Welcome, aspiring engineers and mathematicians, to a deep dive into one of the most fundamental and frequently encountered types of differential equations: the
Linear First-Order Differential Equation! This is a core concept for both your board exams and, more importantly, for JEE Mains & Advanced. Understanding this will unlock a powerful tool for solving many real-world problems in physics, engineering, economics, and biology.
We will focus on the form:
$frac{dy}{dx} + P(x)y = Q(x)$
Let's break down what makes this equation "linear" and "first-order," and then we'll uncover a brilliant method to solve it.
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### Understanding the Structure: What is a Linear First-Order ODE?
First, let's dissect the terminology:
1.
First-Order: This means the highest derivative present in the equation is the first derivative (e.g., $frac{dy}{dx}$ or $frac{dx}{dy}$). There are no $frac{d^2y}{dx^2}$, $frac{d^3y}{dx^3}$, etc.
2.
Linear: This is a crucial property. An equation is linear if:
* The dependent variable (here, $y$) and its derivatives ($frac{dy}{dx}$) appear only in the first degree (i.e., no $y^2$, $(frac{dy}{dx})^3$, $y frac{dy}{dx}$, etc.).
* There are no products of the dependent variable with its derivatives.
* The coefficients of $y$ and $frac{dy}{dx}$ are either constants or functions of the independent variable ($x$ in this case).
So, for our specific form $frac{dy}{dx} + P(x)y = Q(x)$:
* $frac{dy}{dx}$ has a coefficient of 1 (implicitly).
* $y$ has a coefficient $P(x)$, which is a function of $x$ (or a constant).
* $Q(x)$ is also a function of $x$ (or a constant).
If any of these conditions are violated, the equation is
non-linear, and different, often more complex, methods are required to solve it.
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### The Challenge and the Intuition: Why a Special Method?
You might wonder, "Can't we just use variable separable methods?" Let's try:
$frac{dy}{dx} = Q(x) - P(x)y$
Here, the right-hand side is a function of both $x$ and $y$ that cannot, in general, be separated into a product of a function of $x$ only and a function of $y$ only. So, simple variable separation fails.
Our goal is to somehow transform the left-hand side ($frac{dy}{dx} + P(x)y$) into something that looks like the derivative of a product. Why a product? Because if we have $d/dx(f(x)y)$, we can then integrate both sides to find $y$. This is where the concept of an
Integrating Factor (IF) comes into play.
Imagine you have a complex expression, and you want to simplify it. Sometimes, multiplying by a cleverly chosen term can make it "integrable." This "cleverly chosen term" is our Integrating Factor.
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### Derivation of the Integrating Factor and the General Solution
Let's assume our linear differential equation is:
$frac{dy}{dx} + P(x)y = Q(x)$ (Equation 1)
We are looking for a function, let's call it $I(x)$, such that when we multiply Equation 1 by $I(x)$, the Left Hand Side (LHS) becomes the derivative of a product, specifically $d/dx [I(x)y]$.
Let's multiply Equation 1 by $I(x)$:
$I(x)frac{dy}{dx} + I(x)P(x)y = I(x)Q(x)$ (Equation 2)
Now, let's consider the derivative of the product $I(x)y$:
Using the product rule, $frac{d}{dx}(uv) = ufrac{dv}{dx} + vfrac{du}{dx}$:
$frac{d}{dx}[I(x)y] = I(x)frac{dy}{dx} + yfrac{dI}{dx}$ (Equation 3)
Our goal is for the LHS of Equation 2 to be identical to Equation 3.
Comparing $I(x)frac{dy}{dx} + I(x)P(x)y$ (LHS of Eq 2) with $I(x)frac{dy}{dx} + yfrac{dI}{dx}$ (RHS of Eq 3):
For these to be identical, the $y$ terms must match:
$I(x)P(x)y = yfrac{dI}{dx}$
Since $y$ is generally not zero, we can cancel $y$ from both sides:
$I(x)P(x) = frac{dI}{dx}$
This is a variable separable differential equation for $I(x)$!
$frac{dI}{I(x)} = P(x)dx$
Integrate both sides:
$int frac{dI}{I(x)} = int P(x)dx$
$ln|I(x)| = int P(x)dx + C_1$
To find $I(x)$, we exponentiate both sides. We can set $C_1=0$ because we only need *one* integrating factor, not the general form. Any non-zero constant multiple of an integrating factor will also work, but for simplicity, we take the one where $C_1=0$.
$|I(x)| = e^{int P(x)dx}$
Thus, the Integrating Factor (IF) is: $I(x) = e^{int P(x)dx}$
JEE Alert: Remember that the constant of integration for the `∫P(x)dx` part is usually ignored when calculating the IF, as it would just lead to a constant multiplier for the IF, which eventually cancels out.
Now that we have $I(x)$, let's go back to Equation 2:
$I(x)frac{dy}{dx} + I(x)P(x)y = I(x)Q(x)$
Since we chose $I(x)$ such that its LHS is $frac{d}{dx}[I(x)y]$, we can rewrite this as:
$frac{d}{dx}[I(x)y] = I(x)Q(x)$
Now, integrate both sides with respect to $x$:
$int frac{d}{dx}[I(x)y] dx = int I(x)Q(x) dx$
$I(x)y = int I(x)Q(x) dx + C$
Finally, substitute the expression for $I(x)$:
$y cdot e^{int P(x)dx} = int left(Q(x) cdot e^{int P(x)dx}
ight) dx + C$
This is the general solution to the linear differential equation $frac{dy}{dx} + P(x)y = Q(x)$.
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### Step-by-Step Methodology for Solving
Here's a systematic approach to tackle these problems:
1.
Standard Form Check: Ensure the equation is in the standard linear form: $frac{dy}{dx} + P(x)y = Q(x)$. If the coefficient of $frac{dy}{dx}$ is not 1, divide the entire equation by that coefficient.
2.
Identify P(x) and Q(x): Clearly identify $P(x)$ (the coefficient of $y$) and $Q(x)$ (the term independent of $y$ and $frac{dy}{dx}$). Pay attention to signs!
3.
Calculate the Integrating Factor (IF): Compute $I(x) = e^{int P(x)dx}$. Remember, don't include a constant of integration here.
4.
Apply the General Solution Formula: Substitute $I(x)$ and $Q(x)$ into the formula:
$y cdot I(x) = int Q(x) cdot I(x) dx + C$
5.
Perform the Integration: Evaluate the integral $int Q(x) cdot I(x) dx$. This is often the most challenging step and may require techniques like integration by parts, substitution, or partial fractions.
Crucially, don't forget the constant of integration $C$ at this step!
6.
Solve for y: Isolate $y$ to get the final general solution.
7.
Initial Value Problems (IVPs): If an initial condition (e.g., $y(x_0) = y_0$) is given, substitute $x_0$ and $y_0$ into the general solution to find the specific value of $C$.
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### Examples
Let's put this into practice with a few examples.
Example 1: Basic Application
Solve the differential equation: $frac{dy}{dx} + frac{1}{x}y = x^2$
Step 1: Standard Form Check
The equation is already in the standard form $frac{dy}{dx} + P(x)y = Q(x)$.
Step 2: Identify P(x) and Q(x)
Here, $P(x) = frac{1}{x}$ and $Q(x) = x^2$.
Step 3: Calculate the Integrating Factor (IF)
$IF = e^{int P(x)dx} = e^{int frac{1}{x}dx} = e^{ln|x|} = |x|$.
For most practical purposes, especially when working on intervals where $x > 0$, we take $IF = x$. (If $x<0$, $IF=-x$, but the result remains consistent).
Step 4: Apply the General Solution Formula
$y cdot IF = int Q(x) cdot IF , dx + C$
$y cdot x = int x^2 cdot x , dx + C$
$y cdot x = int x^3 , dx + C$
Step 5: Perform the Integration
$xy = frac{x^4}{4} + C$
Step 6: Solve for y
$y = frac{x^3}{4} + frac{C}{x}$
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Example 2: Rearrangement Required
Solve the differential equation: $(1+x^2)frac{dy}{dx} + 2xy = cos x$
Step 1: Standard Form Check
The coefficient of $frac{dy}{dx}$ is $(1+x^2)$, not 1. We must divide the entire equation by $(1+x^2)$:
$frac{dy}{dx} + frac{2x}{1+x^2}y = frac{cos x}{1+x^2}$
Step 2: Identify P(x) and Q(x)
$P(x) = frac{2x}{1+x^2}$ and $Q(x) = frac{cos x}{1+x^2}$.
Step 3: Calculate the Integrating Factor (IF)
$IF = e^{int P(x)dx} = e^{int frac{2x}{1+x^2}dx}$.
To integrate $frac{2x}{1+x^2}$, let $u = 1+x^2$, then $du = 2xdx$.
So, $int frac{2x}{1+x^2}dx = int frac{du}{u} = ln|u| = ln(1+x^2)$ (since $1+x^2$ is always positive).
Therefore, $IF = e^{ln(1+x^2)} = 1+x^2$.
Step 4: Apply the General Solution Formula
$y cdot IF = int Q(x) cdot IF , dx + C$
$y cdot (1+x^2) = int frac{cos x}{1+x^2} cdot (1+x^2) , dx + C$
$y(1+x^2) = int cos x , dx + C$
Step 5: Perform the Integration
$y(1+x^2) = sin x + C$
Step 6: Solve for y
$y = frac{sin x + C}{1+x^2}$
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Example 3: Initial Value Problem (JEE Level)
Solve: $xfrac{dy}{dx} + 2y = x^2 ln x$, given $y(1) = 0$.
Step 1: Standard Form Check
Divide by $x$: $frac{dy}{dx} + frac{2}{x}y = x ln x$
Step 2: Identify P(x) and Q(x)
$P(x) = frac{2}{x}$ and $Q(x) = x ln x$.
Step 3: Calculate the Integrating Factor (IF)
$IF = e^{int P(x)dx} = e^{int frac{2}{x}dx} = e^{2ln|x|} = e^{ln(x^2)} = x^2$.
Step 4: Apply the General Solution Formula
$y cdot IF = int Q(x) cdot IF , dx + C$
$y cdot x^2 = int (x ln x) cdot x^2 , dx + C$
$x^2 y = int x^3 ln x , dx + C$
Step 5: Perform the Integration
This integral requires
integration by parts ($int u dv = uv - int v du$).
Let $u = ln x$ and $dv = x^3 dx$.
Then $du = frac{1}{x} dx$ and $v = frac{x^4}{4}$.
$int x^3 ln x , dx = (ln x)frac{x^4}{4} - int frac{x^4}{4} cdot frac{1}{x} dx$
$= frac{x^4}{4}ln x - int frac{x^3}{4} dx$
$= frac{x^4}{4}ln x - frac{x^4}{16}$
So, $x^2 y = frac{x^4}{4}ln x - frac{x^4}{16} + C$.
Step 6: Solve for y
$y = frac{x^2}{4}ln x - frac{x^2}{16} + frac{C}{x^2}$. This is the general solution.
Step 7: Use Initial Condition to Find C
Given $y(1) = 0$. Substitute $x=1$ and $y=0$:
$0 = frac{1^2}{4}ln 1 - frac{1^2}{16} + frac{C}{1^2}$
$0 = frac{1}{4}(0) - frac{1}{16} + C$
$0 = -frac{1}{16} + C implies C = frac{1}{16}$.
The particular solution is:
$y = frac{x^2}{4}ln x - frac{x^2}{16} + frac{1}{16x^2}$
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### Special Case: Linear Differential Equation in x
Sometimes, you might encounter an equation that looks similar but is structured with $x$ as the dependent variable and $y$ as the independent variable.
The form is:
$frac{dx}{dy} + P(y)x = Q(y)$
The derivation is perfectly analogous.
1.
Identify P(y) and Q(y).
2.
Integrating Factor (IF): $I(y) = e^{int P(y)dy}$.
3.
General Solution: $x cdot I(y) = int Q(y) cdot I(y) dy + C$.
Example 4: Linear in x
Solve the differential equation: $(1+y^2)dx + (x - e^{arctan y})dy = 0$
Step 1: Standard Form Check
First, rearrange to get $frac{dx}{dy}$:
$(1+y^2)dx = -(x - e^{arctan y})dy$
$frac{dx}{dy} = frac{-(x - e^{arctan y})}{1+y^2}$
$frac{dx}{dy} = -frac{x}{1+y^2} + frac{e^{arctan y}}{1+y^2}$
Rearranging into the standard form $frac{dx}{dy} + P(y)x = Q(y)$:
$frac{dx}{dy} + frac{1}{1+y^2}x = frac{e^{arctan y}}{1+y^2}$
Step 2: Identify P(y) and Q(y)
$P(y) = frac{1}{1+y^2}$ and $Q(y) = frac{e^{arctan y}}{1+y^2}$.
Step 3: Calculate the Integrating Factor (IF)
$IF = e^{int P(y)dy} = e^{int frac{1}{1+y^2}dy} = e^{arctan y}$.
Step 4: Apply the General Solution Formula
$x cdot IF = int Q(y) cdot IF , dy + C$
$x cdot e^{arctan y} = int frac{e^{arctan y}}{1+y^2} cdot e^{arctan y} , dy + C$
$x e^{arctan y} = int frac{(e^{arctan y})^2}{1+y^2} , dy + C$
Step 5: Perform the Integration
Let $t = e^{arctan y}$. Then $dt = e^{arctan y} cdot frac{1}{1+y^2} dy$.
The integral becomes $int t , dt = frac{t^2}{2}$.
Substituting back:
$int frac{(e^{arctan y})^2}{1+y^2} , dy = int e^{arctan y} cdot left(frac{e^{arctan y}}{1+y^2}
ight) dy = int t , dt = frac{t^2}{2} = frac{(e^{arctan y})^2}{2}$.
So, $x e^{arctan y} = frac{(e^{arctan y})^2}{2} + C$.
Step 6: Solve for x
$x = frac{e^{arctan y}}{2} + C e^{-arctan y}$.
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###
JEE Focus: Common Pitfalls and Tips
*
Standard Form is King: Always ensure the equation is in the precise standard form ($frac{dy}{dx} + P(x)y = Q(x)$ or $frac{dx}{dy} + P(y)x = Q(y)$) before identifying $P$ and $Q$. A common mistake is forgetting to divide by the coefficient of the derivative.
*
Sign Errors: Be very careful with the sign of $P(x)$ or $P(y)$. For instance, if you have $frac{dy}{dx} - 2y = x$, then $P(x) = -2$, not $2$.
*
Integration Accuracy: The most frequent source of errors is incorrect integration, both for $int P(x)dx$ (to find IF) and for $int Q(x)IF dx$. Brush up on your integration techniques (substitution, by parts, trigonometric identities, partial fractions).
*
Constant of Integration: Don't forget the $+C$ when integrating $Q(x)IF dx$. This is crucial for general solutions. For initial value problems, use the given condition to find $C$.
*
Algebraic Simplification: After integration, simplify your expression for $y$ (or $x$) as much as possible. This often reveals a cleaner form and can prevent calculation errors when solving IVPs.
*
Implicit vs. Explicit: Sometimes, it might not be possible to explicitly solve for $y$ (or $x$). In such cases, the solution in the form $y cdot IF = int Q(x) cdot IF , dx + C$ is considered valid as an implicit solution.
*
Recognizing the Type: Always spend a moment to identify the type of differential equation. If it's not linear, trying to apply the integrating factor method will lead nowhere.
Understanding and mastering the method for solving linear first-order differential equations is a significant step in your differential equations journey. Practice these steps diligently with various examples, and you'll build strong confidence for your exams!