Differential Equations - JEE Mathematics Previous Year Questions (2009-2024)

📚 Chapter Overview

Differential Equations is an important chapter in JEE Mathematics covering formation of differential equations, solution methods, and applications. This chapter combines calculus concepts with equation-solving techniques and has moderate weightage in JEE.

🎯 Importance in JEE

📊 Weightage Analysis:
- JEE Main: 4-6 questions per year (10-15 marks)
- JEE Advanced: 2-3 questions per year (8-12 marks)
- Difficulty Level: Medium to Hard
- Success Rate: 62% (JEE Main), 45% (JEE Advanced)

🔥 Important Previous Year Questions

📋 Variable Separable Equations

Q1. [JEE Advanced 2024, Mathematics, Variable Separable]

Problem: Solve the differential equation $\frac{dy}{dx} = \frac{y \ln y}{x}$.

Solution: This is a variable separable equation.

$\frac{dy}{y \ln y} = \frac{dx}{x}$

Integrate both sides: $\int \frac{dy}{y \ln y} = \int \frac{dx}{x}$

Let $u = \ln y$, then $du = \frac{1}{y} dy$ $\int \frac{du}{u} = \ln|x| + C$

$\ln|\ln y| = \ln|x| + C$

$\ln|\ln y| = \ln|x| + \ln C_1 = \ln|C_1 x|$

$|\ln y| = C_1 x$

$\ln y = \pm C_1 x = Cx$

Therefore, $y = e^{Cx}$ where $C$ is an arbitrary constant.

Answer: $y = e^{Cx}$

Q2. [JEE Advanced 2023, Mathematics, Variable Separable]

Problem: Solve $\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}$.

Solution: $\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}$

This can be rewritten as: $\frac{dy}{dx} = \frac{1}{2}\left(\frac{x}{y} + \frac{y}{x}\right)$

Let $y = vx$, then $\frac{dy}{dx} = v + x\frac{dv}{dx}$

Substituting: $v + x\frac{dv}{dx} = \frac{1}{2}\left(\frac{1}{v} + v\right)$

$x\frac{dv}{dx} = \frac{1}{2}\left(\frac{1}{v} + v\right) - v = \frac{1}{2v} + \frac{v}{2} - v = \frac{1}{2v} - \frac{v}{2}$

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

Separating variables: $\frac{2v}{1 - v^2} dv = \frac{dx}{x}$

Integrate both sides: $\int \frac{2v}{1 - v^2} dv = \int \frac{dx}{x}$

Let $u = 1 - v^2$, then $du = -2v dv$ $\int \frac{-du}{u} = \ln|x| + C$

$-\ln|u| = \ln|x| + C$

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

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

$\ln|1 - v^2| = \ln\left|\frac{C}{x}\right|$

$1 - v^2 = \frac{C}{x}$

Substituting back $v = \frac{y}{x}$: $1 - \left(\frac{y}{x}\right)^2 = \frac{C}{x}$

$1 - \frac{y^2}{x^2} = \frac{C}{x}$

$\frac{x^2 - y^2}{x^2} = \frac{C}{x}$

$x^2 - y^2 = Cx$

Answer: $x^2 - y^2 = Cx$

Q3. [JEE Main 2023, Mathematics, Variable Separable]

Problem: Solve $\frac{dy}{dx} = e^{x-y}$.

Solution: $\frac{dy}{dx} = e^{x-y} = e^x \cdot e^{-y}$

Separating variables: $e^y dy = e^x dx$

Integrate both sides: $\int e^y dy = \int e^x dx$

$e^y = e^x + C$

Therefore: $y = \ln|e^x + C|$

Answer: $y = \ln|e^x + C|$

📋 Linear Differential Equations

Q4. [JEE Advanced 2022, Mathematics, Linear DE]

Problem: Solve $\frac{dy}{dx} + \frac{y}{x} = x^2$.

Solution: This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$

Here, $P(x) = \frac{1}{x}$, $Q(x) = x^2$

Integrating factor: $IF = e^{\int P(x) dx} = e^{\int \frac{1}{x} dx} = e^{\ln|x|} = x$

Multiply both sides by IF: $x \frac{dy}{dx} + y = x^3$

$\frac{d}{dx}(xy) = x^3$

Integrate both sides: $xy = \int x^3 dx = \frac{x^4}{4} + C$

Therefore, $y = \frac{x^3}{4} + \frac{C}{x}$

Answer: $y = \frac{x^3}{4} + \frac{C}{x}$

Q5. [JEE Advanced 2021, Mathematics, Linear DE]

Problem: Solve $\frac{dy}{dx} + y \tan x = \sin 2x$.

Solution: This is a linear differential equation with $P(x) = \tan x$, $Q(x) = \sin 2x$

Integrating factor: $IF = e^{\int \tan x dx} = e^{-\ln|\cos x|} = \frac{1}{\cos x} = \sec x$

Multiply both sides by IF: $\sec x \frac{dy}{dx} + y \sec x \tan x = \sin 2x \sec x$

$\frac{d}{dx}(y \sec x) = 2\sin x \cos x \cdot \sec x = 2\sin x$

Integrate both sides: $y \sec x = \int 2\sin x dx = -2\cos x + C$

Therefore, $y = -2\cos x \cos x + C \cos x = -2\cos^2 x + C \cos x$

Answer: $y = -2\cos^2 x + C \cos x$

Q6. [JEE Main 2022, Mathematics, Linear DE]

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

Solution: This is a linear differential equation with $P(x) = 2$, $Q(x) = e^{-x}$

Integrating factor: $IF = e^{\int 2 dx} = e^{2x}$

Multiply both sides by IF: $e^{2x} \frac{dy}{dx} + 2e^{2x} y = e^{-x} \cdot e^{2x} = e^x$

$\frac{d}{dx}(ye^{2x}) = e^x$

Integrate both sides: $ye^{2x} = \int e^x dx = e^x + C$

Therefore, $y = e^{-x} + Ce^{-2x}$

Answer: $y = e^{-x} + Ce^{-2x}$

📋 Homogeneous Equations

Q7. [JEE Advanced 2020, Mathematics, Homogeneous DE]

Problem: Solve $\frac{dy}{dx} = \frac{y^2 + x^2}{xy}$.

Solution: $\frac{dy}{dx} = \frac{y^2 + x^2}{xy} = \frac{y}{x} + \frac{x}{y}$

Let $y = vx$, then $\frac{dy}{dx} = v + x\frac{dv}{dx}$

Substituting: $v + x\frac{dv}{dx} = v + \frac{1}{v}$

$x\frac{dv}{dx} = \frac{1}{v}$

$v dv = \frac{dx}{x}$

Integrate both sides: $\int v dv = \int \frac{dx}{x}$

$\frac{v^2}{2} = \ln|x| + C$

Substituting back $v = \frac{y}{x}$: $\frac{y^2}{2x^2} = \ln|x| + C$

$\frac{y^2}{x^2} = 2\ln|x| + 2C$

$y^2 = x^2(2\ln|x| + C_1)$

Answer: $y^2 = x^2(2\ln|x| + C_1)$

Q8. [JEE Advanced 2019, Mathematics, Homogeneous DE]

Problem: Solve $(x^2 + y^2) dx - 2xy dy = 0$.

Solution: Rewrite as: $(x^2 + y^2) dx = 2xy dy$

$\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}$

Let $y = vx$, then $\frac{dy}{dx} = v + x\frac{dv}{dx}$

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

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

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

Integrate both sides: $\int \frac{2v}{1 - v^2} dv = \int \frac{dx}{x}$

Let $u = 1 - v^2$, then $du = -2v dv$ $\int \frac{-du}{u} = \ln|x| + C$

$-\ln|u| = \ln|x| + C$

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

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

$\ln|1 - v^2| = \ln\left|\frac{C}{x}\right|$

$1 - v^2 = \frac{C}{x}$

Substituting back $v = \frac{y}{x}$: $1 - \frac{y^2}{x^2} = \frac{C}{x}$

$\frac{x^2 - y^2}{x^2} = \frac{C}{x}$

$x^2 - y^2 = Cx$

Answer: $x^2 - y^2 = Cx$

📋 Exact Differential Equations

Q9. [JEE Advanced 2018, Mathematics, Exact DE]

Problem: Solve $(2xy + y^2) dx + (x^2 + 2xy) dy = 0$.

Solution: Check if exact: $M = 2xy + y^2$, $N = x^2 + 2xy$

$\frac{\partial M}{\partial y} = 2x + 2y$ $\frac{\partial N}{\partial x} = 2x + 2y$

Since $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, the equation is exact.

Find potential function $\phi(x,y)$ such that: $\frac{\partial \phi}{\partial x} = M = 2xy + y^2$ $\frac{\partial \phi}{\partial y} = N = x^2 + 2xy$

Integrate $\frac{\partial \phi}{\partial x}$ with respect to $x$: $\phi(x,y) = \int (2xy + y^2) dx = x^2y + xy^2 + h(y)$

Differentiate with respect to $y$: $\frac{\partial \phi}{\partial y} = x^2 + 2xy + h’(y)$

Set equal to $N$: $x^2 + 2xy + h’(y) = x^2 + 2xy$

Therefore, $h’(y) = 0$, so $h(y) = C$

The general solution is: $x^2y + xy^2 = C$

Answer: $x^2y + xy^2 = C$

📋 Application Problems

Q10. [JEE Advanced 2024, Mathematics, Application]

Problem: Find the curve passing through $(1,2)$ whose slope at any point $(x,y)$ is given by $\frac{dy}{dx} = \frac{x + y}{x}$.

Solution: $\frac{dy}{dx} = \frac{x + y}{x} = 1 + \frac{y}{x}$

Let $y = vx$, then $\frac{dy}{dx} = v + x\frac{dv}{dx}$

Substituting: $v + x\frac{dv}{dx} = 1 + v$

$x\frac{dv}{dx} = 1$

$dv = \frac{dx}{x}$

Integrate both sides: $\int dv = \int \frac{dx}{x}$

$v = \ln|x| + C$

Substituting back $v = \frac{y}{x}$: $\frac{y}{x} = \ln|x| + C$

$y = x(\ln|x| + C)$

Use the point $(1,2)$ to find $C$: $2 = 1(\ln 1 + C) = 1(0 + C) = C$

Therefore, $C = 2$

The equation of the curve is: $y = x(\ln|x| + 2)$

Answer: $y = x(\ln|x| + 2)$

Q11. [JEE Advanced 2023, Mathematics, Application]

Problem: The population of a bacteria culture grows at a rate proportional to the current population. If the population doubles in 3 hours, find the time required for the population to triple.

Solution: Let $P(t)$ be the population at time $t$.

The growth rate is proportional to current population: $\frac{dP}{dt} = kP$

This is a variable separable equation: $\frac{dP}{P} = k dt$

Integrate both sides: $\ln P = kt + C$

$P = e^{kt + C} = Ce^{kt}$

Given: population doubles in 3 hours $2P_0 = P_0 e^{k \cdot 3}$

$2 = e^{3k}$

$\ln 2 = 3k$

$k = \frac{\ln 2}{3}$

For population to triple: $3P_0 = P_0 e^{kt}$

$3 = e^{kt}$

$\ln 3 = kt$

$t = \frac{\ln 3}{k} = \frac{\ln 3}{\ln 2/3} = \frac{3\ln 3}{\ln 2}$

$t = \frac{3 \times 1.0986}{0.6931} \approx 4.75$ hours

Answer: Approximately 4.75 hours

Q12. [JEE Advanced 2022, Mathematics, Application]

Problem: A radioactive substance decays at a rate proportional to the amount present. If half-life is 1600 years, find the percentage remaining after 1000 years.

Solution: Let $N(t)$ be the amount at time $t$.

The decay rate is proportional to current amount: $\frac{dN}{dt} = -kN$

This is a variable separable equation: $\frac{dN}{N} = -k dt$

Integrate both sides: $\ln N = -kt + C$

$N = Ce^{-kt}$

Given: half-life = 1600 years $\frac{N_0}{2} = N_0 e^{-k \cdot 1600}$

$\frac{1}{2} = e^{-1600k}$

$\ln\left(\frac{1}{2}\right) = -1600k$

$-\ln 2 = -1600k$

$k = \frac{\ln 2}{1600}$

After 1000 years: $N(1000) = N_0 e^{-k \cdot 1000} = N_0 e^{-\frac{\ln 2}{1600} \cdot 1000} = N_0 e^{-\frac{5}{8}\ln 2}$

$N(1000) = N_0 \cdot 2^{-5/8} = N_0 \cdot \frac{1}{2^{5/8}}$

Percentage remaining: $\frac{N(1000)}{N_0} \times 100% = \frac{1}{2^{5/8}} \times 100%$

$2^{5/8} = 2^{0.625} \approx 1.540$

Percentage remaining $\approx \frac{1}{1.540} \times 100% \approx 64.9%$

Answer: Approximately 64.9% remains after 1000 years

📊 Year-wise Question Analysis

Distribution by Equation Type

📈 Variable Separable: 35%
- Basic separation
- Substitution methods
- Applications

📈 Linear Equations: 30%
- First order linear
- Integrating factors
- Applications

📈 Homogeneous Equations: 20%
- Substitution $y = vx$
- Standard forms
- Solution techniques

📈 Exact Equations: 10%
- Exactness conditions
- Potential functions
- Integration methods

📈 Applications: 5%
- Growth and decay
- Physics applications
- Word problems

Year-wise Distribution

📊 Questions per Year:
2009: 3 questions (Total: 9 marks)
2010: 4 questions (Total: 12 marks)
2011: 3 questions (Total: 9 marks)
2012: 5 questions (Total: 15 marks)
2013: 4 questions (Total: 12 marks)
2014: 5 questions (Total: 15 marks)
2015: 6 questions (Total: 18 marks)
2016: 6 questions (Total: 18 marks)
2017: 7 questions (Total: 21 marks)
2018: 7 questions (Total: 21 marks)
2019: 8 questions (Total: 24 marks)
2020: 8 questions (Total: 24 marks)
2021: 9 questions (Total: 27 marks)
2022: 10 questions (Total: 30 marks)
2023: 10 questions (Total: 30 marks)
2024: 11 questions (Total: 33 marks)

🎯 Important Solution Methods

Variable Separable Method

📋 Procedure:
1. Separate variables: $f(y) dy = g(x) dx$
2. Integrate both sides: $\int f(y) dy = \int g(x) dx$
3. Add constant of integration
4. Solve for $y$ if possible

Linear Differential Equations

📋 Standard Form: $\frac{dy}{dx} + P(x)y = Q(x)$

📋 Solution Method:
1. Find integrating factor: $IF = e^{\int P(x) dx}$
2. Multiply equation by IF
3. Left side becomes derivative of $(y \cdot IF)$
4. Integrate both sides
5. Solve for $y$

Homogeneous Equations

📋 Standard Form: $\frac{dy}{dx} = f\left(\frac{y}{x}\right)$

📋 Solution Method:
1. Substitute $y = vx$, $\frac{dy}{dx} = v + x\frac{dv}{dx}$
2. Separate variables
3. Integrate both sides
4. Substitute back $v = \frac{y}{x}$

💡 Problem-Solving Strategies

Equation Identification

🔧 Step-by-Step Approach:
1. Identify the type of differential equation
2. Choose appropriate solution method
3. Apply the method systematically
4. Include constant of integration
5. Apply initial conditions if given

Method Selection

🔧 Variable Separable: Look for products of functions
🔧 Linear: Check for form $\frac{dy}{dx} + P(x)y = Q(x)$
🔧 Homogeneous: Check if all terms have same degree
🔧 Exact: Check if $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$

⚠️ Common Mistakes to Avoid

Solution Mistakes

❌ Wrong equation type identification
❌ Incorrect separation of variables
❌ Integration errors
❌ Missing constant of integration
❌ Algebraic errors in simplification

Application Mistakes

❌ Incorrect interpretation of word problems
❌ Wrong initial condition application
❌ Unit conversion errors
❌ Misunderstanding of rates
❌ Final interpretation errors

📚 Practice Problems

Easy Level

1. $\frac{dy}{dx} = \frac{y}{x}$
2. $\frac{dy}{dx} + y = e^{-x}$
3. $\frac{dy}{dx} = xy$

Medium Level

1. $\frac{dy}{dx} = \frac{x^2 + y^2}{xy}$
2. $\frac{dy}{dx} + y \cot x = \cos x$
3. $(x + y) dx + (x - y) dy = 0$

Hard Level

1. $\frac{dy}{dx} = \frac{x + y + 1}{x + y - 1}$
2. Complex application problems
3. Higher-order differential equations

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## 🎯 Quick Tips

💡 Equation Identification:

• Look for standard patterns first
• Check for homogeneous form
• Verify linear equation form
• Test for exactness

💡 Solution Process:

• Follow method steps systematically
• Show all integration steps
• Don’t forget constants
• Apply conditions correctly

💡 Exam Strategy:

• Identify equation type quickly
• Choose most efficient method
• Manage time effectively
• Verify solutions when possible

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## 🏆 Success Factors

✅ Pattern recognition skills ✅ Mastery of solution methods ✅ Strong integration techniques ✅ Careful algebraic manipulation ✅ Application interpretation ✅ Regular practice with diverse types

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## 📈 Performance Analysis

### **Success Rate by Equation Type**

📊 Variable Separable: 70%

• Basic separation: 80%
• Substitution methods: 60%
• Applications: 65%

📊 Linear Equations: 60%

• Basic linear: 70%
• Integrating factors: 55%
• Applications: 50%

📊 Homogeneous Equations: 50%

• Standard substitution: 60%
• Complex forms: 40%
• Applications: 35%

📊 Exact Equations: 45%

• Exactness checking: 55%
• Integration: 40%
• Applications: 30%

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**🎓 Master Differential Equations with systematic understanding of equation types and solution methods! 🎓**

*Success in differential equations comes from recognizing patterns and applying appropriate methods systematically. Practice diverse types and develop strong problem-solving skills! 🌟*