Linear Differential Equations
Introduction
A Linear Differential Equation of first order has a specific standard form and is solved using the powerful Integrating Factor (IF) method. This is one of the most important techniques in JEE differential equations.
Definition
A differential equation is linear if it can be written in the form:
$$\boxed{\frac{dy}{dx} + P(x) \cdot y = Q(x)}$$
where $P(x)$ and $Q(x)$ are functions of $x$ only (or constants).
Key Features:
- $y$ and $\frac{dy}{dx}$ appear to first power only
- No products like $y \cdot \frac{dy}{dx}$
- No powers like $y^2$ or $\left(\frac{dy}{dx}\right)^2$
Standard Form
Form 1: In $x$
$$\boxed{\frac{dy}{dx} + P(x)y = Q(x)}$$Form 2: In $y$ (when $x$ is dependent variable)
$$\boxed{\frac{dx}{dy} + P(y)x = Q(y)}$$Linear vs Non-Linear
Linear Examples
- $\frac{dy}{dx} + 3y = x^2$ ✓
- $\frac{dy}{dx} + xy = \sin x$ ✓
- $x\frac{dy}{dx} + y = x^3$ ✓ (can be written as $\frac{dy}{dx} + \frac{y}{x} = x^2$)
Non-Linear Examples
- $\frac{dy}{dx} + y^2 = x$ ✗ ($y^2$ appears)
- $y\frac{dy}{dx} + x = 0$ ✗ (product of $y$ and $\frac{dy}{dx}$)
- $\left(\frac{dy}{dx}\right)^2 + y = 0$ ✗ (square of derivative)
Integrating Factor Method
Definition of Integrating Factor
Integrating Factor (IF):
$$\boxed{IF = e^{\int P(x)dx}}$$
where $P(x)$ is the coefficient of $y$ in standard form.
Solution Formula
The general solution of $\frac{dy}{dx} + P(x)y = Q(x)$ is:
$$\boxed{y \cdot IF = \int Q(x) \cdot IF \, dx + C}$$
or equivalently:
$$\boxed{y \cdot e^{\int P dx} = \int Q \cdot e^{\int P dx} dx + C}$$
Step-by-Step Solution Method
Step 1: Write the DE in standard form $\frac{dy}{dx} + Py = Q$
Step 2: Identify $P(x)$ and $Q(x)$
Step 3: Find the Integrating Factor: $IF = e^{\int P(x)dx}$
- Don’t add constant in this integration!
Step 4: Multiply both sides of DE by IF
Step 5: Recognize LHS as $\frac{d}{dx}(y \cdot IF)$
Step 6: Integrate both sides
Step 7: Solve for $y$
Why Integrating Factor Works
Starting with: $\frac{dy}{dx} + Py = Q$
Multiply by $IF = e^{\int P dx}$:
$$e^{\int P dx} \frac{dy}{dx} + P e^{\int P dx} y = Q e^{\int P dx}$$Notice that:
$$\frac{d}{dx}\left(y \cdot e^{\int P dx}\right) = e^{\int P dx}\frac{dy}{dx} + y \cdot P e^{\int P dx}$$(by product rule)
So the LHS becomes:
$$\frac{d}{dx}(y \cdot IF) = Q \cdot IF$$Now integrate both sides!
Interactive Demo: Visualize Linear DE Solutions
Plot solution curves for first-order linear differential equations.
Memory Tricks
🎯 Standard Form Recognition
“dy/dx PLUS Py equals Q”:
- Must have $+$ sign (convert $-$ to $+$ if needed)
- $P$ multiplies $y$
- $Q$ is on right side
🎯 IF Formula Memory
“I F equals E to the Integral P”:
$$IF = e^{\int P \, dx}$$Mnemonic: “Integrate P to get IF”
🎯 Solution Pattern
“Y-IF equals Integral Q-IF”:
$$y \cdot IF = \int Q \cdot IF \, dx + C$$Remember:
- LHS: $y$ times IF
- RHS: Integral of ($Q$ times IF)
Common Mistakes to Avoid
❌ Mistake 1: Adding Constant in IF
Wrong: $IF = e^{\int P dx + C}$ ✗
Correct: $IF = e^{\int P dx}$ (NO constant here!) ✓
Why: The constant will cancel out anyway, and including it makes the solution complicated.
❌ Mistake 2: Wrong Sign in Standard Form
Wrong: $\frac{dy}{dx} - 2y = x$ → $P = 2$ ✗
Correct: Standard form needs $+$ sign
- Rewrite as: $\frac{dy}{dx} + (-2)y = x$
- So $P = -2$ ✓
❌ Mistake 3: Forgetting to Multiply Q by IF
Wrong: $y \cdot IF = \int Q \, dx$ ✗
Correct: $y \cdot IF = \int (Q \cdot IF) \, dx$ ✓
❌ Mistake 4: Treating Non-Linear as Linear
Wrong: $\frac{dy}{dx} + y^2 = x$ is linear ✗
Correct: This is Bernoulli equation (contains $y^2$), not linear ✓
Solved Examples
Example 1: Basic Linear DE (JEE Main)
Solve: $\frac{dy}{dx} + y = e^x$
Solution:
Step 1: Already in standard form: $\frac{dy}{dx} + 1 \cdot y = e^x$
Step 2: $P(x) = 1$, $Q(x) = e^x$
Step 3: $IF = e^{\int 1 \, dx} = e^x$
Step 4: Multiply by IF:
$$e^x \frac{dy}{dx} + e^x y = e^x \cdot e^x = e^{2x}$$Step 5: LHS = $\frac{d}{dx}(y \cdot e^x)$
Step 6: Integrate:
$$y \cdot e^x = \int e^{2x} dx = \frac{e^{2x}}{2} + C$$Step 7: Solve for $y$:
$$\boxed{y = \frac{e^x}{2} + Ce^{-x}}$$Example 2: With Variable Coefficient (JEE Main)
Solve: $\frac{dy}{dx} + y\tan x = \sec x$
Solution:
Standard form: $\frac{dy}{dx} + (\tan x)y = \sec x$
$P = \tan x$, $Q = \sec x$
$IF = e^{\int \tan x \, dx} = e^{\ln|\sec x|} = \sec x$
Multiply by IF:
$$\sec x \frac{dy}{dx} + \sec x \tan x \cdot y = \sec^2 x$$ $$\frac{d}{dx}(y \sec x) = \sec^2 x$$Integrate:
$$y \sec x = \tan x + C$$ $$\boxed{y = \sin x + C\cos x}$$Example 3: With Initial Condition (JEE Main)
Solve: $\frac{dy}{dx} + \frac{y}{x} = x^2$ with $y(1) = 1$
Solution:
$P = \frac{1}{x}$, $Q = x^2$
$IF = e^{\int \frac{1}{x}dx} = e^{\ln|x|} = x$ (taking $x > 0$)
Multiply by IF:
$$x\frac{dy}{dx} + y = x^3$$ $$\frac{d}{dx}(xy) = x^3$$Integrate:
$$xy = \frac{x^4}{4} + C$$ $$y = \frac{x^3}{4} + \frac{C}{x}$$Using $y(1) = 1$:
$$1 = \frac{1}{4} + C \Rightarrow C = \frac{3}{4}$$ $$\boxed{y = \frac{x^3}{4} + \frac{3}{4x}}$$Example 4: Requires Rearrangement (JEE Advanced)
Solve: $x\frac{dy}{dx} - y = x^2\sin x$
Solution:
Step 1: Convert to standard form:
$$\frac{dy}{dx} - \frac{y}{x} = x\sin x$$Rewrite with $+$ sign:
$$\frac{dy}{dx} + \left(-\frac{1}{x}\right)y = x\sin x$$$P = -\frac{1}{x}$, $Q = x\sin x$
$IF = e^{\int -\frac{1}{x}dx} = e^{-\ln|x|} = \frac{1}{x}$
Multiply by IF:
$$\frac{1}{x}\frac{dy}{dx} - \frac{y}{x^2} = \sin x$$ $$\frac{d}{dx}\left(\frac{y}{x}\right) = \sin x$$Integrate:
$$\frac{y}{x} = -\cos x + C$$ $$\boxed{y = -x\cos x + Cx}$$Example 5: Trigonometric Coefficients (JEE Advanced)
Solve: $\cos^2 x \frac{dy}{dx} + y = \tan x$
Solution:
Convert to standard form:
$$\frac{dy}{dx} + \frac{y}{\cos^2 x} = \frac{\tan x}{\cos^2 x}$$ $$\frac{dy}{dx} + y\sec^2 x = \tan x \sec^2 x$$$P = \sec^2 x$, $Q = \tan x \sec^2 x$
$IF = e^{\int \sec^2 x \, dx} = e^{\tan x}$
Multiply by IF:
$$e^{\tan x}\frac{dy}{dx} + y e^{\tan x}\sec^2 x = \tan x \sec^2 x \cdot e^{\tan x}$$ $$\frac{d}{dx}(y e^{\tan x}) = \tan x \sec^2 x \cdot e^{\tan x}$$For RHS, let $u = \tan x$, then $du = \sec^2 x \, dx$:
$$\int \tan x \sec^2 x \cdot e^{\tan x} dx = \int u e^u du$$Using integration by parts: $\int ue^u du = ue^u - e^u + C = e^u(u - 1) + C$
So:
$$y e^{\tan x} = e^{\tan x}(\tan x - 1) + C$$ $$\boxed{y = \tan x - 1 + Ce^{-\tan x}}$$Practice Problems
Level 1: JEE Main Basics
Problem 1.1: Solve $\frac{dy}{dx} + 2y = 4$.
Solution
$P = 2$, $Q = 4$
$IF = e^{\int 2 \, dx} = e^{2x}$
$y \cdot e^{2x} = \int 4e^{2x} dx = 2e^{2x} + C$
$y = 2 + Ce^{-2x}$
Problem 1.2: Solve $\frac{dy}{dx} - y = 1$.
Solution
$\frac{dy}{dx} + (-1)y = 1$
$P = -1$, $Q = 1$
$IF = e^{-x}$
$y \cdot e^{-x} = \int e^{-x} dx = -e^{-x} + C$
$y = -1 + Ce^x$
Problem 1.3: Solve $\frac{dy}{dx} + \frac{y}{x} = 1$ where $x > 0$.
Solution
$P = \frac{1}{x}$, $Q = 1$
$IF = e^{\ln x} = x$
$y \cdot x = \int x \, dx = \frac{x^2}{2} + C$
$y = \frac{x}{2} + \frac{C}{x}$
Level 2: JEE Main Advanced
Problem 2.1: Solve $\frac{dy}{dx} + y\cot x = 2\cos x$ with $y\left(\frac{\pi}{2}\right) = 0$.
Solution
$P = \cot x$, $Q = 2\cos x$
$IF = e^{\int \cot x \, dx} = e^{\ln|\sin x|} = \sin x$
$y\sin x = \int 2\cos x \sin x \, dx = \int \sin 2x \, dx = -\frac{\cos 2x}{2} + C$
Using $y(\pi/2) = 0$: $0 = -\frac{\cos \pi}{2} + C = \frac{1}{2} + C$, so $C = -\frac{1}{2}$
$y\sin x = -\frac{\cos 2x}{2} - \frac{1}{2}$
$y = -\frac{\cos 2x + 1}{2\sin x} = -\frac{2\cos^2 x}{2\sin x} = -\frac{\cos^2 x}{\sin x}$
Problem 2.2: Solve $(1 + x^2)\frac{dy}{dx} + 2xy = \frac{1}{1 + x^2}$.
Solution
$\frac{dy}{dx} + \frac{2x}{1 + x^2}y = \frac{1}{(1 + x^2)^2}$
$P = \frac{2x}{1 + x^2}$, $Q = \frac{1}{(1 + x^2)^2}$
$IF = e^{\int \frac{2x}{1+x^2}dx} = e^{\ln(1+x^2)} = 1 + x^2$
$y(1 + x^2) = \int \frac{1 + x^2}{(1 + x^2)^2} dx = \int \frac{1}{1 + x^2} dx = \tan^{-1}x + C$
$y = \frac{\tan^{-1}x + C}{1 + x^2}$
Problem 2.3: Solve $x\frac{dy}{dx} + 2y = x^2\log x$.
Solution
$\frac{dy}{dx} + \frac{2}{x}y = x\log x$
$P = \frac{2}{x}$, $Q = x\log x$
$IF = e^{\int \frac{2}{x}dx} = e^{2\ln x} = x^2$
$yx^2 = \int x^3\log x \, dx$
Using integration by parts: $\int x^3\log x \, dx = \frac{x^4\log x}{4} - \int \frac{x^4}{4} \cdot \frac{1}{x} dx = \frac{x^4\log x}{4} - \frac{x^4}{16} + C$
$y = \frac{x^2\log x}{4} - \frac{x^2}{16} + \frac{C}{x^2}$
Level 3: JEE Advanced
Problem 3.1: Solve $\frac{dy}{dx} = \frac{y\log y - y}{x}$.
Solution
Rewrite: $\frac{dy}{dx} = \frac{y(\log y - 1)}{x}$
This is NOT linear in $y$! It’s linear in $x$ as a function of $y$.
Write: $\frac{dx}{dy} = \frac{x}{y(\log y - 1)}$
$\frac{dx}{dy} - \frac{x}{y(\log y - 1)} = 0$
Actually, better approach: $\frac{dy}{dx} - \frac{y}{x} = -\frac{y\log y}{x}$
This is Bernoulli type. Let’s use different approach.
$x\frac{dy}{dx} = y\log y - y$
$x\frac{dy}{dx} - y\log y + y = 0$
$\frac{1}{y\log y - y}\frac{dy}{dx} = \frac{1}{x}$
Actually: $\frac{dx}{dy} = \frac{x}{y\log y - y}$
$\frac{dx}{dy} - \frac{x}{y(\log y - 1)} = 0$
Linear in $x$: $\frac{dx}{dy} + \left(-\frac{1}{y(\log y - 1)}\right)x = 0$
$P = -\frac{1}{y(\log y - 1)}$
$IF = e^{\int -\frac{1}{y(\log y-1)}dy}$
Let $u = \log y - 1$, $du = \frac{dy}{y}$:
$\int -\frac{1}{y(\log y-1)}dy = -\int \frac{du}{u} = -\ln|u| = -\ln|\log y - 1|$
$IF = e^{-\ln|\log y - 1|} = \frac{1}{\log y - 1}$
$x \cdot \frac{1}{\log y - 1} = C$
$x = C(\log y - 1)$
Problem 3.2: Solve $\frac{dy}{dx} + \frac{y}{x\log x} = \frac{1}{x}$.
Solution
$P = \frac{1}{x\log x}$, $Q = \frac{1}{x}$
$IF = e^{\int \frac{1}{x\log x}dx}$
Let $u = \log x$, $du = \frac{dx}{x}$:
$\int \frac{1}{x\log x}dx = \int \frac{du}{u} = \ln|u| = \ln|\log x|$
$IF = e^{\ln|\log x|} = \log x$ (assuming $x > 1$)
$y\log x = \int \frac{\log x}{x} dx$
Let $v = \log x$, $dv = \frac{dx}{x}$:
$\int \frac{\log x}{x}dx = \int v \, dv = \frac{v^2}{2} = \frac{(\log x)^2}{2} + C$
$y = \frac{\log x}{2} + \frac{C}{\log x}$
Problem 3.3: A curve passes through $(1, 1)$ and at any point $(x, y)$ on it, the slope of tangent is $\frac{y}{x} + \sec\left(\frac{y}{x}\right)$. Find the curve.
Solution
$\frac{dy}{dx} = \frac{y}{x} + \sec\left(\frac{y}{x}\right)$
This is homogeneous! Not linear.
Use $y = vx$: $v + x\frac{dv}{dx} = v + \sec v$
$x\frac{dv}{dx} = \sec v$
$\cos v \, dv = \frac{dx}{x}$
$\sin v = \ln|x| + C$
Using $(1, 1)$: $\sin(1) = 0 + C$, so $C = \sin 1$
$\sin\left(\frac{y}{x}\right) = \ln|x| + \sin 1$
When to Use Linear DE Method
Use this method when:
- DE can be written as $\frac{dy}{dx} + Py = Q$
- $y$ and $\frac{dy}{dx}$ appear to first power only
- No products of $y$ and derivatives
- $P$ and $Q$ are functions of $x$ only
Comparison with Other Methods
| Method | When to Use | Form |
|---|---|---|
| Variable Separable | Can separate $x$ and $y$ | $\frac{dy}{dx} = f(x)g(y)$ |
| Homogeneous | Function of $\frac{y}{x}$ | $\frac{dy}{dx} = F(y/x)$ |
| Linear | First power of $y$ and $y'$ | $\frac{dy}{dx} + Py = Q$ |
Cross-References
- Variable Separable: Alternative method → Variable Separable
- Homogeneous DE: Another first-order method → Homogeneous DE
- Order and Degree: Classification of DEs → Order and Degree
- Formation of DE: Creating DEs from equations → Formation of DE
- Applications: Real-world uses → Applications of DE
- Integration: Required for solving → Indefinite Integrals
- Integration Techniques: For complex integrals → Integration Techniques
- Inverse Trigonometry: For inverse trig solutions → Inverse Trig
Quick Revision Points
- Standard form: $\frac{dy}{dx} + Py = Q$
- IF = $e^{\int P \, dx}$ (NO constant here!)
- Solution: $y \cdot IF = \int Q \cdot IF \, dx + C$
- LHS becomes $\frac{d}{dx}(y \cdot IF)$
- Linear means first power of $y$ and $y'$ only
- Remember to multiply $Q$ by IF before integrating
Last updated: November 18, 2025