Differential Equations: JEE Main Previous Year Questions (2021-2026)

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JEE Main Previous Year Questions (2021-2026): 
Differential Equations  
  
 
(January 2026) 
 
Q1: Let y = y(x) be the solution of the differential equation 
then 6y(p/6) - 8y(p/4) is 
equal to : 
A: -3p 
B: 3p 
C: -p 
D: p 
Answer: C 
Explanation: 
The given equation is: 
 
To get it into the standard form dy/dx + P(x)y = Q(x), we divide the entire equation by x: 
 
 
Finding the Integrating Factor (I.F.) 
The integrating factor is calculated as: 
 
The solution to the differential equation is: 
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JEE Main Previous Year Questions (2021-2026): 
Differential Equations  
  
 
(January 2026) 
 
Q1: Let y = y(x) be the solution of the differential equation 
then 6y(p/6) - 8y(p/4) is 
equal to : 
A: -3p 
B: 3p 
C: -p 
D: p 
Answer: C 
Explanation: 
The given equation is: 
 
To get it into the standard form dy/dx + P(x)y = Q(x), we divide the entire equation by x: 
 
 
Finding the Integrating Factor (I.F.) 
The integrating factor is calculated as: 
 
The solution to the differential equation is: 
 
Applying the Initial Condition 
 
 
Q2: Let y = y(x) be the solution of the differential equation 
x dy/dx - sin 2y = x³(2 - x³) cos² y, x ? 0. 
If y(2) = 0, then tan(y(1)) is equal to 
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JEE Main Previous Year Questions (2021-2026): 
Differential Equations  
  
 
(January 2026) 
 
Q1: Let y = y(x) be the solution of the differential equation 
then 6y(p/6) - 8y(p/4) is 
equal to : 
A: -3p 
B: 3p 
C: -p 
D: p 
Answer: C 
Explanation: 
The given equation is: 
 
To get it into the standard form dy/dx + P(x)y = Q(x), we divide the entire equation by x: 
 
 
Finding the Integrating Factor (I.F.) 
The integrating factor is calculated as: 
 
The solution to the differential equation is: 
 
Applying the Initial Condition 
 
 
Q2: Let y = y(x) be the solution of the differential equation 
x dy/dx - sin 2y = x³(2 - x³) cos² y, x ? 0. 
If y(2) = 0, then tan(y(1)) is equal to 
A  
B  
C 3/4 
D 7/4 
Answer: D 
Explanation: 
Given equation is: 
 
Equation (1) becomes linear differential equation. 
 
 
Page 4


JEE Main Previous Year Questions (2021-2026): 
Differential Equations  
  
 
(January 2026) 
 
Q1: Let y = y(x) be the solution of the differential equation 
then 6y(p/6) - 8y(p/4) is 
equal to : 
A: -3p 
B: 3p 
C: -p 
D: p 
Answer: C 
Explanation: 
The given equation is: 
 
To get it into the standard form dy/dx + P(x)y = Q(x), we divide the entire equation by x: 
 
 
Finding the Integrating Factor (I.F.) 
The integrating factor is calculated as: 
 
The solution to the differential equation is: 
 
Applying the Initial Condition 
 
 
Q2: Let y = y(x) be the solution of the differential equation 
x dy/dx - sin 2y = x³(2 - x³) cos² y, x ? 0. 
If y(2) = 0, then tan(y(1)) is equal to 
A  
B  
C 3/4 
D 7/4 
Answer: D 
Explanation: 
Given equation is: 
 
Equation (1) becomes linear differential equation. 
 
 
Solution of linear differential equation: 
 
 
Q3: Let y = y(x) be the solution of the differential equation x 4 dy + (4x³y + 2 sin 
x)dx = 0, x > 0, y(p/2) = 0.  
Then p 4y(p/3) is equal to: 
A: 92 
B: 72 
C: 64 
D: 81 
Answer: D 
Explanation: 
x4dy + (4x³y + 2 sin x)dx = 0 
dy/dx + 4x³/x 4 y = -2 sin x/x 4 
dy/dx + 4/x y = -2 sin x/x 4 
Page 5


JEE Main Previous Year Questions (2021-2026): 
Differential Equations  
  
 
(January 2026) 
 
Q1: Let y = y(x) be the solution of the differential equation 
then 6y(p/6) - 8y(p/4) is 
equal to : 
A: -3p 
B: 3p 
C: -p 
D: p 
Answer: C 
Explanation: 
The given equation is: 
 
To get it into the standard form dy/dx + P(x)y = Q(x), we divide the entire equation by x: 
 
 
Finding the Integrating Factor (I.F.) 
The integrating factor is calculated as: 
 
The solution to the differential equation is: 
 
Applying the Initial Condition 
 
 
Q2: Let y = y(x) be the solution of the differential equation 
x dy/dx - sin 2y = x³(2 - x³) cos² y, x ? 0. 
If y(2) = 0, then tan(y(1)) is equal to 
A  
B  
C 3/4 
D 7/4 
Answer: D 
Explanation: 
Given equation is: 
 
Equation (1) becomes linear differential equation. 
 
 
Solution of linear differential equation: 
 
 
Q3: Let y = y(x) be the solution of the differential equation x 4 dy + (4x³y + 2 sin 
x)dx = 0, x > 0, y(p/2) = 0.  
Then p 4y(p/3) is equal to: 
A: 92 
B: 72 
C: 64 
D: 81 
Answer: D 
Explanation: 
x4dy + (4x³y + 2 sin x)dx = 0 
dy/dx + 4x³/x 4 y = -2 sin x/x 4 
dy/dx + 4/x y = -2 sin x/x 4 
 
 
Q4: If y = y(x) satisfies the differential equation  
cos y dy = (1 + 2 sin y)dx, x > 0 and y(256) = p/2, y(49) = a, then 2 sin a is equal to 
: 
A: 2v2 - 1 
B: v2 - 1 
C: 2(v2 - 1) 
D: 3(v2 - 1) 
Answer: A 
Explanation: