degree of 0, while `x` and `y` have a degree of 1. Thus, the numerator `(x + y + 1)` is not a homogeneous function, making the entire differential equation non-homogeneous. Applying `y=vx` here would lead to an intractable equation.
βœ… Correct:
For the equation `dy/dx = (x^2 + y^2) / (2xy)`:
To check homogeneity, replace `x` with `Ξ»x` and `y` with `Ξ»y`:
f(Ξ»x, Ξ»y) = ((Ξ»x)^2 + (Ξ»y)^2) / (2(Ξ»x)(Ξ»y))
= (Ξ»^2x^2 + Ξ»^2y^2) / (2Ξ»^2xy)
= Ξ»^2(x^2 + y^2) / (Ξ»^2(2xy))
= (x^2 + y^2) / (2xy) = f(x,y)
Since `f(Ξ»x, Ξ»y) = f(x,y)`, the function is homogeneous of degree zero. Only then should the substitution `y=vx` be applied:
v + x(dv/dx) = (x^2 + (vx)^2) / (2x(vx))
v + x(dv/dx) = (x^2 + v^2x^2) / (2vx^2)
v + x(dv/dx) = (1 + v^2) / (2v), which is separable.
πŸ’‘ Prevention Tips:
  • Thorough Definition Recall: Ensure a crystal-clear understanding of what constitutes a homogeneous function and a homogeneous differential equation.
  • Systematic Check: Before any substitution, always perform the homogeneity test (either `f(Ξ»x, Ξ»y)` or checking degrees of `M` and `N`). Treat this as a mandatory first step.
  • Practice Classification: Solve problems that require classifying differential equations (homogeneous, linear, exact, separable) before solving them. This builds an instinct for identifying types.
JEE_Advanced
Important Sign Error

❌ Sign Error in Variable Separation and Integration

Students frequently make sign errors when rearranging terms to separate variables or during the integration step while solving homogeneous differential equations. This often occurs after substituting `y = vx` (or `x = vy`) and simplifying, leading to incorrect signs for terms involving `v` or `x`, especially before integration.
πŸ’­ Why This Happens:
This mistake typically arises from:
  • Careless algebraic manipulation, particularly when transposing terms across the equality sign.
  • Incorrectly handling the negative sign when integrating forms like ∫ (1/(a-x)) dx, which should yield -ln|a-x|.
  • Rushing through calculations, overlooking subtle sign changes.
βœ… Correct Approach:
Always perform algebraic steps meticulously. When separating variables, double-check the sign of each term before moving it. During integration, pay close attention to the integrand's structure, especially denominators involving subtraction (e.g., (a-x) versus (x-a)), as they can introduce a negative sign in the logarithmic term. For JEE Advanced, precision in every step is crucial.
πŸ“ Examples:
❌ Wrong:
Consider the homogeneous differential equation: dy/dx = (y-x)/x
Substitute y = vx, so dy/dx = v + x(dv/dx).
v + x(dv/dx) = (vx - x)/x = v - 1
x(dv/dx) = -1
Separate variables: dv = -dx/x

Wrong Integration: Student mistakenly writes ∫ dv = ∫ dx/x (missing the negative sign).
v = ln|x| + C
y/x = ln|x| + C
y = x(ln|x| + C)
βœ… Correct:
Using the same steps until variable separation: dv = -dx/x

Correct Integration:
∫ dv = ∫ (-1/x) dx
v = -ln|x| + C
Substitute back v = y/x:
y/x = -ln|x| + C
y = x(C - ln|x|)
πŸ’‘ Prevention Tips:
  • Double Check Signs: After each algebraic step, pause and verify the signs of all terms.
  • Parentheses Use: Use parentheses extensively, especially when dealing with negative signs or complex expressions, to avoid distribution errors.
  • Recall Integration Formulas: Be precise with integration formulas. Remember ∫ (1/(a-bx)) dx = (-1/b)ln|a-bx| + C.
  • Step-by-Step Verification: For JEE Advanced, it's wise to quickly re-derive the signs mentally, or on scratch paper, if any doubt arises.
JEE_Advanced
Important Approximation

❌ Incorrect Simplification by Premature Approximation in Homogeneous DEs

Students frequently make the mistake of prematurely simplifying complex expressions involving v (where v = y/x) by assuming certain terms are negligible or can be approximated, even when no specific limiting condition (e.g., x → 0, x → ∞) is given. This leads to an incorrect separable form and, consequently, an erroneous solution to the differential equation.
πŸ’­ Why This Happens:
  • Lack of Rigorous Algebraic Skills: Students might struggle with precise algebraic manipulation required for complex fractions or terms after the y=vx substitution.
  • Misinterpretation of Problem Context: They might apply approximation techniques (like (1+x)n ≈ 1+nx for small x) without a valid basis or an explicit instruction in the problem statement.
  • Forcing Standard Forms: An attempt to manipulate the integrand into a 'simpler' or known integral form by unjustifiably dropping terms.
  • Confusing Exact vs. Approximate Solutions: Not distinguishing between the need for an exact solution of the DE and situations where approximations are explicitly required for evaluating specific values or behaviors of the solution.
βœ… Correct Approach:
The core of solving homogeneous differential equations involves an exact substitution (y=vx) and precise algebraic manipulation to separate variables without any approximation. Every term must be carried forward accurately until the equation is in the form g(v) dv = h(x) dx. Approximation techniques should only be employed if the problem explicitly asks for an approximate solution under specific conditions (e.g., for small x, or as x → ∞), and typically after the exact general solution has been found.
πŸ“ Examples:
❌ Wrong:
Consider dy/dx = (x2 + y2) / (x2 - xy). After substituting y=vx, we get v + x dv/dx = (1+v2) / (1-v).
A common mistake: A student might incorrectly simplify (1+v2) / (1-v) to 1+v (thinking v2 is negligible or 1-v ≈ 1) to get v + x dv/dx ≈ 1+v. This is an unjustified approximation that will lead to an incorrect solution.
βœ… Correct:
For dy/dx = (x2 + y2) / (x2 - xy):
1. Substitute y=vx and dy/dx = v + x dv/dx.
2. v + x dv/dx = (x2 + (vx)2) / (x2 - x(vx)) = (x2(1+v2)) / (x2(1-v)) = (1+v2) / (1-v).
3. Exactly rearrange terms for separation:
x dv/dx = (1+v2) / (1-v) - v = (1+v2 - v(1-v)) / (1-v) = (1+v2 - v + v2) / (1-v) = (1 - v + 2v2) / (1-v).
4. The exact separable form is ∫(1-v)/(1 - v + 2v2) dv = ∫dx/x. No approximation is made here during the transformation or separation.
πŸ’‘ Prevention Tips:
  • Master Algebraic Simplification: Practice complex algebraic manipulations to ensure accuracy post-substitution.
  • Understand Problem Requirements: Differentiate between questions requiring exact solutions and those asking for approximate behaviors under specific conditions.
  • Verify Steps: Always double-check each step of simplification before moving to integration.
  • Avoid Hasty Assumptions: Do not drop terms or simplify expressions based on intuitive 'smallness' unless explicitly justified by the problem statement or a valid mathematical limit.
JEE_Advanced
Important Conceptual

❌ Incorrect Substitution and Algebraic Errors in Homogeneous DEs

Students often make crucial mistakes during the substitution phase for homogeneous differential equations, either by failing to identify the correct substitution (`y=vx` vs. `x=vy`) or by committing algebraic errors immediately after substitution, leading to an incorrect separable form. Another common error is failing to correctly differentiate `y=vx` (or `x=vy`).
πŸ’­ Why This Happens:
This generally stems from a lack of clear understanding of what defines a homogeneous differential equation and the purpose of the substitution. Students might rush into a substitution without verifying if the equation is indeed homogeneous or without considering which substitution simplifies the problem more effectively. Algebraic carelessness in simplifying the expression `v + x(dv/dx)` or `v + y(dv/dy)` and subsequent variable separation is also a major culprit.
βœ… Correct Approach:
  • Step 1: Verify Homogeneity: First, express the equation in the form dy/dx = f(x, y). Then, check if f(Ξ»x, Ξ»y) = f(x, y). If so, it's homogeneous. You can also check if it can be written as dy/dx = g(y/x) or dx/dy = h(x/y).
  • Step 2: Choose Correct Substitution: If dy/dx = g(y/x), substitute y = vx. This implies dy/dx = v + x(dv/dx). If dx/dy = h(x/y), substitute x = vy. This implies dx/dy = v + y(dv/dy). Tip for JEE Advanced: Sometimes one substitution simplifies much faster than the other. Look for terms like (x+y)/(x-y) where y=vx is better.
  • Step 3: Separate Variables: After substitution, the equation must transform into a variable separable form involving v and x (or v and y).
  • Step 4: Integrate and Resubstitute: Integrate both sides and finally replace v with y/x (or x/y) to get the solution in terms of x and y.
πŸ“ Examples:
❌ Wrong:

Consider the DE: (x - y) dy/dx = x + y.

Incorrect: A student might try to write dy/dx = (x+y)/(x-y) and then directly integrate or incorrectly apply substitution like: y=vx, so dy/dx = (x+vx)/(x-vx) = (1+v)/(1-v). But then instead of setting v + x(dv/dx) = (1+v)/(1-v), they might directly integrate v or just (1+v)/(1-v) without handling the x(dv/dx) term, or make algebraic errors in separating variables after correctly setting up the equation.

For example, a common mistake is: x(dv/dx) = (1+v)/(1-v) - v = (1+v-v+v^2)/(1-v) = (1+v^2)/(1-v). Then, incorrectly writing (1-v)/(1+v^2) dv = dx instead of (1-v)/(1+v^2) dv = dx/x.

βœ… Correct:

For the DE: (x - y) dy/dx = x + y

Correct:

  1. Rewrite as: dy/dx = (x + y) / (x - y). This is homogeneous as f(Ξ»x, Ξ»y) = (Ξ»x + Ξ»y) / (Ξ»x - Ξ»y) = Ξ»(x+y)/Ξ»(x-y) = f(x,y).
  2. Substitute y = vx, so dy/dx = v + x(dv/dx).
  3. Equating the expressions for dy/dx:
    v + x(dv/dx) = (x + vx) / (x - vx) = (1 + v) / (1 - v)
  4. Isolate x(dv/dx):
    x(dv/dx) = (1 + v) / (1 - v) - v = (1 + v - v(1 - v)) / (1 - v) = (1 + v - v + v^2) / (1 - v) = (1 + v^2) / (1 - v)
  5. Separate variables:
    (1 - v) / (1 + v^2) dv = dx / x
  6. Integrate both sides:
    ∫ (1 - v) / (1 + v^2) dv = ∫ dx / x
    ∫ 1/(1+v^2) dv - ∫ v/(1+v^2) dv = ∫ dx/x
    tan⁻¹(v) - (1/2)ln|1+v^2| = ln|x| + C
  7. Resubstitute v = y/x:
    tan⁻¹(y/x) - (1/2)ln|1+(y/x)²| = ln|x| + C
πŸ’‘ Prevention Tips:
  • Always check for homogeneity first, before applying any substitution.
  • Memorize the derivative of y = vx: dy/dx = v + x(dv/dx) (and similarly for x = vy). This is a frequent point of error.
  • Practice algebraic manipulation involved in separating variables after substitution. This is where most students stumble in JEE Advanced problems.
  • For CBSE, simply understanding the process is often sufficient, but for JEE, accuracy in calculation is paramount.
  • When solving, always ensure the final answer is in terms of the original variables x and y, replacing v correctly.
JEE_Advanced
Important Other

❌ <span style='color: #FF0000;'>Incorrect Algebraic Manipulation After Homogeneous Substitution</span>

Students frequently identify a differential equation as homogeneous and apply the substitution y = vx (or x = vy) correctly, but then make errors in the subsequent algebraic steps. These errors often involve incorrectly isolating terms, simplifying expressions, or separating variables, which prevents them from reaching a form suitable for integration.
πŸ’­ Why This Happens:
  • Lack of Algebraic Proficiency: Careless handling of fractions, signs, and common denominators after substituting y = vx and dy/dx = v + x(dv/dx).
  • Misunderstanding Variable Separation: Not clearly understanding the goal of transforming the equation into f(v)dv = g(x)dx, leading to haphazard rearrangement.
  • Rushing Steps: Attempting to combine multiple algebraic steps mentally, increasing the chance of errors.
βœ… Correct Approach:

Once the substitution y = vx (and thus dy/dx = v + x(dv/dx)) is made:

  1. Substitute Systematically: Replace all y with vx and dy/dx with v + x(dv/dx) in the original equation.
  2. Simplify Carefully: Cancel common terms (usually powers of x) to obtain an equation solely in terms of v and x.
  3. Isolate x(dv/dx): Move the v term to the right-hand side, typically by subtraction: x(dv/dx) = F(v) - v.
  4. Combine Terms: Perform algebraic operations (e.g., finding a common denominator) to simplify F(v) - v into a single function of v.
  5. Separate Variables: Rearrange the equation to have all v terms (and dv) on one side and all x terms (and dx) on the other side: [1 / (F(v) - v)] dv = (1/x) dx.
  6. Integrate and Substitute Back: Integrate both sides and finally replace v with y/x to get the solution in terms of x and y.
πŸ“ Examples:
❌ Wrong:

Consider dy/dx = (y - x) / (y + x).

Substitute y = vx, dy/dx = v + x(dv/dx):

v + x(dv/dx) = (vx - x) / (vx + x)

v + x(dv/dx) = x(v - 1) / x(v + 1)

v + x(dv/dx) = (v - 1) / (v + 1)

Common Mistake:

Student incorrectly writes x(dv/dx) = (v - 1) / (v + 1) + v (instead of subtraction)

x(dv/dx) = (v - 1 + v(v + 1)) / (v + 1) = (v - 1 + v^2 + v) / (v + 1) = (v^2 + 2v - 1) / (v + 1)

This leads to an incorrect separable form.

βœ… Correct:

Continuing from v + x(dv/dx) = (v - 1) / (v + 1):

x(dv/dx) = (v - 1) / (v + 1) - v

x(dv/dx) = (v - 1 - v(v + 1)) / (v + 1)

x(dv/dx) = (v - 1 - v^2 - v) / (v + 1)

x(dv/dx) = (-1 - v^2) / (v + 1)

x(dv/dx) = -(1 + v^2) / (v + 1)

Now, separate variables:

(v + 1) / (1 + v^2) dv = - (1/x) dx

∫ [(v / (1 + v^2)) + (1 / (1 + v^2))] dv = ∫ -(1/x) dx

(1/2)ln|1 + v^2| + arctan(v) = -ln|x| + C

Substitute v = y/x:

(1/2)ln|1 + (y/x)^2| + arctan(y/x) = -ln|x| + C

(1/2)ln|(x^2 + y^2)/x^2| + arctan(y/x) = -ln|x| + C

(1/2)[ln|x^2 + y^2| - ln|x^2|] + arctan(y/x) = -ln|x| + C

(1/2)ln|x^2 + y^2| - ln|x| + arctan(y/x) = -ln|x| + C

(1/2)ln(x^2 + y^2) + arctan(y/x) = C

πŸ’‘ Prevention Tips:
  • CBSE vs. JEE Advanced: For JEE Advanced, the algebraic complexity is often higher. Practice a variety of problems to build strong manipulation skills. CBSE focuses more on the basic application of the method.
  • Step-by-Step Approach: Break down the algebraic simplification and separation into small, manageable steps. Avoid doing too much in your head.
  • Double-Check Signs: Pay close attention to positive and negative signs, especially when moving terms across the equality or combining fractions.
  • Verify Separability: After each major algebraic step, mentally check if the equation is progressing towards the f(v)dv = g(x)dx form. If not, re-evaluate.
  • Practice, Practice, Practice: The only way to master the algebraic transformations is through consistent practice with various homogeneous differential equations.
JEE_Advanced
Important Sign Error

❌ Sign Errors in Variable Separation for Homogeneous Differential Equations

A common and critical error in solving homogeneous differential equations is the mishandling of negative signs during the variable separation step, particularly after substituting `y = vx` and simplifying the equation to the form `x dv/dx = f(v)`. Students often drop a negative sign or incorrectly distribute it, leading to an entirely wrong integral and final solution.
πŸ’­ Why This Happens:
This mistake primarily stems from:
  • Carelessness: Rushing through algebraic manipulations.
  • Weak Algebraic Skills: Inability to correctly factor out negative signs or distribute them across terms.
  • Misinterpreting Terms: Incorrectly identifying the sign of a term when moving it from one side of the equation to another, or when dividing/multiplying.
  • Lack of Double-Checking: Not verifying each step, especially the separation of variables, before proceeding to integration.
βœ… Correct Approach:
After substituting `y=vx` and simplifying `dy/dx` to `v + x dv/dx = F(v)`, meticulously rearrange the terms to isolate `x dv/dx = F(v) - v`. Then, carefully perform algebraic operations to simplify `F(v) - v` into a single rational function of `v`. When separating variables, ensure that the sign of the entire `f(v)` term is carried correctly to the `dv` side. Use parentheses extensively to group terms and avoid sign ambiguities.
πŸ“ Examples:
❌ Wrong:
Consider the equation: dy/dx = (y-x)/(y+x)
After substitution `y=vx` and `dy/dx = v + x dv/dx`:
v + x dv/dx = (vx-x)/(vx+x) = (v-1)/(v+1)
x dv/dx = (v-1)/(v+1) - v
x dv/dx = (v-1 - v(v+1))/(v+1)
x dv/dx = (v-1 - v^2 - v)/(v+1)
x dv/dx = (-1 - v^2)/(v+1)
x dv/dx = -(1 + v^2)/(v+1)

Wrong Separation: (v+1)/(1+v^2) dv = dx/x (Negative sign is dropped)
This will lead to an incorrect integration and solution.
βœ… Correct:
Continuing from the wrong example, after simplifying to:
x dv/dx = -(1 + v^2)/(v+1)

Correct Separation:
(v+1)/(1+v^2) dv = -dx/x (The negative sign is correctly carried to the `dx/x` side or kept on the `dv` side).
Now, integrate both sides:
∫ ((v)/(1+v^2) + (1)/(1+v^2)) dv = ∫ -1/x dx
(1/2)ln|1+v^2| + tan&Inverse;v = -ln|x| + C
Substitute back `v=y/x` to get the general solution.
πŸ’‘ Prevention Tips:
  • Step-by-Step Algebra: Perform each algebraic step carefully, especially when combining terms or moving them across the equals sign.
  • Use Parentheses: Employ parentheses to clearly define terms and their signs, especially when distributing or factoring.
  • Isolate the Negative: If a negative sign applies to an entire expression, factor it out or ensure it's carried with the correct term during separation.
  • Verification: Before integrating, mentally (or physically) re-check the separated form of the variables to ensure all signs are correct.
  • Practice: Solve a variety of problems focusing on algebraic manipulation to build proficiency and reduce error rates.
JEE_Main
Important Other

❌ Incorrect Algebraic Manipulation During Variable Separation

Students often correctly identify a homogeneous differential equation and apply the initial substitution (e.g., y = vx). However, a common pitfall occurs immediately after, during the algebraic manipulation required to separate the variables v and x. This includes errors in combining fractions, simplifying terms, or isolating dv and dx, leading to an incorrect or unintegrable form.
πŸ’­ Why This Happens:
This mistake primarily stems from a combination of:
  • Weak algebraic foundations: Difficulty in handling rational expressions, finding common denominators, or factoring.
  • Carelessness: Rushing calculations, especially with signs or combining terms.
  • Lack of systematic approach: Not following a clear step-by-step process for variable separation, leading to jumbled terms.
  • Insufficient practice: Not enough exposure to the various algebraic forms that arise after substitution in homogeneous DEs.
βœ… Correct Approach:
The key is systematic and careful algebraic manipulation after substituting y = vx and dy/dx = v + x(dv/dx).
  1. Substitute: Replace y with vx and dy/dx with v + x(dv/dx).
  2. Isolate x(dv/dx): Rearrange the equation to get x(dv/dx) on one side, typically by subtracting v from the right-hand side.
  3. Combine Terms: If the right-hand side is a rational function of v, combine the terms into a single fraction using a common denominator.
  4. Separate Variables: Move all terms involving v and dv to one side and all terms involving x and dx to the other. Ensure dv is in the numerator on one side and dx in the numerator on the other.
  5. Integrate: Integrate both sides.
  6. Substitute Back: Replace v with y/x (or x/y) for the final solution.
πŸ“ Examples:
❌ Wrong:
Consider the equation dy/dx = (x^2 + y^2) / (2xy).
After substituting y = vx and dy/dx = v + x(dv/dx), we get:
v + x(dv/dx) = (1 + v^2) / (2v)
A common error here is in subtracting v or combining fractions:
❌ x(dv/dx) = (1 + v^2) / (2v) + v (incorrect sign)
❌ x(dv/dx) = (1 + v^2 - v) / (2v) (incorrect common denominator and multiplication)
These errors lead to an incorrect expression for dv/dx, making subsequent integration impossible or incorrect.
βœ… Correct:
Continuing from the previous example: v + x(dv/dx) = (1 + v^2) / (2v)
1. Isolate x(dv/dx):
    x(dv/dx) = (1 + v^2) / (2v) - v
2. Combine Terms:
    x(dv/dx) = (1 + v^2 - 2v*v) / (2v)
    x(dv/dx) = (1 + v^2 - 2v^2) / (2v)
    x(dv/dx) = (1 - v^2) / (2v)
3. Separate Variables:
    (2v / (1 - v^2)) dv = (1/x) dx
From this point, the integration proceeds correctly, leading to the solution x^2 - y^2 = Ax.
πŸ’‘ Prevention Tips:
  • Algebraic Review: Regularly practice simplifying rational expressions and manipulating algebraic equations, especially finding common denominators.
  • Step-by-Step Calculation: Avoid combining too many algebraic steps. Write out each intermediate step clearly.
  • Check Signs: Be extremely careful with positive and negative signs during subtraction and multiplication.
  • Verify Denominators: When combining fractions, always ensure you're using the correct common denominator and multiplying numerators appropriately.
  • Practice Integration Readiness: After separation, briefly check if the resulting integrals are in a recognizable form. If not, re-examine the separation step.
JEE_Main
Important Approximation

❌ Incorrect Algebraic Simplification and Integration Post-Substitution

Students frequently make errors in algebraically simplifying the differential equation after applying the substitution y = vx (or x = vy). This leads to incorrect separation of variables, or a wrong integrand, resulting in an 'approximated' (incorrect) solution rather than the precise analytical one. Common mistakes include algebraic blunders, incorrect factorization, and errors in the integration of the separated terms.
πŸ’­ Why This Happens:
  • Algebraic Oversight: Carelessness in simplifying expressions involving v after substituting y=vx and dy=v dx + x dv.
  • Faulty Variable Separation: Incorrectly grouping dx terms with x and dv terms with v, or failing to isolate the variables completely.
  • Integration Errors: Incorrect application of standard integration formulas, especially for functions involving 1/x (forgetting ln|x| or absolute values), or mishandling complex integrands through substitution or partial fractions.
  • Missing Constant of Integration: Forgetting to add the arbitrary constant + C, which results in a particular solution instead of the general solution.
βœ… Correct Approach:
The correct approach demands meticulous algebraic manipulation and precise integration:
  1. Systematic Substitution: Clearly substitute y=vx and dy=v dx + x dv into the given homogeneous differential equation.
  2. Careful Simplification: Factor out x^n terms and simplify the resulting expression to clearly isolate terms containing dx and dv. This is where JEE Main questions often test algebraic rigor.
  3. Exact Variable Separation: Rearrange the equation such that all v terms (and dv) are on one side, and all x terms (and dx) are on the other. Ensure no x term remains with dv and no v term with dx.
  4. Accurate Integration: Integrate both sides correctly. Remember to use ln|x| for ∫ (1/x) dx and correctly apply substitution method for integrands involving v. For CBSE, basic integration is sufficient; for JEE, more complex integration techniques may be required. Always add the constant of integration C.
πŸ“ Examples:
❌ Wrong:
Consider (x^2 + y^2) dx - 2xy dy = 0.
After substitution y=vx and simplification, we get (1 - v^2) dx = 2vx dv.
Wrong Separation/Integration Attempt:
dx = 2vx dv / (1 - v^2) (Variables not separated)
Or, integrating ∫ (2v / (1 - v^2)) dv as ln(1 - v^2) (missing negative sign and absolute value).
βœ… Correct:
Continuing from (1 - v^2) dx = 2vx dv:
Correct Separation:
dx/x = 2v dv / (1 - v^2)
Correct Integration:
∫ dx/x = ∫ (2v / (1 - v^2)) dv
Let u = 1 - v^2, then du = -2v dv, so 2v dv = -du.
∫ dx/x = ∫ (-du/u)
ln|x| = -ln|1 - v^2| + ln|C| (using ln|C| for the constant is often beneficial in JEE)
ln|x| = ln|C / (1 - v^2)|
x = C / (1 - v^2)
Substituting back v = y/x:
x = C / (1 - (y/x)^2) = C / ((x^2 - y^2)/x^2) = C x^2 / (x^2 - y^2)
1 = Cx / (x^2 - y^2) or x^2 - y^2 = Cx.
πŸ’‘ Prevention Tips:
  • Step-by-Step Verification: After each algebraic step, pause and verify the simplification before proceeding.
  • Check Variable Isolation: Ensure that all terms involving one variable are perfectly isolated before integration.
  • Master Integration Formulas: Have all basic integration formulas, especially those for 1/x, 1/(a^2-x^2), etc., at your fingertips. Pay attention to constants and absolute values.
  • Always Add 'C': Never forget the constant of integration. For JEE, sometimes expressing it as ln|C| or tan⁻¹(C) can simplify the final solution form.
  • Practice Diverse Problems: Work through a variety of problems to become proficient in handling different types of algebraic expressions and integrands.
JEE_Main
Important Unit Conversion

❌ Incorrect Differentiation During Substitution (y = vx or x = vy)

A common pitfall in solving homogeneous differential equations involves making errors during the substitution step. Students often correctly identify the need to substitute `y = vx` (or `x = vy`), but then incorrectly differentiate `y = vx` with respect to `x`. The mistake stems from treating `v` as a constant, or misapplying the product rule, leading to an incorrect expression for `dy/dx` (or `dx/dy`). This fundamental error prevents the successful separation of variables.
πŸ’­ Why This Happens:
This mistake primarily occurs due to a misunderstanding that `v` is not a constant, but rather a function of `x` (specifically, `v = y/x`). Students might be accustomed to differentiating expressions where variables are independent, leading them to incorrectly assume `dv/dx = 0`. Additionally, a rushed application or a weak grasp of the product rule of differentiation contributes to this error.
βœ… Correct Approach:
When performing the substitution `y = vx` in a homogeneous differential equation, always remember that `v` is a function of `x`. Therefore, differentiate `y = vx` with respect to `x` using the product rule. The correct differentiation is:
`dy/dx = d/dx (vx)`
`dy/dx = v * (d/dx x) + x * (d/dx v)`
`dy/dx = v * 1 + x * (dv/dx)`
`dy/dx = v + x(dv/dx)`
A similar approach applies if using `x = vy`, where `dx/dy = v + y(dv/dy)`.
πŸ“ Examples:
❌ Wrong:
Consider the equation `dy/dx = (x+y)/x`.
  • Incorrect Substitution: A student might substitute `y = vx` and then incorrectly state `dy/dx = v` (assuming `v` is a constant).
  • Resulting Equation: `v = (x + vx)/x => v = 1 + v => 0 = 1`. This contradiction indicates a severe error.
βœ… Correct:
Consider the equation `dy/dx = (x+y)/x`.
  • Correct Substitution: Let `y = vx`. Differentiate correctly: `dy/dx = v + x(dv/dx)`.
  • Substituting into the DE:
    `v + x(dv/dx) = (x + vx)/x`
    `v + x(dv/dx) = x(1 + v)/x`
    `v + x(dv/dx) = 1 + v`
    `x(dv/dx) = 1`
    This is now a separable differential equation (`dv = dx/x`), which can be easily integrated.
πŸ’‘ Prevention Tips:
  • Master the Product Rule: Ensure you are proficient in applying the product rule for differentiation.
  • Always Treat 'v' as a Function of 'x' (or 'y'): Explicitly remind yourself that `v = y/x` (or `v = x/y`).
  • Memorize the Standard Derivative: For JEE Main, it's highly beneficial to have `dy/dx = v + x(dv/dx)` (and `dx/dy = v + y(dv/dy)`) firmly memorized for this specific type of problem.
  • Verify Intermediate Steps: If your equation simplifies to an illogical result (e.g., `0 = 1`), immediately recheck your differentiation and substitution steps.
JEE_Main
Important Other

❌ Incorrect Application of Substitution `y = vx` and Subsequent Variable Separation

A frequent error students make is the incorrect application of the standard substitution `y = vx` (or `x = vy`) for homogeneous differential equations, or making algebraic mistakes during the subsequent separation of variables. This often leads to an equation that is not variable separable, or an incorrect integral.
πŸ’­ Why This Happens:
This mistake primarily stems from:
  • Algebraic Errors: Miscalculation when differentiating `y = vx` to `dy/dx = v + x dv/dx`.
  • Incorrect Substitution: Substituting `y/x` as `v` but failing to replace `dy/dx` correctly, or forgetting to substitute `y=vx` in all `y` terms.
  • Poor Manipulation: Difficulty in isolating `dv` and `dx` terms after substitution, leading to `v` and `x` remaining on the same side of an integral.
  • Lack of Practice: Insufficient practice with algebraic simplification after substitution can lead to confusion.
βœ… Correct Approach:
For a homogeneous differential equation `dy/dx = f(y/x)` (or `dx/dy = g(x/y)`):
  1. Confirm it's homogeneous.
  2. Substitute `y = vx` and its derivative `dy/dx = v + x dv/dx` into the equation.
  3. Carefully rearrange the terms to separate `v` and `x` variables: bring all `v` terms with `dv` and all `x` terms with `dx`.
  4. Integrate both sides.
  5. Replace `v` with `y/x` in the final solution to express it in terms of `y` and `x`.
πŸ“ Examples:
❌ Wrong:
Consider `(x^2 + y^2)dx = 2xy dy`. Students might write `dy/dx = (x^2 + y^2) / (2xy)`. After `y = vx` and `dy/dx = v + x dv/dx`, they incorrectly substitute `v` for `y` in the `dy/dx` expression: `y/x + x dv/dx` instead of `v + x dv/dx`. Or they might perform incorrect algebraic steps like:
`v + x dv/dx = (x^2 + (vx)^2) / (2x(vx))`
`v + x dv/dx = (x^2 + v^2x^2) / (2vx^2)`
`v + x dv/dx = (1 + v^2) / (2v)`
`x dv/dx = (1 + v^2) / (2v) + v` (Wrong sign here, should be -v)
This leads to a completely different integral.
βœ… Correct:
For `(x^2 + y^2)dx = 2xy dy`, we have `dy/dx = (x^2 + y^2) / (2xy)`.
Substitute `y = vx` and `dy/dx = v + x dv/dx`.
`v + x dv/dx = (x^2 + (vx)^2) / (2x(vx))`
`v + x dv/dx = (x^2 + v^2x^2) / (2vx^2)`
`v + x dv/dx = (1 + v^2) / (2v)`
`x dv/dx = (1 + v^2) / (2v) - v` (Correct subtraction)
`x dv/dx = (1 + v^2 - 2v^2) / (2v)`
`x dv/dx = (1 - v^2) / (2v)`
Now, separate variables:
`(2v / (1 - v^2)) dv = (1 / x) dx`
Integrate both sides: `∫(2v / (1 - v^2)) dv = ∫(1 / x) dx`
`-ln|1 - v^2| = ln|x| + C`
`-ln|1 - (y/x)^2| = ln|x| + C`
This is the general solution.
πŸ’‘ Prevention Tips:
  • Double-Check Homogeneity: Always ensure the equation is homogeneous before applying the substitution method.
  • Memorize Derivatives: Clearly remember `dy/dx = v + x dv/dx` for `y = vx`.
  • Step-by-Step Algebra: Perform algebraic manipulations meticulously. Avoid skipping steps, especially when combining terms or taking common denominators.
  • Isolate Terms: After substitution, focus on isolating all `v` terms with `dv` and all `x` terms with `dx` on separate sides of the equation.
  • Practice: Solve numerous problems to build confidence and speed in algebraic simplification. This is crucial for both CBSE and JEE exams.
CBSE_12th
Important Approximation

❌ Approximating Algebraic Simplifications and Integrations

After applying the substitution (y=vx or x=vy) to transform a homogeneous differential equation into a separable form, students frequently make errors in the subsequent algebraic simplification (e.g., combining terms, factoring, partial fractions) or during the integration step. These errors effectively lead to an "approximation" of the correct algebraic form or the integral, resulting in an incorrect general solution. This reflects a misunderstanding that exact analytical solutions require precise manipulation, not just "close enough" steps.
πŸ’­ Why This Happens:
  • Haste and Carelessness: Rushing through algebraic steps or integration leads to oversights.
  • Weak Foundation in Algebra/Calculus: Inadequate skills in manipulating rational functions, performing partial fractions, or applying standard integration techniques are common.
  • Lack of Precision: Students sometimes do not realize that every step in finding an exact analytical solution must be mathematically accurate, not an estimation.
βœ… Correct Approach:
Always perform algebraic simplifications meticulously and apply integration techniques precisely. Double-check each step, especially after separating variables and before integrating. Remember that even a minor algebraic or integration error can lead to a completely different (and incorrect) final solution.
πŸ“ Examples:
❌ Wrong:

Consider the equation: dy/dx = (x+y)/x

Substitute y=vx, so dy/dx = v + x(dv/dx).

v + x(dv/dx) = (x+vx)/x = 1+v

x(dv/dx) = 1

Separate variables: dv = (1/x) dx

Wrong Integration: A common error is to integrate ∫(1/x) dx incorrectly as x instead of ln|x|.

v = x + C

Substitute back v=y/x:

y/x = x + C

y = x2 + Cx (This is an incorrect general solution due to the integration error).

βœ… Correct:

Using the same equation: dy/dx = (x+y)/x

Substitute y=vx, so dy/dx = v + x(dv/dx).

v + x(dv/dx) = (x+vx)/x = 1+v

x(dv/dx) = 1

Separate variables: dv = (1/x) dx

Correct Integration: Apply standard integration formulas precisely.

∫dv = ∫(1/x) dx

v = ln|x| + C

Substitute back v=y/x:

y/x = ln|x| + C

y = x(ln|x| + C) (This is the correct general solution).

πŸ’‘ Prevention Tips:
  • Master Basic Integration: Ensure strong proficiency in integrating common functions (e.g., power rule, logarithmic, trigonometric, partial fractions). This is crucial for both CBSE and JEE.
  • Practice Algebraic Manipulation: Develop strong skills in simplifying complex rational expressions and factoring, as these are critical after substitution.
  • Step-by-Step Verification: After each significant algebraic simplification or integration, pause and review your work for potential errors.
  • Refer to Formulas: If unsure about an integral, always refer to standard integration formulas rather than guessing.
CBSE_12th
Important Sign Error

❌ Sign Errors in Variable Separation and Integration

Students frequently make critical sign errors during the algebraic manipulation required to separate variables or while performing the subsequent integration. This is particularly common when dealing with terms in the denominator of the form `(a - bx)` or `(b - ax)` where the coefficient of the variable is negative.
πŸ’­ Why This Happens:
  • Careless Algebraic Manipulation: Mistakes occur when transposing terms, factoring out negative signs, or simplifying expressions, leading to an incorrect sign for a variable term.
  • Incorrect Application of Integration Formulas: Forgetting the chain rule when integrating functions like `1/(ax+b)`, especially when `a` is negative. For instance, `∫1/(1-v) dv` is often incorrectly integrated as `ln|1-v|` instead of `-ln|1-v|`.
  • Rushing Calculations: Not double-checking each step, particularly the signs of terms during variable separation and the constant multiplier during integration.
βœ… Correct Approach:
Be meticulous with every algebraic step. When performing integration, always apply the correct formula for `∫1/(ax+b) dx = (1/a)ln|ax+b| + C`. Pay extremely close attention to the sign and value of 'a'. If `a` is negative, the `(1/a)` term will introduce a negative sign.
πŸ“ Examples:
❌ Wrong:

Consider a step in solving a homogeneous differential equation, where after substitution `y=vx` and simplification, you arrive at:

x dv/dx = v / (1-v)

Separating variables:

∫ (1-v)/v dv = ∫ 1/x dx

Now, integrating the LHS, a common error is in the term `∫ 1/v dv - ∫ dv` (assuming `(1-v)/v` is split as `1/v - 1`).

Or, if the separation was `∫ v/(1-v) dv = ∫ 1/x dx`. A frequent error in integrating `∫ v/(1-v) dv` is:

∫ (-(1-v) + 1)/(1-v) dv = ∫ (-1 + 1/(1-v)) dv = -v + ln|1-v| + C

The bolded `ln|1-v|` is incorrect due to missing the negative coefficient from `(1-v)`.

βœ… Correct:

Following the correct integration for `∫ v/(1-v) dv` from the above scenario:

∫ v/(1-v) dv = ∫ (-(1-v) + 1)/(1-v) dv
                 = ∫ (-1 + 1/(1-v)) dv
                 = -v + ∫ 1/(1-v) dv

Now, integrate `1/(1-v)`. Here, `a = -1` (coefficient of `v`).

                 = -v + (1/-1)ln|1-v| + C
                 = -v - ln|1-v| + C

This careful application of the integration formula avoids the sign error. Remember to substitute `v = y/x` back into the solution.

πŸ’‘ Prevention Tips:
  • Step-by-Step Verification: After each algebraic manipulation (like transposing terms or factoring), quickly re-check the signs.
  • Coefficient Check: When integrating `1/(ax+b)`, always explicitly identify `a`. If `a` is negative (e.g., in `1-v`, `a = -1`), ensure `(1/a)` is correctly applied.
  • Use Parentheses: Employ parentheses generously during substitution and variable separation to avoid ambiguity, especially with negative terms.
  • CBSE & JEE Importance: Sign errors are severely penalized in both CBSE (loss of marks for the final answer and sometimes intermediate steps) and JEE (incorrect final answer means zero marks). Precision is paramount.
CBSE_12th
Important Unit Conversion

❌ Misapplication of Substitution Method in Homogeneous Differential Equations

Students often struggle with the correct application of the substitution `y=vx` (or `x=vy`) and its subsequent differentiation. Unit conversions are not applicable to solving homogeneous differential equations in the CBSE/JEE syllabus. Common pitfalls include incorrect differentiation of `y=vx` (e.g., `dy/dx = dv/dx` instead of the product rule) or algebraic errors during variable separation.
πŸ’­ Why This Happens:
  • Lack of understanding of the product rule for differentiation when `v` is a function of `x`.
  • Carelessness in algebraic manipulation after substitution, leading to difficulties in separating variables.
  • Rushing through calculations without double-checking each step.
βœ… Correct Approach:
  1. Identify Homogeneous: Check if `dy/dx = f(x,y)` can be written as `f(y/x)`.
  2. Substitute: Use `y = vx` (or `x = vy` if appropriate).
  3. Differentiate: Apply the product rule: `dy/dx = v + x(dv/dx)`.
  4. Separate Variables: Substitute `y=vx` and `dy/dx` into the original equation and rearrange to isolate `v` and `x` terms.
  5. Integrate: Integrate both sides with respect to their respective variables.
  6. Resubstitute: Replace `v` with `y/x` (or `x/y`) to obtain the general solution.
πŸ“ Examples:
❌ Wrong:
Consider `dy/dx = (x^2 + y^2) / (2xy)`. Let `y = vx`.
Student incorrectly assumes `dy/dx = dv/dx`.
Substituting: `dv/dx = (x^2 + (vx)^2) / (2x(vx)) = (1 + v^2) / (2v)`.
Mistake: The derivative of `y=vx` is wrong. This fundamental error prevents correct separation and integration.
βœ… Correct:
Consider `dy/dx = (x^2 + y^2) / (2xy)`. Let `y = vx`.
Then `dy/dx = v + x(dv/dx)`.
Substituting: `v + x(dv/dx) = (x^2 + v^2x^2) / (2v x^2) = (1 + v^2) / (2v)`.
Separating variables:
`x(dv/dx) = (1 + v^2) / (2v) - v = (1 + v^2 - 2v^2) / (2v) = (1 - v^2) / (2v)`.
Correct Separation: `(2v / (1 - v^2)) dv = (1/x) dx`. This sets up for proper integration.
πŸ’‘ Prevention Tips:
  • Master Differentiation: Firmly know the product rule: `d/dx(vx) = v + x(dv/dx)`.
  • Practice Algebra: Accurately simplify expressions and separate variables.
  • Verify Steps: Double-check algebraic manipulations before proceeding.
  • Understand Concept: Recognize that the substitution transforms a homogeneous equation into a variable separable form.
CBSE_12th
Important Formula

❌ Incorrect Substitution of dy/dx or Improper Variable Separation after Substitution

Students often correctly identify a homogeneous differential equation and even choose the substitution y = vx (or x = vy). However, a common mistake arises in two critical steps:
1. Incorrectly substituting for dy/dx: Forgetting that if y = vx, then dy/dx is not simply v or dv/dx, but rather derived using the product rule: dy/dx = v + x(dv/dx).
2. Improperly separating variables: After the substitution, failing to rearrange the terms correctly to achieve the form f(v) dv = g(x) dx, leading to difficulties in integration. This is crucial for CBSE exam success.
πŸ’­ Why This Happens:
This mistake stems from a lack of rigorous understanding of the substitution process. Students might:
  • Rely on rote memorization without grasping the derivation of dy/dx.
  • Rush through the algebraic manipulation, leading to errors in collecting terms of v and x.
  • Confuse homogeneous equations with variable separable or linear equations, expecting a simpler separation.
βœ… Correct Approach:
The correct approach involves a clear, step-by-step process:
  1. Identify Homogeneous DE: Ensure all terms have the same degree.
  2. Choose Substitution: If dy/dx is given, substitute y = vx. If dx/dy is given, substitute x = vy.
  3. Derive dy/dx (or dx/dy):
     If y = vx, then differentiate with respect to x using the product rule: dy/dx = v(1) + x(dv/dx).
     If x = vy, then differentiate with respect to y: dx/dy = v(1) + y(dv/dy).
  4. Substitute into DE: Replace y with vx (or x with vy) and dy/dx with v + x(dv/dx).
  5. Separate Variables: After substitution, simplify the equation and rearrange all terms involving v with dv on one side and all terms involving x with dx on the other side. This step is often where algebraic errors occur.
  6. Integrate: Integrate both sides to find the general solution.
πŸ“ Examples:
❌ Wrong:
Consider dy/dx = (x+y)/x. A common wrong step:
Let y=vx. Then dy/dx = v (incorrect!)
Substituting: v = (x+vx)/x = 1+v
This leads to 0 = 1, which is absurd, indicating a fundamental error in substitution.
βœ… Correct:
For dy/dx = (x+y)/x:
1. Substitute y = vx.
2. Differentiate: dy/dx = v + x(dv/dx).
3. Substitute into the DE:
v + x(dv/dx) = (x + vx)/x
v + x(dv/dx) = x(1+v)/x
v + x(dv/dx) = 1+v
4. Separate variables:
x(dv/dx) = 1+v - v
x(dv/dx) = 1
dv = dx/x (Variables are now correctly separated).
5. Integrate: ∫dv = ∫(1/x)dx => v = ln|x| + C.
6. Resubstitute v = y/x: y/x = ln|x| + C, or y = x(ln|x| + C).
πŸ’‘ Prevention Tips:
  • Practice Derivative: Thoroughly practice differentiating y = vx to get dy/dx = v + x(dv/dx). Make it second nature.
  • Systematic Steps: Always follow the standard sequence of steps: Substitution of y, then substitution of dy/dx, then algebraic simplification, then variable separation.
  • Algebraic Precision: Be meticulous with algebraic manipulations, especially when isolating terms of v and x for separation. A single sign error or misplaced term can derail the entire solution.
  • CBSE Exam Tip: Clearly show all steps of substitution and variable separation. Examiners award marks for these intermediate steps, even if the final integration has a minor error.
CBSE_12th
Important Calculation

❌ Algebraic Errors in Variable Separation After Substitution

Students frequently make critical algebraic errors when separating variables after applying the standard substitution for homogeneous differential equations, i.e., y = vx (or x = vy). These errors typically occur in the simplification process after substituting `dy/dx = v + x(dv/dx)` (or `dx/dy = v + x(dv/dy)`) and involve incorrect combination of terms, sign mistakes, or misapplication of common denominators when dealing with fractions involving 'v'.
πŸ’­ Why This Happens:
This mistake often stems from a lack of careful algebraic manipulation, especially when the expression for `dv/dx` (or `dv/dy`) becomes a complex rational function of `v`. Students might rush the simplification, leading to errors in finding common denominators, distributing terms, or transposing elements across the equals sign. A weak foundation in basic algebraic fractions can contribute significantly.
βœ… Correct Approach:
After substituting `y = vx` and `dy/dx = v + x(dv/dx)` into the homogeneous equation, the next step is to isolate `x(dv/dx)` and simplify the right-hand side. This simplification often involves combining 'v' terms, which usually requires finding a common denominator for rational expressions. Meticulously perform all algebraic steps:
  1. Isolate `x(dv/dx)`.
  2. Combine `v` terms on the right-hand side by finding a common denominator.
  3. Simplify the resulting expression.
  4. Then, separate variables by bringing all `v` terms (with `dv`) to one side and all `x` terms (with `dx`) to the other.
CBSE Tip: Show all intermediate algebraic steps clearly to earn full marks, even if they seem trivial.
πŸ“ Examples:
❌ Wrong:
Consider the equation `dy/dx = (y-x)/(y+x)`. After substituting `y=vx` and `dy/dx = v + x(dv/dx)`, we get:
`v + x(dv/dx) = (vx-x)/(vx+x) = (v-1)/(v+1)`
`x(dv/dx) = (v-1)/(v+1) - v`
A common calculation error here would be:
`x(dv/dx) = (v-1 - v) / (v+1)` (Incorrectly assuming the common denominator applies to 'v' without multiplying 'v' by '(v+1)')
`x(dv/dx) = -1 / (v+1)`
βœ… Correct:
Using the same equation `dy/dx = (y-x)/(y+x)` and the step `x(dv/dx) = (v-1)/(v+1) - v`:
The correct algebraic manipulation is:
`x(dv/dx) = (v-1 - v(v+1)) / (v+1)` (Correctly finding the common denominator and multiplying 'v' by '(v+1)')
`x(dv/dx) = (v-1 - v^2 - v) / (v+1)`
`x(dv/dx) = (-v^2 - 1) / (v+1)`
This leads to the separable form: `(v+1)/(-v^2-1) dv = dx/x`.
πŸ’‘ Prevention Tips:
  • Practice Algebraic Manipulation: Regularly solve problems involving fractions and polynomial simplification.
  • Step-by-Step Approach: Avoid skipping steps, especially during the simplification phase. Write down each algebraic operation clearly.
  • Double-Check Signs and Terms: Pay close attention to positive/negative signs and ensure all terms are correctly transposed and combined.
  • Common Denominator Mastery: Ensure you are proficient in finding and using common denominators for rational expressions.
  • JEE Tip: While speed is essential, accuracy in these algebraic steps is paramount. Develop mental math skills for simpler manipulations but don't hesitate to write down complex steps.
CBSE_12th
Important Conceptual

❌ Incorrect Differentiation of Substitution <span style='color: #FF4500;'>y=vx</span>

Students frequently err in differentiating y=vx with respect to x (finding dy/dx). This crucial step often leads to an incorrect transformed equation and subsequent errors in variable separation and integration.
πŸ’­ Why This Happens:
This stems from either forgetting the product rule of differentiation or misapplying it, treating v as a constant instead of a function of x. Algebraic simplification errors after substitution are also common.
βœ… Correct Approach:
  • Identify the homogeneous differential equation (expressible as dy/dx = f(y/x)).
  • Substitute y = vx (or x = vy if more convenient).
  • Crucial Step: Differentiate y = vx using the product rule to get dy/dx = v + x(dv/dx).
  • Substitute both into the original DE, then simplify and carefully separate variables.
  • Integrate both sides and finally replace v with y/x for the general solution.
πŸ“ Examples:
❌ Wrong:

Problem: Solve dy/dx = (x + y)/x

Student's Incorrect Steps:

  1. Substitutes y = vx.
  2. Mistake: Incorrectly assumes dy/dx = dv/dx (treating 'v' as a constant).
  3. Substituting into the DE: dv/dx = (x + vx)/x = 1 + v.
  4. Separating and integrating: ∫ dv/(1+v) = ∫ dx &implies; ln|1+v| = x + C.
  5. Incorrect Final Solution: ln|1 + y/x| = x + C.
βœ… Correct:

Problem: Solve dy/dx = (x + y)/x

Correct Steps:

  1. Recognize as homogeneous. Substitute y = vx.
  2. Correct Differentiation: Using the product rule, dy/dx = d/dx(vx) = v(1) + x(dv/dx) = v + x(dv/dx).
  3. Substitute into the DE: v + x(dv/dx) = (x + vx)/x = x(1 + v)/x = 1 + v.
  4. Simplify: x(dv/dx) = 1 + v - v = 1.
  5. Separate variables: dv = dx/x.
  6. Integrate both sides: ∫ dv = ∫ (1/x) dx &implies; v = ln|x| + C.
  7. Replace v = y/x back into the solution: y/x = ln|x| + C.
  8. Final (Correct) Solution: y = x(ln|x| + C).
πŸ’‘ Prevention Tips:
  • Master Product Rule: Always correctly derive and use dy/dx = v + x(dv/dx) when y=vx. Practice this derivation until it's second nature.
  • Algebraic Precision: Be meticulous in simplifying expressions and separating variables. Avoid rushing through algebraic steps.
  • Recheck Substitution: After substituting y=vx and dy/dx = v + x(dv/dx), take a moment to verify the algebraic simplification of the resulting equation before proceeding to integration.
CBSE_12th
Important Conceptual

❌ Misidentifying a Homogeneous Equation or Incorrect Substitution

Students often fail to correctly identify if a given differential equation is homogeneous before applying the standard substitution method. A differential equation of the form dy/dx = f(x,y) is homogeneous if f(Ξ»x, Ξ»y) = f(x,y) for any non-zero constant Ξ». If not homogeneous, applying y=vx (or x=vy) will lead to an unseparable equation, wasting valuable exam time. Even when correctly identified, common errors include incorrect differentiation of y=vx (forgetting the product rule) or mistakes in separating variables after substitution.
πŸ’­ Why This Happens:
This mistake stems from a conceptual misunderstanding of what constitutes a homogeneous differential equation. Students might rush to apply the method without verification or make algebraic errors during differentiation (e.g., writing dy/dx = dv/dx instead of v + x(dv/dx)). A lack of practice with the separation of variables step also contributes to errors.
βœ… Correct Approach:
To correctly solve a homogeneous differential equation:
  • Verify Homogeneity: First, express the equation in the form dy/dx = f(x,y). Then, check if f(Ξ»x, Ξ»y) = f(x,y). If this condition is met, it's homogeneous.
  • Apply Substitution: Substitute y = vx into the equation. Consequently, replace dy/dx with v + x(dv/dx) (using the product rule for differentiation).
  • Separate Variables: The equation should transform into a separable form in terms of v and x. Rearrange to isolate all v terms with dv and all x terms with dx.
  • Integrate: Integrate both sides with respect to their respective variables.
  • Back-Substitute: Finally, replace v with y/x to express the general solution in terms of x and y.
πŸ“ Examples:
❌ Wrong:

Equation: dy/dx = (x+y+1)/(x-y)

Student's Mistake: Incorrectly assumes it's homogeneous and substitutes y=vx.

dy/dx = v + x(dv/dx)
v + x(dv/dx) = (x+vx+1)/(x-vx)
v + x(dv/dx) = (1+v+1/x)/(1-v)

This equation cannot be separated into functions of only v and only x (due to the 1/x term). This indicates the initial assumption of homogeneity was incorrect, leading to a dead end.

βœ… Correct:

Equation: dy/dx = (x+y)/(x-y)

1. Verify Homogeneity: Let f(x,y) = (x+y)/(x-y). Then f(Ξ»x, Ξ»y) = (Ξ»x+Ξ»y)/(Ξ»x-Ξ»y) = Ξ»(x+y)/Ξ»(x-y) = (x+y)/(x-y) = f(x,y). It is homogeneous.

2. Substitute: Let y=vx, so dy/dx = v + x(dv/dx).

v + x(dv/dx) = (x+vx)/(x-vx) = x(1+v)/x(1-v) = (1+v)/(1-v)

3. Separate Variables:
x(dv/dx) = (1+v)/(1-v) - v = (1+v - v(1-v))/(1-v) = (1+v-v+v^2)/(1-v) = (1+v^2)/(1-v)
(1-v)/(1+v^2) dv = (1/x) dx

4. Integrate:
∫ (1/(1+v^2)) dv - ∫ (v/(1+v^2)) dv = ∫ (1/x) dx
arctan(v) - (1/2)ln|1+v^2| = ln|x| + C

5. Back-Substitute: Replace v with y/x.
arctan(y/x) - (1/2)ln|1+(y/x)^2| = ln|x| + C

πŸ’‘ Prevention Tips:
  • Always Verify: Before any substitution, rigorously check the homogeneity condition f(Ξ»x, Ξ»y) = f(x,y). This is a critical first step for JEE.
  • Memorize and Understand: Be thorough with the substitution y=vx and its derivative dy/dx = v + x(dv/dx). Understand the product rule application.
  • Practice Separation: Work through multiple problems to master the algebraic manipulation required to separate variables after substitution.
  • Don't Forget Back-Substitution: The final answer must be in terms of x and y. Remember to replace v with y/x.
JEE_Main
Important Calculation

❌ Algebraic Errors During Variable Separation and Incorrect Integration

Students frequently make algebraic mistakes after substituting y=vx and dy/dx = v + x(dv/dx) into a homogeneous differential equation. This often leads to errors in isolating x(dv/dx) and then incorrectly separating the variables v and x. A common error involves incorrect sign manipulation or misplacing terms. Subsequently, if variables are separated incorrectly, or even if correctly separated, students might apply an incorrect integration formula for functions of v, especially for rational functions, leading to an incorrect general solution.
πŸ’­ Why This Happens:
  • Carelessness: Lack of attention to detail during algebraic simplification and rearrangement of terms.
  • Sign Errors: Mistakes in handling positive and negative signs when transposing terms across the equals sign.
  • Weak Algebraic Foundation: Inadequate practice with complex algebraic manipulations.
  • Integration Formula Confusion: Misremembering or confusing similar-looking integration formulas, e.g., ∫ dv/(1+v2) vs. ∫ dv/(v2-1).
  • Exam Pressure: Rushing under time constraints can exacerbate these errors.
βœ… Correct Approach:
  • Systematic Simplification: After substitution, meticulously simplify the equation to the form x(dv/dx) = G(v), ensuring all algebraic steps and signs are correct.
  • Careful Separation: Isolate all terms involving v with dv on one side and all terms involving x with dx on the other side: dv/G(v) = dx/x. Double-check this step.
  • Correct Integration: Identify and apply the precise integration formula for ∫ dv/G(v). Be vigilant about the form of the integrand and constants.
  • Final Substitution: Remember to substitute back v = y/x to express the solution in terms of x and y.
πŸ“ Examples:
❌ Wrong:
Consider the equation: dy/dx = (y2 - x2) / (2xy).
After substituting y=vx and dy/dx = v + x(dv/dx), we correctly get:
v + x(dv/dx) = (v2x2 - x2) / (2x(vx)) = (v2 - 1) / (2v)
x(dv/dx) = (v2 - 1) / (2v) - v = (v2 - 1 - 2v2) / (2v) = (-1 - v2) / (2v) = -(1 + v2) / (2v)
Common Mistake: A student might incorrectly write x(dv/dx) = (1 + v2) / (2v), missing the crucial negative sign. This leads to:
∫ (2v / (1 + v2)) dv = ∫ (1 / x) dx
ln|1 + v2| = ln|x| + ln|C|
ln|1 + v2| = ln|Cx|
1 + (y/x)2 = Cx
(x2 + y2) / x2 = Cx
x2 + y2 = Cx3 (This is incorrect due to the missing negative sign).
βœ… Correct:
Following the correct derivation from the wrong example:
x(dv/dx) = -(1 + v2) / (2v)
∫ (2v / (1 + v2)) dv = ∫ (-1 / x) dx
ln|1 + v2| = -ln|x| + ln|C|
ln|1 + v2| + ln|x| = ln|C|
ln|x(1 + v2)| = ln|C|
x(1 + (y/x)2) = C
x((x2 + y2) / x2) = C
(x2 + y2) / x = C
x2 + y2 = Cx (This is the correct general solution).
πŸ’‘ Prevention Tips:
  • JEE Tip: Always write down intermediate steps clearly, especially when rearranging terms and changing signs.
  • Practice Aloud: Verbally stating each algebraic step can help catch errors.
  • CBSE Focus: Ensure you are comfortable with common integration formulas for rational functions. Mistakes here are heavily penalized.
  • Self-Check: After separating variables, quickly review if the terms on each side are solely dependent on their respective variables.
  • Regularly revise standard integrals, particularly those involving ln, tan-1, and partial fractions for more complex cases.
JEE_Main
Important Formula

❌ Incorrect Differentiation of the Substitution `y = vx` or `x = vy`

A frequent error in solving homogeneous differential equations is the improper differentiation of the substitution. When using `y = vx`, students often forget or incorrectly apply the product rule to find `dy/dx`. Similarly, for `x = vy`, the differentiation with respect to `y` (i.e., `dx/dy`) is often miscalculated. This leads to an incorrect expression for the derivative, making the subsequent variable separation and integration steps erroneous.
πŸ’­ Why This Happens:
This mistake primarily stems from a lack of careful application of the product rule of differentiation. Students often differentiate `v` and `x` separately without considering their product, or treat `v` as a constant, which it is not (as `v` is a function of `x` when `y=vx`). Rushing through the substitution step is another common cause.
βœ… Correct Approach:
For a homogeneous differential equation of the form `dy/dx = f(y/x)`, the correct substitution is `y = vx`. Differentiating both sides with respect to `x` using the product rule gives:
`dy/dx = v * (d/dx(x)) + x * (d/dx(v))`
`dy/dx = v + x (dv/dx)`.
This derived expression for `dy/dx` must then be substituted into the original differential equation.

Alternatively, for `dx/dy = g(x/y)`, the substitution `x = vy` leads to:
`dx/dy = v + y (dv/dy)`.
πŸ“ Examples:
❌ Wrong:
Given `dy/dx = (x+y)/(x-y)`. Students might mistakenly substitute `y=vx` and then write `dy/dx = dv/dx`, or `dy/dx = v`, ignoring the product rule. This leads to a simplified but incorrect equation like `dv/dx = (x+vx)/(x-vx)` which cannot be solved correctly by variable separation.
βœ… Correct:
Consider the same equation: `dy/dx = (x+y)/(x-y)`.
1. Substitute `y = vx`.
2. Differentiate `y = vx` correctly with respect to `x`: `dy/dx = v + x (dv/dx)`.
3. Substitute both `y = vx` and `dy/dx = v + x(dv/dx)` into the original equation:
   `v + x (dv/dx) = (x + vx) / (x - vx)`
   `v + x (dv/dx) = x(1 + v) / x(1 - v)`
   `v + x (dv/dx) = (1 + v) / (1 - v)`
This is the correct form, ready for variable separation. This step is crucial for JEE Main problems.
πŸ’‘ Prevention Tips:
  • Always write down the full differentiation step: `d/dx(uv) = u(dv/dx) + v(du/dx)`.
  • Treat `v` as a function of `x` (or `y`) when differentiating.
  • Practice the substitution `y=vx` and `x=vy` multiple times to internalize the correct derivative.
  • For JEE Main, speed and accuracy in this step are vital, as subsequent steps depend entirely on it.
  • Verify your `dy/dx` (or `dx/dy`) expression before proceeding to separate variables.
JEE_Main
Critical Approximation

❌ Misidentifying a Non-Homogeneous Differential Equation as Homogeneous

Students often incorrectly classify a non-homogeneous differential equation as homogeneous, leading them to apply the substitution y = vx or x = vy, which is only valid for homogeneous equations. This fundamental error results in an unsolvable or incorrectly solved equation, wasting valuable exam time.
πŸ’­ Why This Happens:
This mistake primarily occurs due to a lack of a clear understanding of the definition of a homogeneous function and a homogeneous differential equation. Students might:
  • Fail to check the degree of each term in the numerator and denominator.
  • Mistakenly assume that any equation with dy/dx = f(x,y) is homogeneous if it looks 'similar' to previous examples.
  • Overlook constant terms or terms of different degrees that make the equation non-homogeneous.
βœ… Correct Approach:
Before attempting to solve, always verify if the differential equation is indeed homogeneous. A first-order differential equation of the form dy/dx = F(x,y) is homogeneous if F(x,y) is a homogeneous function of degree zero. This means that if we replace x with Ξ»x and y with Ξ»y, then F(Ξ»x, Ξ»y) = Ξ»^0 F(x,y) = F(x,y). Alternatively, check if the degree of every term in the numerator and denominator is the same.
πŸ“ Examples:
❌ Wrong:
Consider the equation: dy/dx = (x + y + 1) / (x + y)
Wrong Approach: A student might immediately substitute y = vx and dy/dx = v + x(dv/dx), thinking it's homogeneous because x+y appears.
v + x(dv/dx) = (x + vx + 1) / (x + vx) = (x(1+v) + 1) / (x(1+v)) = 1 + 1/(x(1+v)).
This does not simplify to a function of v alone, indicating it's not homogeneous and the substitution is incorrect.
βœ… Correct:
For the equation dy/dx = (x + y + 1) / (x + y):
Correct Identification: Let F(x,y) = (x + y + 1) / (x + y).
Replace x with Ξ»x and y with Ξ»y:
F(Ξ»x, Ξ»y) = (Ξ»x + Ξ»y + 1) / (Ξ»x + Ξ»y) = (Ξ»(x + y) + 1) / (Ξ»(x + y)).
Since F(Ξ»x, Ξ»y) β‰  F(x,y) (due to the '+1' in the numerator not being multiplied by Ξ»), the function F(x,y) is not homogeneous. Therefore, the substitution y = vx is invalid.

Tip: For equations like dy/dx = (ax + by + c) / (Ax + By + C) where c or C is non-zero, it is generally not homogeneous. These require different methods like reduction to homogeneous form (if ab' - a'b β‰  0) or direct substitution (if ab' - a'b = 0). This is a crucial distinction for both CBSE and JEE.
πŸ’‘ Prevention Tips:
  • Strictly Apply Definition: Before any substitution, explicitly check if F(Ξ»x, Ξ»y) = F(x,y) or if all terms in numerator/denominator have the same degree.
  • Look for Constants: The presence of constant terms (e.g., +1, -2) in a linear combination of x and y often makes the equation non-homogeneous.
  • Practice Classification: Solve multiple problems focusing solely on classifying differential equations (homogeneous, linear, variable separable) before attempting full solutions.
  • Recognize Standard Forms: Familiarize yourself with the standard forms of homogeneous equations, and also those that can be reduced to homogeneous form, to avoid confusion.
CBSE_12th
Critical Other

❌ Incorrect Application of Substitution and Variable Separation in Homogeneous DEs

A very common and critical mistake students make when solving homogeneous differential equations is the improper application of the substitution method (y = vx or x = vy) and subsequent errors during variable separation.

  • Error in differentiation: Students often forget to apply the product rule when differentiating y = vx with respect to x, incorrectly writing dy/dx = dv/dx instead of dy/dx = v + x dv/dx. Similarly, for x = vy, dx/dy = v + y dv/dy.
  • Algebraic manipulation errors: After substituting dy/dx and simplifying the right-hand side, students frequently make algebraic mistakes while isolating terms involving v and dv on one side and x and dx on the other. This often leads to non-separable forms or incorrect integration steps.
πŸ’­ Why This Happens:

These mistakes typically stem from:

  • Lack of conceptual clarity: Not fully understanding why the product rule is necessary for the substitution derivatives.
  • Rushing through steps: Students often try to combine multiple algebraic steps in their head, leading to errors.
  • Weak algebraic foundations: Difficulty in rearranging terms, finding common denominators, or factoring expressions correctly.
  • Insufficient practice: Not solving enough problems to internalize the step-by-step process reliably.
βœ… Correct Approach:

The correct approach involves a systematic application of the substitution and careful algebraic manipulation:

  1. Identify Homogeneous DE: First, ensure the given differential equation is homogeneous.
  2. Choose Substitution: If dy/dx is easily obtainable, use y = vx and dy/dx = v + x dv/dx. If dx/dy is easier, use x = vy and dx/dy = v + y dv/dy.
  3. Substitute and Simplify: Replace y with vx (or x with vy) and dy/dx (or dx/dy) with its derived form. Simplify the right-hand side.
  4. Separate Variables: Carefully isolate all terms involving v and dv on one side and all terms involving x and dx on the other. This often requires algebraic manipulation, such as subtracting v from both sides, finding a common denominator, and cross-multiplication.
  5. Integrate: Integrate both sides independently.
  6. Resubstitute: Replace v with y/x (or x/y) to get the solution in terms of x and y.
πŸ“ Examples:
❌ Wrong:

Consider the equation: (x + y) dy - (x - y) dx = 0. This can be written as dy/dx = (x - y) / (x + y).

Wrong Step: Student applies y = vx, but incorrectly writes dy/dx = dv/dx.

dv/dx = (x - vx) / (x + vx) = x(1 - v) / x(1 + v) = (1 - v) / (1 + v)

Then attempts to integrate ∫ (1 + v) / (1 - v) dv = ∫ dx/x. This is fundamentally incorrect as the substitution for dy/dx was wrong.

βœ… Correct:

Using the same equation: dy/dx = (x - y) / (x + y).

Correct Approach: Apply y = vx, so dy/dx = v + x dv/dx.

v + x dv/dx = (x - vx) / (x + vx)

v + x dv/dx = (1 - v) / (1 + v)

Now, separate variables:

x dv/dx = (1 - v) / (1 + v) - v

x dv/dx = (1 - v - v(1 + v)) / (1 + v)

x dv/dx = (1 - v - v - v^2) / (1 + v)

x dv/dx = (1 - 2v - v^2) / (1 + v)

∫ (1 + v) / (1 - 2v - v^2) dv = ∫ dx/x

This form is now correctly separated and ready for integration (e.g., by substitution for the denominator on the left side).

πŸ’‘ Prevention Tips:
  • Memorize the derivative: Always remember that if y = vx, then dy/dx = v + x dv/dx (using the product rule). Write it down explicitly in every problem.
  • Step-by-step algebra: Do not rush the algebraic manipulation. Write out each step clearly when simplifying and separating variables. Use separate lines for each transformation.
  • Practice, Practice, Practice: Solve a variety of homogeneous differential equations. The more you practice, the more intuitive the algebraic steps become.
  • Check your separation: Before integrating, mentally (or physically) verify that all terms involving v are with dv and all terms involving x are with dx, and there are no mixed terms.
  • CBSE Exam Tip: Showing clear steps for substitution and variable separation fetches marks even if the final integration is slightly off. Errors in these initial steps, however, can lead to zero marks for the entire solution.
CBSE_12th
Critical Sign Error

❌ Critical Sign Errors in `f(v) - v` Simplification

A common and critical error arises when students simplify the expression `f(v) - v` (derived from `dy/dx = f(v)`) after substituting `y = vx` (or `x = vy`) into a homogeneous differential equation. These sign errors, typically occurring during fractional combination or the distribution of negative signs, lead to an incorrect separable form and, consequently, a completely wrong final solution in the CBSE 12th examination.
πŸ’­ Why This Happens:
  • Haste: Rushing through algebraic steps.
  • Neglecting distribution: Not carefully distributing negative signs when subtracting `v` from `f(v)`.
  • Algebraic slips: Basic errors in combining terms or handling fractions.
βœ… Correct Approach:
  1. Substitute `y = vx` (or `x = vy`) and the respective derivative.
  2. Rearrange the equation to isolate `x(dv/dx) = f(v) - v`.
  3. Carefully combine `f(v)` and `v` using a common denominator. Pay extreme attention to the signs, especially during distribution.
  4. Ensure all terms have their correct signs before separating variables for integration.
πŸ“ Examples:
❌ Wrong:

Consider `dy/dx = (x+y)/(x-y)`. Substituting `y=vx` and `dy/dx = v + x(dv/dx)` gives:

`v + x(dv/dx) = (x+vx)/(x-vx) = (1+v)/(1-v)`

Wrong Step (Common Sign Error):

`x(dv/dx) = (1+v)/(1-v) - v = (1+v - v(1-v))/(1-v) = (1+v - v - v^2)/(1-v) = (1-v^2)/(1-v)` (Incorrectly taking `-v(-v)` as `-v^2` instead of `+v^2`).

βœ… Correct:

Using the same equation `dy/dx = (x+y)/(x-y)`:

`v + x(dv/dx) = (1+v)/(1-v)`

Correct Step:

`x(dv/dx) = (1+v)/(1-v) - v`

`x(dv/dx) = (1+v - v(1-v))/(1-v)`

`x(dv/dx) = (1+v - v + v^2)/(1-v)` (Note the correct `+v^2` term here)

`x(dv/dx) = (1+v^2)/(1-v)`

πŸ’‘ Prevention Tips:
  • Work Methodically: Perform one algebraic operation at a time. Avoid mental shortcuts when dealing with signs.
  • Explicit Sign Distribution: Always write out the distribution of negative signs (e.g., `-(A-B)` becomes `-A+B`) to prevent slips.
  • Verify Algebra: After simplifying `f(v) - v`, quickly recheck the entire algebraic manipulation for any sign errors.
  • Consistent Practice: Solve a variety of problems, focusing on the algebraic accuracy, to build speed and reduce errors.
CBSE_12th
Critical Unit Conversion

❌ Incorrect Conversion of Variables (y=vx) and its Differentiation

Students frequently make critical errors during the initial 'conversion' of a homogeneous differential equation by substituting y = vx (or x = vy) and subsequently differentiating it. While not a literal 'unit conversion' involving physical quantities, this step is a fundamental transformation of the equation's structure. Mistakes here are equivalent to an incorrect 'conversion' of the problem into a solvable form, leading to completely wrong solutions. This is a critical error for CBSE 12th exams as it's the defining step for this type of differential equation.
πŸ’­ Why This Happens:
This error primarily stems from two reasons:

  • Forgetting the Product Rule: When differentiating y = vx with respect to x, students often forget to apply the product rule, treating v as a constant or differentiating only one term.

  • Algebraic Errors: Incorrectly substituting y = vx and dy/dx = v + x(dv/dx) into the original equation, or making algebraic mistakes during the subsequent simplification and separation of variables.

βœ… Correct Approach:
The correct approach involves a precise application of the substitution and differentiation:

  1. Identify that the differential equation is homogeneous (all terms have the same degree).

  2. Substitute y = vx. This implies that v = y/x (a dimensionless ratio if x and y have the same units, conceptually simplifying the problem's 'units').

  3. Differentiate y = vx with respect to x using the product rule:

    dy/dx = d/dx (vx) = v(d/dx x) + x(d/dx v) = v(1) + x(dv/dx)

    So, dy/dx = v + x(dv/dx).

  4. Substitute both y = vx and dy/dx = v + x(dv/dx) into the original homogeneous differential equation. This 'converts' the equation into a separable variable form in terms of v and x.

πŸ“ Examples:
❌ Wrong:

Consider dy/dx = (x^2 + y^2) / (xy). Student attempts:


Let y = vx. Then dy/dx is mistakenly taken as dv/dx or v, instead of using the product rule.


Incorrect Step: dy/dx = dv/dx


Resulting incorrect substitution: dv/dx = (x^2 + (vx)^2) / (x(vx)), which will lead to a completely unresolvable or wrong equation.

βœ… Correct:

For dy/dx = (x^2 + y^2) / (xy):


1. Substitute y = vx.


2. Differentiate: dy/dx = v + x(dv/dx) (Applying product rule correctly).


3. Substitute both into the equation:


v + x(dv/dx) = (x^2 + (vx)^2) / (x(vx))


v + x(dv/dx) = (x^2 + v^2x^2) / (vx^2)


v + x(dv/dx) = x^2(1 + v^2) / (vx^2)


v + x(dv/dx) = (1 + v^2) / v


This is the correct 'converted' equation, now separable: x(dv/dx) = (1 + v^2)/v - v.

πŸ’‘ Prevention Tips:

  • Master the Product Rule: Practice differentiating y = vx until it's second nature.

  • Understand the 'Why': Grasp that v is a function of x (v = y/x), not a constant, which necessitates the product rule.

  • Algebraic Precision: Be extremely careful with algebraic simplification after substitution, especially cancelling terms and combining fractions.

  • Practice Regularly: Solve multiple problems, focusing on this initial setup step to build confidence and accuracy.


For JEE Advanced: While the core mistake remains the same, be prepared for more complex algebraic expressions post-substitution. The fundamental differentiation rule remains crucial.

CBSE_12th
Critical Calculation

❌ Incorrect Differentiation of Substitution Term (y = vx)

Students frequently make a critical error when differentiating the substitution y = vx (or x = vy) with respect to x (or y). They often forget to apply the product rule, leading to an incorrect expression for dy/dx (or dx/dy). This fundamental mistake propagates through the entire solution, rendering the final answer incorrect. This is a critical calculation understanding error.
πŸ’­ Why This Happens:
This error commonly occurs due to:
  • Lack of attention to the product rule: Students treat v as a constant, or simply write dv/dx instead of v + x(dv/dx).
  • Rushing the differentiation step: Under exam pressure, this crucial step is often hurried.
  • Misunderstanding that v is a function of x: For y = vx, v = y/x, so v is indeed a function of x, requiring the product rule.
βœ… Correct Approach:
When substituting y = vx into a homogeneous differential equation, always remember to differentiate it with respect to x using the product rule:
If y = vx, then dy/dx = d/dx(vx) = v(d/dx(x)) + x(d/dx(v)) = v(1) + x(dv/dx) = v + x(dv/dx).
Similarly, if x = vy, then dx/dy = v + y(dv/dy). This correctly transformed expression for dy/dx (or dx/dy) is essential for proceeding with variable separation, which is the next step in solving homogeneous differential equations.
πŸ“ Examples:
❌ Wrong:

Consider the equation: x dy/dx = y + x cos(y/x)

Let y = vx, so dy/dx = dv/dx (INCORRECT DIFFERENTIATION)

Substituting into the equation:

x (dv/dx) = vx + x cos(vx/x)

x (dv/dx) = vx + x cos(v)

dv/dx = v + cos(v) (This is an incorrect transformed equation, leading to a wrong solution.)

βœ… Correct:

Consider the equation: x dy/dx = y + x cos(y/x)

This is a homogeneous differential equation.

Let y = vx.

Differentiating with respect to x using the product rule:
dy/dx = d/dx(vx) = v(1) + x(dv/dx) = v + x(dv/dx) (CORRECT DIFFERENTIATION)

Substitute y = vx and dy/dx = v + x(dv/dx) into the original equation:

x (v + x(dv/dx)) = vx + x cos(vx/x)

xv + x2(dv/dx) = vx + x cos(v)

Subtract xv from both sides:

x2(dv/dx) = x cos(v)

x(dv/dx) = cos(v)

Now, separate variables correctly:

dv/cos(v) = dx/x

sec(v) dv = dx/x

Integrate both sides: ∫ sec(v) dv = ∫ dx/x

ln|sec(v) + tan(v)| = ln|x| + C

Replace v with y/x:

ln|sec(y/x) + tan(y/x)| = ln|x| + C (Correct Final Solution)

πŸ’‘ Prevention Tips:
  • Always write down the differentiation step: Do not perform it mentally. Explicitly show d/dx(vx) = v(1) + x(dv/dx).
  • Understand 'v' as a function of 'x': Recognize that v = y/x, which means v changes with x, necessitating the product rule.
  • Practice differentiation rules: Regular practice of product rule, quotient rule, and chain rule will strengthen your foundational skills.
  • Cross-check your substitution: After substituting dy/dx, carefully check if the v term cancels out or if the equation simplifies as expected. If not, re-evaluate your dy/dx expression. This is a common point where CBSE students lose significant marks.
CBSE_12th
Critical Formula

❌ Incorrect Substitution for Homogeneous Differential Equations

A frequent and critical error in solving homogeneous differential equations is the improper application or differentiation of the standard substitution. Students either choose the wrong form of substitution (e.g., trying x = vy when dy/dx is easily expressible as f(y/x)) or, more commonly, make mistakes when differentiating the chosen substitution (e.g., differentiating y = vx to obtain dy/dx = dv/dx or dy/dx = x(dv/dx) instead of the correct form). This leads to an incorrect transformed equation and an entirely wrong solution.
πŸ’­ Why This Happens:
This mistake stems from a lack of understanding of the product rule of differentiation and the chain rule's application. When y = vx, v is not a constant but a function of x (i.e., v = y/x). Students often forget to apply the product rule or incorrectly treat v as a constant during differentiation, leading to an incorrect expression for dy/dx or dx/dy.
βœ… Correct Approach:
For a homogeneous differential equation of the form dy/dx = f(y/x), the correct substitution is y = vx. You must then differentiate this with respect to x using the product rule:
d/dx (y) = d/dx (vx)
dy/dx = v * d/dx (x) + x * d/dx (v)
dy/dx = v + x(dv/dx)
Similarly, for dx/dy = g(x/y), the substitution is x = vy, leading to dx/dy = v + y(dv/dy).
πŸ“ Examples:
❌ Wrong:
Consider the substitution y = vx.
Incorrect differentiation: A common mistake is to write dy/dx = dv/dx (treating x as a constant) or dy/dx = x(dv/dx) (forgetting the v term). If dy/dx = (x^2 + y^2) / (xy), substituting y=vx and dy/dx = dv/dx would incorrectly transform the equation into dv/dx = (x^2 + (vx)^2) / (x(vx)), leading to a completely erroneous solution.
βœ… Correct:
Given y = vx.
Applying the product rule correctly:
dy/dx = (d/dx)(vΒ·x)
dy/dx = vΒ·(d/dx)(x) + xΒ·(d/dx)(v)
dy/dx = vΒ·1 + xΒ·(dv/dx)
Therefore, the substitution for dy/dx is v + x(dv/dx). For the example dy/dx = (x^2 + y^2) / (xy), the correct substitution would be:
v + x(dv/dx) = (x^2 + (vx)^2) / (x(vx))
v + x(dv/dx) = (x^2 + v^2x^2) / (vx^2)
v + x(dv/dx) = (1 + v^2) / v. This is now a separable differential equation.
πŸ’‘ Prevention Tips:
  • Understand the Derivation: Don't just memorize the formula; understand how dy/dx = v + x(dv/dx) is derived using the product rule.
  • Practice Differentiation: Regularly practice differentiating expressions like y = f(x)g(x) to reinforce the product rule.
  • JEE & CBSE Alert: In both JEE and CBSE exams, this is a fundamental step. Errors here guarantee zero marks for the solution, even if subsequent integration steps are correct. Pay extra attention.
  • Double Check: Always re-verify your differentiation step before proceeding with variable separation.
CBSE_12th
Critical Conceptual

❌ Incomplete `dy/dx` Substitution in Homogeneous Differential Equations

Students often correctly identify a differential equation as homogeneous but then make a critical error by failing to apply the complete substitution. They might substitute y = vx (or x = vy) into the right-hand side, but crucially forget to replace dy/dx on the left-hand side with v + x(dv/dx) (or dx/dy = v + y(dv/dy)). This omission fundamentally alters the equation, making it non-separable and leading to an incorrect solution.

πŸ’­ Why This Happens:

  • Conceptual Gap: Misunderstanding that dy/dx itself must be transformed using the product rule, not just y.

  • Exam Pressure: Rushing under pressure often leads to overlooking the derivative part of the substitution.

  • Algebraic Complexity: Even if the substitution is correctly applied, students might struggle with the subsequent algebraic manipulation required to separate the variables v and x.

βœ… Correct Approach:

To correctly solve a homogeneous differential equation (e.g., dy/dx = f(y/x)):



  1. Apply Full Substitution: Simultaneously substitute y = vx AND dy/dx = v + x(dv/dx) into the equation.

  2. Separate Variables: The equation will transform into v + x(dv/dx) = f(v). Rearrange this algebraically to isolate dv terms with v and dx terms with x, typically yielding dv / (f(v) - v) = dx / x.

  3. Integrate & Re-substitute: Integrate both sides, then finally replace v with y/x to express the solution in terms of x and y.

πŸ“ Examples:
❌ Wrong:

Consider the equation: dy/dx = (x+y)/x


Incorrect Method: A student substitutes y=vx only on the RHS, getting dy/dx = (x+vx)/x = 1+v. They then mistakenly try to integrate this directly (e.g., ∫dy = ∫(1+v)dx), which is invalid since v is a function of x and not a constant, thus failing to achieve proper variable separation.

βœ… Correct:

For the equation: dy/dx = (x+y)/x (which simplifies to dy/dx = 1 + y/x):



  1. Substitute y = vx AND dy/dx = v + x(dv/dx).

  2. The equation becomes: v + x(dv/dx) = 1 + v

  3. Simplify: x(dv/dx) = 1

  4. Separate variables: dv = dx/x

  5. Integrate: ∫dv = ∫(1/x) dxv = ln|x| + C

  6. Re-substitute v=y/x: y/x = ln|x| + Cy = x(ln|x| + C).

πŸ’‘ Prevention Tips:

  • Memorize Both Parts: Always associate y = vx with its derivative dy/dx = v + x(dv/dx) and apply them simultaneously.

  • Systematic Practice: Follow the full methodical stepsβ€”Substitute → Simplify → Separate → Integrate → Re-substituteβ€”without skipping any.

  • Algebraic Vigilance: Pay meticulous attention to algebraic rearrangement. Even small errors here can completely derail the solution. (JEE Tip: Speed comes from accurate, practiced steps, not from risky shortcuts.)

JEE_Main
Critical Other

❌ Algebraic Errors Post-Substitution in Homogeneous ODEs

Students often correctly identify a differential equation as homogeneous and apply the standard substitution, usually y = vx (or x = vy). However, a critical mistake arises during the subsequent algebraic manipulation. After substituting y = vx and dy/dx = v + x(dv/dx), students frequently make errors in simplifying the resulting expression. This can lead to a non-separable form, incorrect separation of variables, or a completely wrong function of v, making the integration step impossible or incorrect. This is particularly critical in JEE Advanced where multi-step problems test careful execution.
πŸ’­ Why This Happens:
  • Hasty Simplification: Rushing through the algebraic steps without paying attention to common factors, proper fraction addition/subtraction, or distribution.
  • Lack of Practice: Insufficient practice in complex algebraic manipulations involving fractions and variables, leading to errors under exam pressure.
  • Overlooking Simplifications: Failing to identify and cancel out common terms or factor expressions correctly, which are crucial for achieving variable separability.
βœ… Correct Approach:
The correct approach involves meticulous and step-by-step algebraic simplification after substituting y = vx and dy/dx = v + x(dv/dx). Always aim to isolate x(dv/dx) on one side and a function of v on the other, ensuring that the variables v and x can be separated. Factor out common terms (e.g., xn) from the numerator and denominator before substitution to simplify the f(y/x) term into f(v). Then, move the v term to the RHS and combine fractions carefully.
πŸ“ Examples:
❌ Wrong:
Consider dy/dx = (x2 + y2) / (xy).
After substituting y = vx and dy/dx = v + x(dv/dx):
v + x(dv/dx) = (x2 + (vx)2) / (x(vx))
v + x(dv/dx) = (x2 + v2x2) / (vx2)
v + x(dv/dx) = (1 + v2) / v
Wrong Step (Common Mistake): x(dv/dx) = 1 + v2 (Students might incorrectly subtract v or misinterpret the RHS). This prevents separation of variables.
βœ… Correct:
Continuing from the same equation: v + x(dv/dx) = (1 + v2) / v
Correct Step: x(dv/dx) = (1 + v2) / v - v
x(dv/dx) = (1 + v2 - v2) / v
x(dv/dx) = 1 / v
This correctly leads to separation of variables: v dv = dx / x, which can be easily integrated.
πŸ’‘ Prevention Tips:
  • Double-Check Algebraic Steps: After substituting and before separating variables, meticulously re-check all additions, subtractions, multiplications, and divisions.
  • Factor Out Carefully: Before substitution, ensure you've factored out the highest power of x from both numerator and denominator to simplify f(y/x) to f(v).
  • Systematic Approach: Always follow a systematic approach: substitution, simplification of f(v), moving v, combining fractions, then separating variables.
  • Practice Fraction Arithmetic: Strong foundational skills in adding/subtracting algebraic fractions are vital for JEE Advanced.
JEE_Advanced
Critical Approximation

❌ Premature and Unjustified Approximation of `y/x` Terms

Students often make a critical mistake by applying small-value approximations (e.g., sin(u) β‰ˆ u, tan(u) β‰ˆ u, e^u β‰ˆ 1 + u) to terms involving y/x (or v after substituting y=vx) while solving homogeneous differential equations. This is incorrect because y/x is a variable that is not necessarily small, and homogeneous equations are generally solved exactly, not approximately, in JEE Advanced.
πŸ’­ Why This Happens:
  • Misapplication of Approximations: Students recall small-angle/value approximations from other chapters (like Limits, Physics, or Series Expansions) but fail to understand their specific conditions of validity (i.e., when the argument approaches zero).
  • Simplification Tendency: An attempt to simplify complex functions of y/x to make the integration step seemingly easier, bypassing the correct (and often more complex) integration of the exact function.
  • Lack of Conceptual Clarity: Not distinguishing between problems where an exact solution is required versus those rare cases where an approximation might be justified under specific, stated conditions.
βœ… Correct Approach:
The standard and correct approach for solving homogeneous differential equations is to use the substitution y = vx (or x = vy), which transforms the equation into a separable form. The subsequent integration must be performed on the exact functions of v that arise, without any premature or unjustified approximations. Homogeneous differential equations in JEE Advanced virtually always demand an exact solution.
πŸ“ Examples:
❌ Wrong:

Problem: Solve the differential equation (x sin(y/x) + y) dx - x dy = 0.

Incorrect Step (Approximation): A student might incorrectly assume y/x is small and substitute sin(y/x) β‰ˆ y/x.

(x(y/x) + y) dx - x dy = 0
(y + y) dx - x dy = 0
2y dx - x dy = 0
2 dx/x = dy/y
Integrating: 2 ln|x| = ln|y| + C
ln(x^2) = ln|y| + C
y = A x^2 (This is an incorrect solution due to the approximation)
βœ… Correct:

Problem: Solve the differential equation (x sin(y/x) + y) dx - x dy = 0.

Correct Approach (Exact Solution): This is a homogeneous DE. Substitute y = vx, so dy = v dx + x dv.

(x sin(v) + vx) dx - x(v dx + x dv) = 0
x sin(v) dx + vx dx - vx dx - x^2 dv = 0
x sin(v) dx - x^2 dv = 0
x sin(v) dx = x^2 dv
Separate variables (assuming x β‰  0 and sin(v) β‰  0):
dx/x = dv/sin(v)
dx/x = csc(v) dv
Integrate both sides:
∫ dx/x = ∫ csc(v) dv
ln|x| = ln|csc(v) - cot(v)| + C
Substitute back v = y/x:
ln|x| = ln|csc(y/x) - cot(y/x)| + C (This is the correct general solution)
πŸ’‘ Prevention Tips:
  • Identify DE Type First: Always correctly classify the differential equation. For homogeneous DEs, immediately recall the standard y=vx substitution method.
  • Understand Approximation Validity: Reiterate that approximations like sin(ΞΈ) β‰ˆ ΞΈ are valid only when ΞΈ is very small. In general DE problems, y/x is a variable and not necessarily small.
  • Prioritize Exact Solutions: Unless the problem explicitly asks for an approximate solution under specific limiting conditions, always aim for the exact solution using established methods.
  • Practice Meticulously: Solve a wide range of homogeneous differential equations carefully, focusing on the correct substitution, variable separation, and exact integration techniques.
JEE_Advanced
Critical Sign Error

❌ Critical Sign Errors in Separation of Variables and Integration

Students frequently make critical sign errors during the solution of homogeneous differential equations. These errors typically occur after the substitution (e.g., y=vx), specifically when rearranging terms to separate variables (v and x) or during the subsequent integration step. A common mistake is misplacing or losing a negative sign in the algebraic manipulation of f(v) - v or when integrating terms like 1/(a-x) versus 1/(x-a).
πŸ’­ Why This Happens:
This error stems from a combination of factors:
  • Haste: Rushing through algebraic steps, especially when moving terms across the equality sign.
  • Algebraic Oversight: Incorrectly factoring out negative signs or misapplying distributive properties.
  • Integral Formula Misapplication: Confusing integral formulas for expressions like ∫ 1/(a-x) dx = -ln|a-x| with ∫ 1/(x-a) dx = ln|x-a|.
  • Complex Expressions: The f(v)-v term can sometimes be complex, increasing the chance of an algebraic slip.
βœ… Correct Approach:
The correct approach demands meticulous attention to algebraic detail at every step. After substituting y=vx (or x=vy) and simplifying dy/dx = v + x(dv/dx) = f(v), carefully isolate x(dv/dx) = f(v) - v. Then, separate the variables with extreme precision, ensuring all negative signs are correctly carried over to their respective sides before integration. Double-check the numerator and denominator signs before applying any integral formulas.
πŸ“ Examples:
❌ Wrong:
Consider the equation: dy/dx = (y-x)/(y+x)
After substituting y=vx, we get:
v + x(dv/dx) = (vx-x)/(vx+x) = (v-1)/(v+1)
x(dv/dx) = (v-1)/(v+1) - v = (v-1 - v(v+1))/(v+1) = (v-1 - v^2 - v)/(v+1) = (-v^2-1)/(v+1) = -(v^2+1)/(v+1)
Wrong Separation: A student might write ∫ (v+1)/(v^2+1) dv = ∫ dx/x, incorrectly missing the negative sign on the RHS.
βœ… Correct:
Following the previous example, the correct separation of variables should be:
x(dv/dx) = -(v^2+1)/(v+1)
Correct Separation: ∫ (v+1)/(v^2+1) dv = ∫ -dx/x
Now integrate both sides:
∫ (v/(v^2+1) + 1/(v^2+1)) dv = -∫ 1/x dx
(1/2)ln|v^2+1| + arctan(v) = -ln|x| + C
Finally, substitute v=y/x back to get the solution in terms of x and y.
πŸ’‘ Prevention Tips:
  • Isolate Carefully: Before separating variables, ensure the f(v) - v term is fully simplified and any negative signs are clear.
  • Step-by-Step Algebra: Avoid skipping steps, especially when dealing with negative signs and fractions. Write down each algebraic manipulation clearly.
  • Verify Integral Formulas: Be mindful of the signs in integration, particularly for logarithmic and inverse trigonometric functions where the argument might be (a-x) instead of (x-a).
  • Check Your Work: After separating variables and setting up the integrals, quickly re-verify the signs and terms before proceeding to integration.
  • Use Parentheses: Employ parentheses liberally to clarify the scope of negative signs and fractional terms.
JEE_Advanced
Critical Unit Conversion

❌ Misinterpreting Homogeneity and Attempting `y=vx` Substitution for Non-Homogeneous Equations

Students frequently misidentify a non-homogeneous differential equation as homogeneous, often by incorrectly assessing the 'degree' of all terms. For an equation to be homogeneous (e.g., of the form `dy/dx = F(x,y)`, where `F(x,y)` is a homogeneous function of degree zero, or `M(x,y)dx + N(x,y)dy = 0`, where `M` and `N` are homogeneous functions of the same degree), all terms in the relevant functions must have the same total degree. A common error is overlooking constant terms or terms with different total powers of `x` and `y`, which leads to an incorrect classification and a failed solution attempt.
πŸ’­ Why This Happens:
This mistake stems from a lack of thorough understanding of the precise definition of a homogeneous function and how to rigorously check its degree. Students might quickly identify `x^2`, `xy`, `y^2` as parts of homogeneous expressions but fail to notice a stray `x` term, `y` term, or a constant (like `+5` or `-1`) which immediately makes the equation non-homogeneous. This oversight is akin to a 'unit mismatch' if one considers the 'degree' of a term as its mathematical dimension.
βœ… Correct Approach:
Before attempting any substitution like `y=vx` or `x=vy`, always verify that the differential equation is indeed homogeneous. For an equation `dy/dx = f(x,y)`, check if `f(tx,ty) = f(x,y)` (i.e., `f` is homogeneous of degree zero). For `M(x,y)dx + N(x,y)dy = 0`, check if `M(tx,ty) = t^n M(x,y)` and `N(tx,ty) = t^n N(x,y)` for the same `n`. If the equation is not homogeneous, the standard substitution method for homogeneous equations will not work, and a different solution technique (e.g., linear, exact, reducible to homogeneous) is required.
πŸ“ Examples:
❌ Wrong:
Consider the equation:
(x^2 + y^2 + x)dx - 2xydy = 0
A student might incorrectly assume this is homogeneous because `x^2`, `y^2`, and `xy` are all terms of degree 2. They might then proceed to substitute `y=vx` and `dy = vdx + xdv`, leading to a complicated non-separable equation and wasted time.
βœ… Correct:
To correctly check the homogeneity of (x^2 + y^2 + x)dx - 2xydy = 0:
Let `M(x,y) = x^2 + y^2 + x` and `N(x,y) = -2xy`.
For `M(x,y)`:
  • Term 1: `x^2` has degree 2.
  • Term 2: `y^2` has degree 2.
  • Term 3: `x` has degree 1.
Since all terms in `M(x,y)` do not have the same degree (2, 2, and 1), `M(x,y)` is NOT a homogeneous function. Therefore, the differential equation (x^2 + y^2 + x)dx - 2xydy = 0 is NOT homogeneous.

Correct Homogeneous Example:
For (x^2 + y^2)dx - 2xydy = 0:
All terms (`x^2`, `y^2`, and `xy`) are of degree 2. Thus, this equation is homogeneous.
πŸ’‘ Prevention Tips:
  • Systematic Degree Check: For every term in `M(x,y)` and `N(x,y)` (or in the numerator/denominator of `dy/dx`), sum the powers of `x` and `y`. All terms must yield the exact same sum for the function to be homogeneous.
  • JEE Advanced vs. CBSE: While CBSE exams often present clearly homogeneous equations, JEE Advanced frequently includes subtle non-homogeneous forms. Always perform this check rigorously to avoid critical errors.
  • Beware of Constant Terms: Any constant term (e.g., `+1`, `-5`) or terms with varying degrees (like `x` in `x^2+y^2+x`) immediately indicate a non-homogeneous equation.
  • Practice Identification: Solve numerous problems focusing solely on identifying the type of differential equation before attempting to solve it.
JEE_Advanced
Critical Formula

❌ Incorrect Substitution Formula or Algebraic Separation

A common critical error in solving homogeneous differential equations is either misapplying the derivative formula for the substitution (e.g., `y=vx`, often forgetting the `v` term) or making fundamental algebraic mistakes during the variable separation step after substitution, which prevents the equation from being correctly integrated.
πŸ’­ Why This Happens:
  • Product Rule Neglect: Forgetting that the derivative of `y = vx` with respect to `x` is `dy/dx = v + x(dv/dx)`, not just `x(dv/dx)`.
  • Algebraic Errors: Carelessness in combining terms or isolating `v` and `x` expressions to achieve the desired `f(v)dv = g(x)dx` form.
  • Lack of Verification: Not systematically checking if the equation is truly variable separable after algebraic manipulation.
βœ… Correct Approach:
  • 1. Confirm Homogeneity: First, ensure the given differential equation is homogeneous (can be expressed as `dy/dx = f(y/x)` or `dx/dy = g(x/y)`).
  • 2. Apply Correct Substitution & Derivative:
            – If `dy/dx = f(y/x)`, substitute `y = vx` and `dy/dx = v + x(dv/dx)`.
            – If `dx/dy = g(x/y)`, substitute `x = vy` and `dx/dy = v + y(dv/dy)`.
  • 3. Systematic Variable Separation: Substitute into the original equation and meticulously rearrange terms. The goal is to separate all `v` terms with `dv` on one side and all `x` terms with `dx` (or `y` terms with `dy`) on the other.
  • 4. Integrate and Resubstitute: Integrate both sides and finally replace `v` with `y/x` (or `x/y`) to get the solution in original variables.
πŸ“ Examples:
❌ Wrong:

Question: Solve the homogeneous differential equation `dy/dx = (y^2 + x^2) / (2xy)`.

Common Wrong Approach:
1. Student correctly identifies `y=vx` but incorrectly writes `dy/dx = x(dv/dx)` (critical mistake: the `v` term is omitted from the product rule).
2. Substituting this into the equation:
  `x(dv/dx) = ((vx)^2 + x^2) / (2x(vx))`
  `x(dv/dx) = (v^2x^2 + x^2) / (2vx^2)`
  `x(dv/dx) = (v^2 + 1) / (2v)`
3. Attempting to separate variables:
  `dv = (v^2 + 1) / (2vx) dx`
This equation is not variable separable because `x` remains on the RHS with `v`. This clearly indicates an error in the initial derivative substitution or subsequent algebra.

βœ… Correct:

Question: Solve the homogeneous differential equation `dy/dx = (y^2 + x^2) / (2xy)`.

Correct Approach (Focus on Substitution and Separation):
1. The equation is homogeneous. We use `y = vx` and its correct derivative: `dy/dx = v + x(dv/dx)`.
2. Substitute these into the given differential equation:
  `v + x(dv/dx) = ((vx)^2 + x^2) / (2x(vx))`
  `v + x(dv/dx) = (v^2x^2 + x^2) / (2vx^2)`
  `v + x(dv/dx) = (x^2(v^2 + 1)) / (x^2(2v))` (Factor out `x^2`)
  `v + x(dv/dx) = (v^2 + 1) / (2v)`
3. Carefully separate the variables:
  `x(dv/dx) = (v^2 + 1) / (2v) - v`
  `x(dv/dx) = (v^2 + 1 - 2v^2) / (2v)`
  `x(dv/dx) = (1 - v^2) / (2v)`
The correctly separated form for integration is: `2v / (1 - v^2) dv = dx / x`.
From this point, both sides can be integrated to find the solution.

πŸ’‘ Prevention Tips:
  • Always explicitly write down the substitution `y = vx` and its derivative `dy/dx = v + x(dv/dx)` before proceeding, for every problem.
  • After performing substitution, critically check if the equation truly becomes variable separable (i.e., all `v` terms with `dv` on one side, and all `x` terms with `dx` on the other). If not, re-evaluate your derivative and algebraic steps.
  • For JEE Advanced, speed combined with accuracy is key. Practice diverse problems to solidify algebraic manipulation skills and quickly recognize correctly separable forms.
JEE_Advanced
Critical Calculation

❌ <span style='color: #FF0000;'>Incorrect Algebraic Manipulation During Variable Separation</span>

A common critical error in solving homogeneous differential equations is making algebraic mistakes during the separation of variables after the substitution `y=vx` (or `x=vy`). This often involves sign errors, incorrect handling of fractions, or improper rearrangement of terms when isolating `v` and `x` expressions. Such errors lead to an entirely wrong integral setup, making the final solution incorrect.
πŸ’­ Why This Happens:
This mistake primarily stems from a lack of carefulness in algebraic simplification and rearrangement. Students, under exam pressure or due to weak fundamental algebraic skills, tend to rush through steps involving combining fractions or distributing negative signs. This leads to an incorrect `f(v) - v` term, which subsequently results in an improper differential form for integration. For JEE Advanced, precision in these steps is paramount.
βœ… Correct Approach:
After substituting `y=vx` (and `dy/dx = v + x(dv/dx)`) into the homogeneous equation `dy/dx = f(y/x)`, you get `v + x(dv/dx) = f(v)`. The critical step is to accurately isolate `x(dv/dx) = f(v) - v` and then separate variables as `dv / (f(v) - v) = dx / x`. Each algebraic step, particularly when combining fractions for `f(v) - v`, must be executed with meticulous attention to signs and common denominators.
πŸ“ Examples:
❌ Wrong:
Consider the equation `dy/dx = (x+2y)/(2x-y)`. Following substitution, we get `v + x(dv/dx) = (1+2v)/(2-v)`.
`x(dv/dx) = (1+2v)/(2-v) - v = (1+2v - v(2-v))/(2-v)`.
A common error: incorrectly simplifying `1+2v - v(2-v)` to `1+2v - 2v - v^2 = 1-v^2`.
This leads to `x(dv/dx) = (1-v^2)/(2-v)`.
The incorrect separation would be `(2-v)/(1-v^2) dv = dx/x`.
βœ… Correct:
Continuing from the same equation `dy/dx = (x+2y)/(2x-y)`:
`x(dv/dx) = (1+2v - v(2-v))/(2-v) = (1+2v - 2v + v^2)/(2-v) = (1+v^2)/(2-v)`.
The correct separation is `(2-v)/(1+v^2) dv = dx/x`.
This correctly separates into `∫(2/(1+v^2) - v/(1+v^2)) dv = ∫dx/x`, leading to `2tan⁻¹(v) - (1/2)ln|1+v^2| = ln|x| + C`.
Then, substitute `v=y/x` back into the solution.
πŸ’‘ Prevention Tips:
  • Systematic Steps: Always perform algebraic rearrangements one step at a time, especially when combining fractions.
  • Sign Check: Double-check the sign of every term after distribution or combining, a major source of errors.
  • Reciprocal Awareness: Ensure the reciprocal is correctly taken when separating `dv/(f(v)-v)` from `dx/x`.
  • Practice: Regular practice with diverse homogeneous equations will build proficiency and reduce calculation errors. For CBSE, focus on fundamental accuracy; for JEE Advanced, speed combined with accuracy is key.
JEE_Advanced
Critical Conceptual

❌ <span style='color: #FF0000;'>Misidentification and Incorrect Substitution in Homogeneous Differential Equations</span>

Students critically err by misidentifying non-homogeneous differential equations as homogeneous, or by incorrectly applying the standard substitution (y=vx or x=vy). This leads to algebraic dead ends and an inability to reach a correct solution, stemming from a fundamental conceptual flaw in method applicability.
πŸ’­ Why This Happens:

  • Conceptual Gap: Weak understanding of homogeneous function definition (f(tx,ty) = f(x,y)).

  • Hasty Application: Direct substitution without verifying homogeneity.

  • Algebraic Errors: Incorrect derivative substitution (v + xdv/dx) or simplification.

βœ… Correct Approach:

  1. Verify Homogeneity: Always check if dy/dx = f(x,y) satisfies f(tx,ty) = f(x,y). (CBSE & JEE)

  2. Apply Substitution:

    • For dy/dx = f(y/x), use y = vx (dy/dx = v + xdv/dx).

    • For dx/dy = g(x/y), use x = vy (dx/dy = v + ydv/dy).



  3. Separate & Integrate: Transform to variable separable form; integrate.

  4. Back-Substitute: Replace 'v' with y/x (or x/y).

πŸ“ Examples:
❌ Wrong:

Equation: dy/dx = (x + y + 2) / (x + y - 1)


Mistake: Applying y = vx. This equation is NOT homogeneous due to constant terms. Direct substitution yields a non-separable form, highlighting a fundamental misidentification of the equation type. This error is critical.

βœ… Correct:

Solve: dy/dx = (x2 - y2) / (2xy)



  1. Verify: Homogeneous.

  2. Substitute: y = vx &Rightarrow v + x(dv/dx) = (1 - v2) / (2v).

  3. x(dv/dx) = (1 - 3v2) / (2v).

  4. Separate & Integrate: ∫ (2v / (1 - 3v2)) dv = ∫ (1/x) dx &Rightarrow -1/3 ln|1 - 3v2| = ln|x| + C.

  5. Back-Substitute: x2 - 3y2 = Cx3.

πŸ’‘ Prevention Tips:

  • Master Definitions: Thoroughly understand what makes a function/equation homogeneous. (JEE Tip: This is often tested via identification).

  • Systematic Approach: Always verify homogeneity *before* applying any substitution.

  • Practice Identification: Solve problems specifically focused on identifying correct differential equation types, including those reducible to homogeneous form.

JEE_Advanced
Critical Calculation

❌ Algebraic Simplification and Variable Separation Errors

Students frequently make critical algebraic errors when simplifying the expression `f(v) - v` after substituting `y=vx` and `dy/dx = v + x(dv/dx)`. These errors directly lead to an incorrect separable form, making the subsequent integration impossible or wrong, thus yielding an incorrect general solution.
πŸ’­ Why This Happens:
This mistake stems from a lack of careful algebraic manipulation, especially with fractions involving variables. Haste during calculations, inadequate practice in combining rational expressions, and overlooking common denominators contribute significantly. Sometimes, students also miss signs or exponents during this crucial step.
βœ… Correct Approach:
After substituting `y=vx` and `dy/dx = v + x(dv/dx)` into the homogeneous differential equation, isolate `x(dv/dx)` on one side. Then, carefully combine the terms on the other side, typically involving `f(v) - v`, by finding a common denominator and simplifying the numerator. This step must be performed meticulously to ensure correct variable separation.
πŸ“ Examples:
❌ Wrong:
Consider the equation: dy/dx = (xΒ² + yΒ²)/(xy)
Substitute `y=vx` and `dy/dx = v + x(dv/dx)`:
v + x(dv/dx) = (xΒ² + (vx)Β²)/(x(vx)) = (xΒ² + vΒ²xΒ²)/(vxΒ²) = (1 + vΒ²)/v
An incorrect simplification might be:
x(dv/dx) = (1 + vΒ²)/v - v = (1 + vΒ² - v)/v (Incorrectly assuming `v` is `v/1` and only subtracting `v` from the numerator instead of `vΒ²` after finding a common denominator).
βœ… Correct:
For the same equation: dy/dx = (xΒ² + yΒ²)/(xy)
Correct substitution and initial simplification as above:
v + x(dv/dx) = (1 + vΒ²)/v
The correct algebraic simplification for `x(dv/dx)` is:
x(dv/dx) = (1 + vΒ²)/v - v
To combine, `v` is `v/1`. The common denominator is `v`:
x(dv/dx) = (1 + vΒ²)/v - (v*v)/v = (1 + vΒ² - vΒ²)/v = 1/v
This leads to the correct separable form: v dv = dx/x, which integrates to vΒ²/2 = ln|x| + C.
πŸ’‘ Prevention Tips:
  • Practice Algebraic Simplification: Regularly practice combining fractions and simplifying rational expressions.
  • Step-by-Step Approach: Break down the `f(v) - v` simplification into smaller, manageable steps. Do not rush this part.
  • Check Common Denominators: Always ensure correct common denominators when combining terms like `(A/B) - C`. Remember `C = C/1`.
  • Self-Correction: After separating variables, quickly check if the resulting integrals look standard or solvable. An overly complicated integral might indicate an earlier algebraic error.
JEE_Main
Critical Formula

❌ Incorrect Substitution for dy/dx after setting y = vx

Students frequently make a critical error in understanding the differentiation step when solving homogeneous differential equations. After correctly identifying the equation as homogeneous and applying the standard substitution y = vx, they often incorrectly replace dy/dx with just dv/dx or v, completely neglecting the product rule of differentiation.
πŸ’­ Why This Happens:
This mistake stems from a fundamental misunderstanding of the chain rule and product rule. Students often perceive 'v' as a constant or overlook that 'v' itself is a function of 'x' (i.e., v = y/x). Rote memorization of the substitution without understanding its derivation leads to this conceptual flaw, especially under exam pressure.
βœ… Correct Approach:
When substituting y = vx into a homogeneous differential equation, it is crucial to understand that both 'v' and 'x' are variables, with 'v' being a function of 'x'. Therefore, to find dy/dx, the product rule of differentiation must be applied correctly: d(vx)/dx = v(dx/dx) + x(dv/dx), which simplifies to dy/dx = v + x(dv/dx). This is the correct formula to substitute.
πŸ“ Examples:
❌ Wrong:
If y = vx, then dy/dx = dv/dx (Incorrect) or dy/dx = v (Incorrect).
βœ… Correct:
Given y = vx, applying the product rule yields dy/dx = v + x(dv/dx). (Correct)
πŸ’‘ Prevention Tips:
  • Understand the Derivation: Always remember that v = y/x, making 'v' a function of 'x'.
  • Practice Product Rule: Regularly practice differentiation using the product rule to strengthen this fundamental skill.
  • Write It Down: Before substitution, explicitly write down y = vx and then dy/dx = v + x(dv/dx) to ensure correct application.
  • Verify Steps: After substitution, check if the equation transforms into a separable form in 'v' and 'x', which is the goal of this method.
JEE_Main
Critical Unit Conversion

❌ Misinterpreting the Role and Dimensionality of Arbitrary Constants in Mathematical Solutions

A common conceptual error, though not a direct 'unit conversion' mistake in the typical sense, arises when students try to apply physical unit reasoning to purely mathematical homogeneous differential equations. They may incorrectly assume that the arbitrary constant of integration (C) must always be dimensionless or have specific physical units, even when the variables 'x' and 'y' are abstract mathematical quantities without specified units. This misunderstanding can lead to confusion, especially when comparing different forms of the general solution or attempting to evaluate boundary conditions.
πŸ’­ Why This Happens:
This confusion often stems from an overgeneralization of principles from physics or chemistry, where units are paramount, to pure mathematics problems. In JEE Mathematics, variables in differential equations are usually abstract, and units are rarely specified. Students might attempt to 'balance units' in an equation where it's unnecessary, leading to conceptual roadblocks or incorrect manipulations. The 'dimension' of the constant 'C' is implicitly determined by the mathematical form of the solution, not by a separate unit conversion step.
βœ… Correct Approach:
In the context of JEE Mathematics, unless explicitly stated, the variables 'x' and 'y' in a homogeneous differential equation (e.g., `dy/dx = f(y/x)`) are abstract quantities. Consequently, the arbitrary constant 'C' is also an abstract real constant. Its 'dimensionality' (if one were to consider it) would simply be whatever is required to maintain mathematical consistency with the rest of the terms in the general solution. For instance, if the solution is `ln|y/x| = C`, then `C` is dimensionless. If the solution is `y = Cx`, then 'C' takes on the 'dimension' of `y/x`. There is typically no unit conversion step involved in solving these equations in JEE.
πŸ“ Examples:
❌ Wrong:
After correctly solving the homogeneous differential equation `(x^2 - y^2)dx + 2xydy = 0` to reach `ln|x| + (1/2)ln|1 + (y/x)^2| = C`, a student might worry that the left-hand side involves logarithms (which are dimensionless) and then mistakenly try to write `C` as `ln|K|` to 'match units', believing `C` *must* be dimensionless. They might waste time trying to justify this or get confused if the problem requires finding `C` from initial conditions where its value doesn't naturally fit a `ln` form.
βœ… Correct:
For the equation `(x^2 - y^2)dx + 2xydy = 0`, the general solution can be expressed as `x^2 + y^2 = Cx`. Here, the student simply understands `C` as an arbitrary real constant. If, hypothetically, `x` and `y` represented lengths (e.g., in meters), then `x^2` and `y^2` would be in `(meters)^2`. For the equation to be dimensionally consistent, `Cx` would also need to be in `(meters)^2`. This would imply that `C` has the 'dimension' of `meters` (since `C * meters = (meters)^2`). In a typical JEE problem, however, such physical interpretations are not required, and `C` is merely a number determined by initial conditions. The focus is on the mathematical steps, not unit analysis.
πŸ’‘ Prevention Tips:
  • JEE Specific & Critical: For homogeneous differential equations in JEE Mathematics, do not overthink units or attempt unit conversions. These concepts are generally not applicable to the mathematical solution process itself.
  • Treat 'x', 'y', and the integration constant 'C' as abstract mathematical variables/constants unless the problem explicitly introduces a physical context with units.
  • Focus purely on the algebraic manipulation, substitution (`y=vx`), separation of variables, and integration techniques.
  • Understand that the form of the arbitrary constant `C` (e.g., `C`, `ln|C|`, `tan⁻¹C`) is often chosen for algebraic convenience during integration, and its ultimate numerical value will be determined by given initial or boundary conditions.
JEE_Main
Critical Sign Error

❌ Sign Errors in Variable Separation or Algebraic Simplification

A frequent critical error in solving homogeneous differential equations (HDEs) is mismanaging signs during the substitution `y=vx` (or `x=vy`), especially when separating variables or simplifying algebraic expressions. This leads to incorrect integration terms and ultimately, a wrong solution. This is particularly prevalent when combining fractions or distributing negative signs.
πŸ’­ Why This Happens:
Students often rush through algebraic manipulations, particularly when terms involve multiple variables and negative signs. Common causes include:
  • Incorrectly distributing a negative sign across a binomial or trinomial.
  • Errors in combining fractions after substituting `dy/dx = v + x(dv/dx)`.
  • Misplacing a negative sign during variable separation, leading to an incorrect integral form (e.g., `dx/x` instead of `-dx/x`).
  • Lack of systematic checking of each step.
βœ… Correct Approach:
Always perform algebraic simplifications meticulously. After substituting `y=vx` and `dy/dx = v + x(dv/dx)`, carefully combine the terms on the RHS. When separating variables, ensure every sign is correctly transferred. For integration, double-check the integrand's sign. Remember, a single sign error can render the entire solution incorrect.
πŸ“ Examples:
❌ Wrong:
Consider the equation: `dy/dx = (x - y)/(x + y)`
Substituting `y = vx`, `dy/dx = v + x(dv/dx)`.
RHS becomes `(x - vx)/(x + vx) = (1 - v)/(1 + v)`.
So, `v + x(dv/dx) = (1 - v)/(1 + v)`
`x(dv/dx) = (1 - v)/(1 + v) - v`
Wrong Step: `x(dv/dx) = (1 - v - v(1 - v))/(1 + v) = (1 - v - v + v^2)/(1 + v) = (1 - 2v + v^2)/(1 + v)` (Incorrect sign in `v(1-v)`)
βœ… Correct:
Consider the equation: `dy/dx = (x - y)/(x + y)`
Substituting `y = vx`, `dy/dx = v + x(dv/dx)`.
RHS becomes `(x - vx)/(x + vx) = (1 - v)/(1 + v)`.
So, `v + x(dv/dx) = (1 - v)/(1 + v)`
`x(dv/dx) = (1 - v)/(1 + v) - v`
Correct Step: `x(dv/dx) = (1 - v - v(1 + v))/(1 + v)`
`x(dv/dx) = (1 - v - v - v^2)/(1 + v)`
`x(dv/dx) = (1 - 2v - v^2)/(1 + v)`
Separating variables: `(1 + v)/(1 - 2v - v^2) dv = dx/x`
Now integrate both sides carefully. Note that the derivative of `1 - 2v - v^2` is `-2 - 2v = -2(1+v)`, which simplifies the LHS integral.
πŸ’‘ Prevention Tips:
  • Double-Check Distribution: Always verify sign distribution when multiplying or opening parentheses, especially after terms like `-(a+b)` or `v(1+v)`.
  • Systematic Fraction Combination: When combining `A/B - C`, write it as `(A - CB)/B` explicitly. Do not skip intermediate steps.
  • Verify Variable Separation: Before integrating, physically trace the negative signs to ensure they are on the correct side of the equation and part of the correct differential.
  • Practice with Negative Coefficients: Solve ample problems where the homogeneous function naturally leads to negative terms, to build familiarity.
  • Review Basic Algebra: A strong foundation in algebraic manipulation is crucial for success in differential equations.
JEE_Main
Critical Approximation

❌ <span style='color: #FF0000;'>Incorrect Algebraic Simplification of <i>f(v) - v</i> or Unwarranted 'Approximation' of Terms</span>

After substituting y = vx and dy/dx = v + x(dv/dx) into a homogeneous differential equation dy/dx = f(y/x), students typically arrive at an expression for x(dv/dx) = f(v) - v. A critical mistake occurs when students make incorrect algebraic simplifications during this step. This can manifest as wrongly cancelling terms, incorrectly finding a common denominator, or, in some cases, erroneously treating certain terms as negligible (an 'approximation' not suitable for exact solutions), leading to a fundamentally altered and incorrect separable differential equation. This is a severe error as it changes the entire problem.
πŸ’­ Why This Happens:
  • Hasty Calculation: Students rush through the algebraic manipulation, leading to careless errors.
  • Weak Algebraic Foundation: Lack of proficiency in fractional arithmetic, polynomial expansion, and factorisation.
  • Misapplication of Concepts: Confusion with concepts like 'dominant terms' or 'ignoring small terms' from limits or Taylor series, which are not applicable for finding exact solutions of differential equations.
  • Attempting to Simplify Too Much: An overzealous attempt to simplify complex expressions can lead to invalid steps.
βœ… Correct Approach:
The correct approach involves meticulous and precise algebraic manipulation at every step. After substituting y = vx, simplify f(v) correctly. Then, when calculating f(v) - v, always take a proper common denominator for all terms, expand diligently, and combine like terms without any arbitrary cancellations or neglect of terms. Every part of the expression must be accounted for accurately.
πŸ“ Examples:
❌ Wrong:
Consider the homogeneous equation: dy/dx = (x^2 + y^2) / (xy)
Substituting y = vx, we get v + x(dv/dx) = (x^2 + (vx)^2) / (x(vx)) = (1 + v^2) / v.
Thus, x(dv/dx) = (1 + v^2) / v - v.
Wrong Step (Mistake): A student might incorrectly simplify this by cancelling v in the denominator with v^2 in the numerator, or by miscombining, leading to something like x(dv/dx) = 1 + v - v = 1. This is an invalid algebraic 'approximation' that results in v dv = dx/x becoming dv = dx/x, which is incorrect.
βœ… Correct:
Using the same equation: dy/dx = (x^2 + y^2) / (xy)
1. Substitute y = vx, dy/dx = v + x(dv/dx).
2. v + x(dv/dx) = (x^2 + (vx)^2) / (x(vx))
3. v + x(dv/dx) = (x^2(1 + v^2)) / (vx^2)
4. v + x(dv/dx) = (1 + v^2) / v
5. Correct Algebraic Step: Separate v from x(dv/dx):
x(dv/dx) = (1 + v^2) / v - v
x(dv/dx) = (1 + v^2 - v*v) / v
x(dv/dx) = (1 + v^2 - v^2) / v
x(dv/dx) = 1 / v
6. This correctly separates variables as v dv = dx/x, which upon integration gives v^2/2 = ln|x| + C.
7. Substituting back v = y/x, the solution is (y/x)^2 / 2 = ln|x| + C or y^2 = 2x^2(ln|x| + C).
πŸ’‘ Prevention Tips:
  • Master Algebraic Skills: Practice fractional arithmetic and polynomial manipulation rigorously.
  • Use Parentheses: Always use parentheses correctly, especially when subtracting expressions or dealing with complex fractions.
  • Common Denominators: Ensure all terms are combined using a correct common denominator before simplifying the numerator.
  • Step-by-Step Verification: After each algebraic step, pause and verify its correctness. Do not rush.
  • No Arbitrary Approximations: For exact solutions, do not ignore or approximate terms, however complex they seem. Every term contributes to the final solution.
  • CBSE vs JEE: While CBSE might be more forgiving of minor algebraic slips, JEE Main demands absolute precision. Critical algebraic errors will lead to completely incorrect answers.
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Solution of a homogeneous differential equation

Subject: Mathematics
Complexity: High
Syllabus: JEE_Main

Content Completeness: 22.2%

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