How to Solve Differential Equations (IIT JEE) – Complete Guide

How to Solve Differential Equations (IIT JEE) – Complete Guide

Introduction

"Mastering differential equations unlocks 10–15 marks in IIT JEE—enough to push you from a 90 to a 100+ percentile. These equations model everything from population growth to electrical circuits, and JEE loves testing them in disguised forms. Today, you’ll learn the exact steps to solve variable separable, linear, exact, homogeneous, and particular solution problems—no guesswork, just a repeatable system."

What You Need To Know First

1. Basic integration techniques (substitution, by-parts, partial fractions).
2. Differentiation rules (product rule, chain rule, implicit differentiation).
3. Algebraic manipulation (factoring, cross-multiplication, solving for constants).

If you’re shaky on any of these, pause and review them first.

Key Vocabulary

Formulas To Know

1. Variable Separable

Formula: [ \frac{dy}{dx} = f(x)g(y) ] Steps: 1. Rewrite as ( \frac{dy}{g(y)} = f(x) \, dx ). 2. Integrate both sides. 3. Solve for ( y ).

MEMORISE THIS (Not given on exam sheet).

2. Linear First-Order DE

Standard Form: [ \frac{dy}{dx} + P(x)y = Q(x) ] Integrating Factor (IF): [ \text{IF} = e^{\int P(x) \, dx} ] Solution: [ y \cdot \text{IF} = \int Q(x) \cdot \text{IF} \, dx + C ]

MEMORISE THIS (Integrating factor method is not on the exam sheet).

3. Exact DE

Condition for Exactness: [ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} ] where ( M(x,y) \, dx + N(x,y) \, dy = 0 ).

Solution Steps: 1. Find ( \psi(x,y) ) such that:
[ \frac{\partial \psi}{\partial x} = M ]
[ \frac{\partial \psi}{\partial y} = N ] 2. Integrate ( M ) w.r.t. ( x ) (treat ( y ) as constant). 3. Differentiate result w.r.t. ( y ) and equate to ( N ). 4. Solve for ( \psi(x,y) = C ).

MEMORISE THIS (Exactness condition is not on the exam sheet).

4. Homogeneous DE

Standard Form: [ \frac{dy}{dx} = f\left(\frac{y}{x}\right) ] Substitution: Let ( v = \frac{y}{x} ) → ( y = vx ) → ( \frac{dy}{dx} = v + x \frac{dv}{dx} ).

MEMORISE THIS (Substitution trick is not on the exam sheet).

5. Particular Solution

Given: General solution + initial condition (e.g., ( y(0) = 2 )). Steps: 1. Substitute initial condition into general solution. 2. Solve for ( C ). 3. Rewrite solution with ( C ) evaluated.

MEMORISE THIS (No formula, but critical for full marks).

Step-by-Step Method

Step 1: Identify the Type

• Variable Separable? → Can you write ( \frac{dy}{dx} = f(x)g(y) )?
• Linear? → Is it ( \frac{dy}{dx} + P(x)y = Q(x) )?
• Exact? → Check ( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} ).
• Homogeneous? → Can you write ( \frac{dy}{dx} = f\left(\frac{y}{x}\right) )?

Step 2: Apply the Correct Method

• Variable Separable: Separate variables → Integrate.
• Linear: Find integrating factor → Multiply → Integrate.
• Exact: Find ( \psi(x,y) ) → Set ( \psi = C ).
• Homogeneous: Substitute ( v = \frac{y}{x} ) → Solve.

Step 3: Solve for ( y )

• Simplify the integrated result.
• If initial conditions are given, find ( C ).

Step 4: Verify

• Differentiate your solution and check if it satisfies the original DE.

Worked Examples

Example 1 – Variable Separable (Basic)

Problem: Solve ( \frac{dy}{dx} = 2xy ), given ( y(0) = 1 ).

Solution: 1. Identify: ( \frac{dy}{dx} = 2x \cdot y ) → Variable separable. 2. Separate:
[ \frac{dy}{y} = 2x \, dx ] 3. Integrate:
[ \int \frac{dy}{y} = \int 2x \, dx ]
[ \ln|y| = x^2 + C ] 4. Solve for ( y ):
[ y = e^{x^2 + C} = e^C \cdot e^{x^2} ]
Let ( e^C = A ) → ( y = A e^{x^2} ). 5. Apply initial condition:
( y(0) = 1 ) → ( 1 = A e^0 ) → ( A = 1 ). 6. Final solution:
[ y = e^{x^2} ]

What we did and why: - Recognized separable form → Split variables → Integrated → Applied initial condition. - Key: Always check if variables can be separated before jumping to other methods.

Example 2 – Linear DE (Medium)

Problem: Solve ( \frac{dy}{dx} + 2y = e^{-x} ).

Solution: 1. Identify: Linear DE of form ( \frac{dy}{dx} + P(x)y = Q(x) ).
Here, ( P(x) = 2 ), ( Q(x) = e^{-x} ). 2. Find Integrating Factor (IF):
[ \text{IF} = e^{\int P(x) \, dx} = e^{\int 2 \, dx} = e^{2x} ] 3. Multiply through by IF:
[ e^{2x} \frac{dy}{dx} + 2e^{2x} y = e^{2x} \cdot e^{-x} ]
[ \frac{d}{dx} \left( y e^{2x} \right) = e^{x} ] 4. Integrate both sides:
[ y e^{2x} = \int e^{x} \, dx = e^{x} + C ] 5. Solve for ( y ):
[ y = e^{-2x} (e^{x} + C) = e^{-x} + C e^{-2x} ]

What we did and why: - Recognized linear form → Found IF → Multiplied → Integrated. - Key: The integrating factor must be applied to both sides of the equation.

Example 3 – Exam-Style (Homogeneous + Particular Solution)

Problem: Solve ( x \frac{dy}{dx} = y + \sqrt{x^2 + y^2} ), given ( y(1) = 0 ).

Solution: 1. Identify: Rewrite as ( \frac{dy}{dx} = \frac{y}{x} + \sqrt{1 + \left(\frac{y}{x}\right)^2} ) → Homogeneous. 2. Substitute ( v = \frac{y}{x} ):
( y = vx ) → ( \frac{dy}{dx} = v + x \frac{dv}{dx} ). 3. Substitute into DE:
[ v + x \frac{dv}{dx} = v + \sqrt{1 + v^2} ]
[ x \frac{dv}{dx} = \sqrt{1 + v^2} ] 4. Separate variables:
[ \frac{dv}{\sqrt{1 + v^2}} = \frac{dx}{x} ] 5. Integrate:
[ \int \frac{dv}{\sqrt{1 + v^2}} = \int \frac{dx}{x} ]
[ \ln|v + \sqrt{1 + v^2}| = \ln|x| + C ] 6. Simplify:
[ v + \sqrt{1 + v^2} = A x ]
(where ( A = e^C )) 7. Back-substitute ( v = \frac{y}{x} ):
[ \frac{y}{x} + \sqrt{1 + \left(\frac{y}{x}\right)^2} = A x ]
[ y + \sqrt{x^2 + y^2} = A x^2 ] 8. Apply initial condition ( y(1) = 0 ):
[ 0 + \sqrt{1 + 0} = A \cdot 1 ] → ( A = 1 ). 9. Final solution:
[ y + \sqrt{x^2 + y^2} = x^2 ]
(Solve for ( y ) if needed: ( y = \frac{x^2 - 1}{2} ))

What we did and why: - Recognized homogeneous form → Used substitution ( v = \frac{y}{x} ) → Separated → Integrated → Applied initial condition. - Key: Always simplify after substitution to avoid messy algebra.

Common Mistakes

Exam Traps

1-Minute Recap

"Listen up—this is your last-minute cheat sheet for differential equations in JEE:

1. Variable Separable? Split ( \frac{dy}{dx} = f(x)g(y) ) → Integrate both sides.
2. Linear? ( \frac{dy}{dx} + P(x)y = Q(x) ) → Find IF = ( e^{\int P \, dx} ) → Multiply → Integrate.
3. Exact? Check ( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} ) → Find ( \psi(x,y) = C ).
4. Homogeneous? Substitute ( v = \frac{y}{x} ) → Solve.
5. Particular Solution? Plug in initial conditions last to find ( C ).

Common traps? - Forgetting ( C ) → Instant zero marks. - Mixing up linear and separable → Waste time. - Ignoring exactness check → Solve the wrong way.

You’ve got this. Now go solve 3 problems tonight—no shortcuts, just the steps. Good luck!