f(tx, ty) = tn f(x, y)
Quick Check: Non-Homogeneous Function
M(x, y) dx + N(x, y) dy = 0
dy/dx = v + x (dv/dx)
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.
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:
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.
A first-order, first-degree differential equation is called homogeneous if it can be written in one of the following forms:
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.
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 $y$.
Let's consolidate the method into a clear, actionable set of steps:
$$frac{dv}{F(v) - v} = frac{dx}{x} quad ext{or} quad frac{dv}{G(v) - v} = frac{dy}{y}$$
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.
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:
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!
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:
Let's break down each part with its corresponding shortcut:
H - Homogeneous Check
V - V-Substitution
S - Separate Variables
I - Integrate
R - Replace V
Common Pitfalls to Avoid (Shortcuts for Error Prevention):
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!
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:
Mastering these steps ensures you can efficiently solve homogeneous differential equations in your exams!
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.
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':
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.dy/dx = F(x, y) can always be rewritten such that dy/dx depends *only* on the ratio y/x (or x/y).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.
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:
dy/dx: Differentiating y = vx with respect to x using the product rule, we get:dy/dx = v + x(dv/dx)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.F(x, y) = G(y/x) becomes G(v).v + x(dv/dx) = G(v)x(dv/dx) = G(v) - vdv / (G(v) - v) = dx / xv 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.
dy/dx can be expressed as a function of y/x.y = vx is the cornerstone of the solution method. Remember to also substitute for dy/dx.Understanding *why* the y = vx substitution works, by recognizing the inherent ratio-dependency of homogeneous equations, makes the solution process intuitive and memorable.
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:
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:
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.
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.
By systematically checking each step β identification, substitution, separation, integration, and back-substitution β you can minimize these common errors and secure full marks in differential equation problems. Practice makes perfect!
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:
Mastering these points will ensure you can confidently tackle homogeneous differential equations in your exams!
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.
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.
Solve the differential equation: $(x^2 + y^2)dx - 2xy dy = 0$
Mastering this methodical approach will significantly improve your speed and accuracy in solving homogeneous differential equations in exams.
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.
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.
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.
A first-order, first-degree differential equation is said to be homogeneous if it can be expressed in the form:
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 $f(tx,ty) = t^n f(x,y)$.
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:
While CBSE also covers homogeneous differential equations, JEE questions often present a higher level of complexity:
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.
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v = ln|x| + C, without replacing v with y/x.y/x = ln|x| + Cy = x(ln|x| + C)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. 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. 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).x=vy, then dx/dy = v + y(dv/dy). Always substitute this full expression into the differential equation. dy/dx = (x^2 + y^2) / (xy).y=vx, we get dy/dx = (x^2 + (vx)^2) / (x(vx)) = (x^2 + v^2x^2) / (vx^2) = (1 + v^2)/v.dv/dx = (1 + v^2)/vv' term from the left-hand side, leading to an incorrect differential equation.dy/dx = (x^2 + y^2) / (xy):y=vx, which simplifies the right-hand side to (1 + v^2)/v.dy/dx with its correct transformation: v + x(dv/dx).v + x(dv/dx) = (1 + v^2)/vx(dv/dx) = (1 + v^2)/v - v = (1 + v^2 - v^2)/v = 1/vv dv = dx/x.dy/dx = v + x(dv/dx); understand its derivation using the product rule.y=vx until it becomes second nature.dy/dx has been replaced by the complete expression v + x(dv/dx) (or dx/dy = v + y(dv/dy)).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.
v as v/1).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.
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) - vIncorrect Calculation:
x(dv/dx) = (v-1 - v)/(v+1) (Mistake: 'v' is incorrectly subtracted only from the numerator)x(dv/dx) = -1/(v+1)Continuing from x(dv/dx) = (v-1)/(v+1) - v:
Correct Calculation:
x(dv/dx) = (v-1)/(v+1) - v/1x(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)dy/dx = dv/dx, dy/dx = v, or miss one of the terms, leading to an incorrect transformed equation. d(uv)/dx = u(dv/dx) + v(du/dx)) yields: dy/dx = v + x(dv/dx). dx/dy = v + y(dv/dy). dy/dx = dv/dxdy/dx = vdy/dx = (d/dx(v)) * x + v * (d/dx(x)) = dv/dx * x + v * 1 = v + x(dv/dx).dy/dx or dx/dy after substitution.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.x, y) in the differential equation represent physical quantities (e.g., distance, time, concentration).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.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:
y(1 meter) = 2: 2 = 1 * (C + ln|1|) => C = 2. (Correctly determined C assuming x in meters)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.Using the same problem:
C: Using the initial condition y(1 meter) = 2 in y = x * (C + ln|x|): 2 = 1 * (C + ln|1|)=> C = 2.y = x * (2 + ln|x|), where x is assumed to be in meters for consistency with how C was derived.x to Consistent Units: To find y when x = 50 cm, convert x to meters: x = 50 cm = 0.5 meters.y = 0.5 * (2 + ln|0.5|). This yields the correct numerical result.∫ -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|.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|). dy/dx = (y^2 - x^2) / (2xy)y=vx and simplification, we get:x(dv/dx) = -(1+v^2)/(2v)2v/(1+v^2) dv = -dx/x∫ 2v/(1+v^2) dv = ∫ dx/x (missing the negative sign on the right side).ln(1+v^2) = ln|x| + C∫ 2v/(1+v^2) dv = ∫ -dx/xln(1+v^2) = -ln|x| + Cln(1+v^2) + ln|x| = Cln|x(1+v^2)| = Cx(1+v^2) = e^C = K (where K is a constant)v=y/x:x(1 + y^2/x^2) = Kx((x^2+y^2)/x^2) = K (x^2+y^2)/x = K or x^2+y^2 = Kx∫ 1/(a-bx) dx.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. y=vx substitution first and tend to stick to it.dy/dx or dx/dy) simplifies the expression and subsequent integration.dy/dx = f(y/x) and dx/dy = g(x/y) forms.f(v) or g(v) after substitution.dy/dx = f(y/x) is simpler, use y = vx and dy/dx = v + x(dv/dx).dx/dy = g(x/y) is simpler, use x = vy and dx/dy = v + y(dv/dy).(x2 + y2) dy = 2xy dx.y=vx:dy/dx = 2xy / (x2 + y2)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)β« (1+v2) / (v(1-v2)) dv = β« (1/x) dx.(x2 + y2) dy = 2xy dx.dx/dy = (x2 + y2) / (2xy).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)β« 2v / (1-v2) dv = β« (1/y) dy.u=1-v2, du=-2v dv), which is much simpler and faster to solve.dy/dx or dx/dy) simplifies the right-hand side of the equation the most.x=vy is the more efficient substitution, not just y=vx.dy/dx = v + x(dv/dx) for y=vx and dx/dy = v + y(dv/dy) for x=vy.y = vx (or x = vy) in a homogeneous differential equation. This can lead to an incorrectly simplified expression, making subsequent integration difficult or erroneous. 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. x(dv/dx) = v^2 + 1/v - v. A student might incorrectly simplify the RHS as:x(dv/dx) = (v^2 + 1 - v) / vx(dv/dx) = (v^2 + 1)/v - v = (v^2 + 1 - v*v)/v = (v^2 + 1 - v^2)/v = 1/vv dv = (1/x) dx1. 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.
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.
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.
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. 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)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)(v + 1)/(v^2 + 1) dv = - (1/x) dx (Ready for correct integration)x (dv/dx) (or its equivalent) is isolated correctly before simplifying.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. 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. 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. dy/dx = (xΒ² + yΒ²) / (xy).y=vx, the RHS becomes: (xΒ² + (vx)Β²) / (x(vx)) = (xΒ² + vΒ²xΒ²) / (vxΒ²).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.dy/dx = (xΒ² + yΒ²) / (xy).y=vx into the RHS: (xΒ² + (vx)Β²) / (x(vx))(xΒ² + vΒ²xΒ²) / (vxΒ²)xΒ² from the numerator: xΒ²(1 + vΒ²) / (vxΒ²)xΒ² terms completely: (1 + vΒ²) / v.v only (i.e., f(v)), demonstrating a complete and correct variable transformation.f(tx, ty) = f(x,y), which implies f(x,y) can be written as g(y/x).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.x or y variables, only v and constants. If any x or y terms remain, a mistake has occurred.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. v is a function of x (when y=vx) or y (when x=vy). It represents a fundamental calculus oversight. y = vx, differentiate both sides with respect to x: dy/dx = d/dx(vx) = (dv/dx)x + v(dx/dx) = v + x(dv/dx)x = vy, differentiate both sides with respect to y: dx/dy = d/dy(vy) = (dv/dy)y + v(dy/dy) = v + y(dv/dy)v is a variable function, not a constant, in these contexts. dy/dx = (x+y)/x.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.dy/dx = (x+y)/x.y = vx.dy/dx: dy/dx = d/dx(vx) = v + x(dv/dx)v + x(dv/dx) = (x + vx)/xv + x(dv/dx) = x(1+v)/xv + x(dv/dx) = 1 + vx(dv/dx) = 1dv = dx/x∫dv = ∫dx/xv = ln|x| + Cv = y/x:y/x = ln|x| + Cy = x(ln|x| + C)dy/dx.v's nature: Remember that v is a function of x (or y), so its derivative is not zero.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. 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. x dv/dx = v - (v^2 + 1)/v. x dv/dx = (v^2 - v^2 - 1)/v = -1/vβ« dv = β« (-1/v) dx/x -1/v with dx/x, leading to wrong integration variable.)x dv/dx = v - (v^2 + 1)/v:x dv/dx = (v^2 - (v^2 + 1))/v = (v^2 - v^2 - 1)/v = -1/vβ« v dv = β« -1/x dx v terms with dv, all x terms with dx, ready for integration.)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.
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.
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.
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.
To correctly identify, let's verify if dy/dx = (x^2 + y^2) / (xy) is homogeneous.
f(x,y) = (x^2 + y^2) / (xy).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.
f(tx, ty) = f(x,y) or if all terms have the same total degree.x or y) in the numerator or denominator immediately indicates a non-homogeneous equation.dy/dx = d(vx)/dx = v(dx/dx) + x(dv/dx) = v + x(dv/dx)dx/dy = d(vy)/dy = v(dy/dy) + y(dv/dy) = v + y(dv/dy)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.
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.
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. 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. 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). 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)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) y = vx explicitly until it becomes second nature.v is a dependent variable (a function of x), not a constant.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. 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.y/x = ln|x/x0| + C, then C must also be dimensionless.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.y/x = C + ln|x| where y and x have the same units, say meters):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.y/x = C + ln|x/x0|.C correctly represents a dimensionless constant.ln, exp, sin, etc.) to ensure they are dimensionless. Introduce reference units (e.g., x/x0) if necessary.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. 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. 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. ∫ dv / (1 - v) = ∫ dx / xln|1 - v| = ln|x| + C1/(1 - v) (where a=1, b=1, so -1/b = -1):-ln|1 - v| = ln|x| + Cln|1/(1 - v)| = ln|x| + C or ln|1/(1 - v)| - ln|x| = C.x or v) in the denominator when integrating 1/(a ± bx) forms.1/(a - bx), think 'minus b in the denominator, so minus 1/b outside'.u = a - bx, then du = -b dx, so dx = du / (-b) to reinforce the sign.dx/dy and using x = vy would be simpler or correct, leading to algebraic complexity and errors in JEE Advanced. 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)).y=vx blindly without considering the alternative x=vy or the structural form of the equation.dx/dy form when appropriate.dy/dx = F(y/x) or dx/dy = G(x/y).dy/dx = F(y/x), use y = vx. Then, differentiate to get dy/dx = v + x(dv/dx).dx/dy = G(x/y), use x = vy. Then, differentiate to get dx/dy = v + y(dv/dy).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.
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/v ⇒ y(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.
dy/dx = F(y/x) or dx/dy = G(x/y) before deciding on the substitution.y=vx for F(y/x) and x=vy for G(x/y). This choice dramatically simplifies the subsequent steps.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)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.∫ (1/(a-x)) dx, which should yield -ln|a-x|.(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. dy/dx = (y-x)/xy = vx, so dy/dx = v + x(dv/dx).v + x(dv/dx) = (vx - x)/x = v - 1x(dv/dx) = -1dv = -dx/x∫ dv = ∫ dx/x (missing the negative sign).v = ln|x| + Cy/x = ln|x| + Cy = x(ln|x| + C)dv = -dx/x∫ dv = ∫ (-1/x) dxv = -ln|x| + Cv = y/x:y/x = -ln|x| + Cy = x(C - ln|x|)∫ (1/(a-bx)) dx = (-1/b)ln|a-bx| + C.dy/dx = (x2 + y2) / (x2 - xy). After substituting y=vx, we get v + x dv/dx = (1+v2) / (1-v). (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.dy/dx = (x2 + y2) / (x2 - xy):y=vx and dy/dx = v + x dv/dx.v + x dv/dx = (x2 + (vx)2) / (x2 - x(vx)) = (x2(1+v2)) / (x2(1-v)) = (1+v2) / (1-v).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).∫(1-v)/(1 - v + 2v2) dv = ∫dx/x. No approximation is made here during the transformation or separation.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).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.v and x (or v and y).v with y/x (or x/y) to get the solution in terms of x and y.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.
For the DE: (x - y) dy/dx = x + y
Correct:
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).y = vx, so dy/dx = v + x(dv/dx).dy/dx:v + x(dv/dx) = (x + vx) / (x - vx) = (1 + v) / (1 - v)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)(1 - v) / (1 + v^2) dv = dx / xβ« (1 - v) / (1 + v^2) dv = β« dx / xβ« 1/(1+v^2) dv - β« v/(1+v^2) dv = β« dx/xtanβ»ΒΉ(v) - (1/2)ln|1+v^2| = ln|x| + Cv = y/x:tanβ»ΒΉ(y/x) - (1/2)ln|1+(y/x)Β²| = ln|x| + Cy = vx: dy/dx = v + x(dv/dx) (and similarly for x = vy). This is a frequent point of error.x and y, replacing v correctly. 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. y = vx and dy/dx = v + x(dv/dx).f(v)dv = g(x)dx, leading to haphazard rearrangement.Once the substitution y = vx (and thus dy/dx = v + x(dv/dx)) is made:
y with vx and dy/dx with v + x(dv/dx) in the original equation.x) to obtain an equation solely in terms of v and x.x(dv/dx): Move the v term to the right-hand side, typically by subtraction: x(dv/dx) = F(v) - v.F(v) - v into a single function of v.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.v with y/x to get the solution in terms of x and y.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.
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
β« [(v / (1 + v^2)) + (1 / (1 + v^2))] dv = β« -(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
f(v)dv = g(x)dx form. If not, re-evaluate.dy/dx = (y-x)/(y+x)v + x dv/dx = (vx-x)/(vx+x) = (v-1)/(v+1)x dv/dx = (v-1)/(v+1) - vx 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)(v+1)/(1+v^2) dv = dx/x (Negative sign is dropped)x dv/dx = -(1 + v^2)/(v+1)(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).∫ ((v)/(1+v^2) + (1)/(1+v^2)) dv = ∫ -1/x dx(1/2)ln|1+v^2| + tan&Inverse;v = -ln|x| + Cy = 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. y = vx and dy/dx = v + x(dv/dx). y with vx and dy/dx with v + x(dv/dx).x(dv/dx): Rearrange the equation to get x(dv/dx) on one side, typically by subtracting v from the right-hand side.v, combine the terms into a single fraction using a common denominator.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.v with y/x (or x/y) for the final solution.dy/dx = (x^2 + y^2) / (2xy).y = vx and dy/dx = v + x(dv/dx), we get:v + x(dv/dx) = (1 + v^2) / (2v)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)dv/dx, making subsequent integration impossible or incorrect.v + x(dv/dx) = (1 + v^2) / (2v)x(dv/dx):x(dv/dx) = (1 + v^2) / (2v) - vx(dv/dx) = (1 + v^2 - 2v*v) / (2v)x(dv/dx) = (1 + v^2 - 2v^2) / (2v)x(dv/dx) = (1 - v^2) / (2v)(2v / (1 - v^2)) dv = (1/x) dxx^2 - y^2 = Ax.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. v after substituting y=vx and dy=v dx + x dv.dx terms with x and dv terms with v, or failing to isolate the variables completely.1/x (forgetting ln|x| or absolute values), or mishandling complex integrands through substitution or partial fractions.+ C, which results in a particular solution instead of the general solution.y=vx and dy=v dx + x dv into the given homogeneous differential equation.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.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.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.(x^2 + y^2) dx - 2xy dy = 0.y=vx and simplification, we get (1 - v^2) dx = 2vx dv.dx = 2vx dv / (1 - v^2) (Variables not separated)∫ (2v / (1 - v^2)) dv as ln(1 - v^2) (missing negative sign and absolute value).(1 - v^2) dx = 2vx dv:dx/x = 2v dv / (1 - v^2)∫ dx/x = ∫ (2v / (1 - v^2)) dvu = 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)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.1/x, 1/(a^2-x^2), etc., at your fingertips. Pay attention to constants and absolute values.ln|C| or tanβ»ΒΉ(C) can simplify the final solution form.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 β«(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).
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.
β«dv = β«(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).
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:
β« (1-v)/v dv = β« 1/x dxNow, integrating the LHS, a common error is in the term `β« 1/v dv - β« dv` (assuming `(1-v)/v` is split as `1/v - 1`).
Or, if the separation was `β« v/(1-v) dv = β« 1/x dx`. A frequent error in integrating `β« v/(1-v) dv` is:
β« (-(1-v) + 1)/(1-v) dv = β« (-1 + 1/(1-v)) dv = -v + ln|1-v| + CThe bolded `ln|1-v|` is incorrect due to missing the negative coefficient from `(1-v)`.
Following the correct integration for `β« v/(1-v) dv` from the above scenario:
β« v/(1-v) dv = β« (-(1-v) + 1)/(1-v) dv = β« (-1 + 1/(1-v)) dv = -v + β« 1/(1-v) dvNow, integrate `1/(1-v)`. Here, `a = -1` (coefficient of `v`).
= -v + (1/-1)ln|1-v| + C = -v - ln|1-v| + CThis careful application of the integration formula avoids the sign error. Remember to substitute `v = y/x` back into the solution.
y = vx (or x = vy). However, a common mistake arises in two critical steps: 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).f(v) dv = g(x) dx, leading to difficulties in integration. This is crucial for CBSE exam success. dy/dx.v and x.dy/dx is given, substitute y = vx. If dx/dy is given, substitute x = vy.y = vx, then differentiate with respect to x using the product rule: dy/dx = v(1) + x(dv/dx).x = vy, then differentiate with respect to y: dx/dy = v(1) + y(dv/dy).y with vx (or x with vy) and dy/dx with v + x(dv/dx).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.dy/dx = (x+y)/x. A common wrong step:y=vx. Then dy/dx = v (incorrect!)v = (x+vx)/x = 1+v0 = 1, which is absurd, indicating a fundamental error in substitution.dy/dx = (x+y)/x:y = vx.dy/dx = v + x(dv/dx).v + x(dv/dx) = (x + vx)/xv + x(dv/dx) = x(1+v)/xv + x(dv/dx) = 1+vx(dv/dx) = 1+v - vx(dv/dx) = 1dv = dx/x (Variables are now correctly separated).β«dv = β«(1/x)dx => v = ln|x| + C.v = y/x: y/x = ln|x| + C, or y = x(ln|x| + C).y = vx to get dy/dx = v + x(dv/dx). Make it second nature.y, then substitution of dy/dx, then algebraic simplification, then variable separation.v and x for separation. A single sign error or misplaced term can derail the entire solution.Problem: Solve dy/dx = (x + y)/x
Student's Incorrect Steps:
Problem: Solve dy/dx = (x + y)/x
Correct Steps:
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. dy/dx = dv/dx instead of v + x(dv/dx)). A lack of practice with the separation of variables step also contributes to errors. dy/dx = f(x,y). Then, check if f(Ξ»x, Ξ»y) = f(x,y). If this condition is met, it's homogeneous.y = vx into the equation. Consequently, replace dy/dx with v + x(dv/dx) (using the product rule for differentiation).v and x. Rearrange to isolate all v terms with dv and all x terms with dx.v with y/x to express the general solution in terms of x and y.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.
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:β« (1/(1+v^2)) dv - β« (v/(1+v^2)) dv = β« (1/x) dxarctan(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
f(Ξ»x, Ξ»y) = f(x,y). This is a critical first step for JEE.y=vx and its derivative dy/dx = v + x(dv/dx). Understand the product rule application.x and y. Remember to replace v with y/x.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. ∫ dv/(1+v2) vs. ∫ dv/(v2-1).x(dv/dx) = G(v), ensuring all algebraic steps and signs are correct.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.∫ dv/G(v). Be vigilant about the form of the integrand and constants.v = y/x to express the solution in terms of x and y.dy/dx = (y2 - x2) / (2xy).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)x(dv/dx) = (1 + v2) / (2v), missing the crucial negative sign. This leads to:∫ (2v / (1 + v2)) dv = ∫ (1 / x) dxln|1 + v2| = ln|x| + ln|C|ln|1 + v2| = ln|Cx|1 + (y/x)2 = Cx(x2 + y2) / x2 = Cxx2 + y2 = Cx3 (This is incorrect due to the missing negative sign).x(dv/dx) = -(1 + v2) / (2v)∫ (2v / (1 + v2)) dv = ∫ (-1 / x) dxln|1 + v2| = -ln|x| + ln|C|ln|1 + v2| + ln|x| = ln|C|ln|x(1 + v2)| = ln|C|x(1 + (y/x)2) = Cx((x2 + y2) / x2) = C(x2 + y2) / x = Cx2 + y2 = Cx (This is the correct general solution).ln, tan-1, and partial fractions for more complex cases.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. dy/dx = f(x,y) is homogeneous if it looks 'similar' to previous examples.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. dy/dx = (x + y + 1) / (x + y)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)).v alone, indicating it's not homogeneous and the substitution is incorrect.dy/dx = (x + y + 1) / (x + y):F(x,y) = (x + y + 1) / (x + y).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)).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.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.F(Ξ»x, Ξ»y) = F(x,y) or if all terms in numerator/denominator have the same degree.+1, -2) in a linear combination of x and y often makes the equation non-homogeneous.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.
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.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.These mistakes typically stem from:
The correct approach involves a systematic application of the substitution and careful algebraic manipulation:
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.y with vx (or x with vy) and dy/dx (or dx/dy) with its derived form. Simplify the right-hand side.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.v with y/x (or x/y) to get the solution in terms of x and y.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 β« (1 + v) / (1 - v) dv = β« dx/x. This is fundamentally incorrect as the substitution for dy/dx was wrong.
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)
β« (1 + v) / (1 - 2v - v^2) dv = β« dx/x
This form is now correctly separated and ready for integration (e.g., by substitution for the denominator on the left side).
y = vx, then dy/dx = v + x dv/dx (using the product rule). Write it down explicitly in every problem.v are with dv and all terms involving x are with dx, and there are no mixed terms.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`).
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)`
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. y = vx with respect to x, students often forget to apply the product rule, treating v as a constant or differentiating only one term.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.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').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)dy/dx = v + x(dv/dx).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.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.
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.
y = vx until it's second nature.v is a function of x (v = y/x), not a constant, which necessitates the product rule.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.
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.)
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)
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. 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. 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)dx/dy = g(x/y), the substitution is x = vy, leading to dx/dy = v + y(dv/dy). y = vx. 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.y = vx.dy/dx = (d/dx)(vΒ·x)dy/dx = vΒ·(d/dx)(x) + xΒ·(d/dx)(v)dy/dx = vΒ·1 + xΒ·(dv/dx)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.dy/dx = v + x(dv/dx) is derived using the product rule.y = f(x)g(x) to reinforce the product rule.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.
dy/dx itself must be transformed using the product rule, not just y.v and x.To correctly solve a homogeneous differential equation (e.g., dy/dx = f(y/x)):
y = vx AND dy/dx = v + x(dv/dx) into the equation.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.v with y/x to express the solution in terms of x and y.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., β«dy = β«(1+v)dx), which is invalid since v is a function of x and not a constant, thus failing to achieve proper variable separation.
For the equation: dy/dx = (x+y)/x (which simplifies to dy/dx = 1 + y/x):
y = vx AND dy/dx = v + x(dv/dx).v + x(dv/dx) = 1 + vx(dv/dx) = 1dv = dx/xβ«dv = β«(1/x) dx ⇒ v = ln|x| + Cv=y/x: y/x = ln|x| + C ⇒ y = x(ln|x| + C).y = vx with its derivative dy/dx = v + x(dv/dx) and apply them simultaneously.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. 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. dy/dx = (x2 + y2) / (xy).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) / vx(dv/dx) = 1 + v2 (Students might incorrectly subtract v or misinterpret the RHS). This prevents separation of variables.v + x(dv/dx) = (1 + v2) / vx(dv/dx) = (1 + v2) / v - vx(dv/dx) = (1 + v2 - v2) / vx(dv/dx) = 1 / vv dv = dx / x, which can be easily integrated.x from both numerator and denominator to simplify f(y/x) to f(v).f(v), moving v, combining fractions, then separating variables.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. y/x to make the integration step seemingly easier, bypassing the correct (and often more complex) integration of the exact function.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. 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 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)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:
β« dx/x = β« 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)y=vx substitution method.sin(ΞΈ) β ΞΈ are valid only when ΞΈ is very small. In general DE problems, y/x is a variable and not necessarily small.(x^2 + y^2 + x)dx - 2xydy = 0(x^2 + y^2 + x)dx - 2xydy = 0:(x^2 + y^2 + x)dx - 2xydy = 0 is NOT homogeneous.(x^2 + y^2)dx - 2xydy = 0: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.
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.
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.
Solve: dy/dx = (x2 - y2) / (2xy)
dy/dx = (xΒ² + yΒ²)/(xy)v + x(dv/dx) = (xΒ² + (vx)Β²)/(x(vx)) = (xΒ² + vΒ²xΒ²)/(vxΒ²) = (1 + vΒ²)/vx(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).dy/dx = (xΒ² + yΒ²)/(xy)v + x(dv/dx) = (1 + vΒ²)/vx(dv/dx) = (1 + vΒ²)/v - vx(dv/dx) = (1 + vΒ²)/v - (v*v)/v = (1 + vΒ² - vΒ²)/v = 1/vv dv = dx/x, which integrates to vΒ²/2 = ln|x| + C.No summary available yet.
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