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Differential Equations

Previous year questions with solutions · 2024 – 2026

88Questions
58Papers covered
5.05%Weightage 2026same as last year
’24
’25
’26
Year-wise statistics questions · papers · average per paper · weightage
YearQuestionsPapersAvg / paperWeightage
202624191.35.05% same as last year
202524191.35.05% ▼ 24.2% vs last year
2024402026.67%

31 questions · newest paper first · page 1 of 2

Ques. 1JEE Main 2026, 8 April (Evening Shift)

Let y=y(x) be the solution of the differential equation x 1-x2 dy+ (y 1-x2-x -1x )dx=0, x (0, 1), x 1-y(x)=1. Then y ( 12 ) equals :

  1. 3- 3
  2. 4- 3
  3. 4- 2 3
  4. 3- 2 3
Ques. 2JEE Main 2026, 6 April (Evening Shift)

Let f: R R be such that f(xy)=f(x)f(y), for all x, y R and f(0) 0. Let g: [1, ) R be a differentiable function such that x2g(x)= 1x (t2f(t)-tg(t) )dt. Then g(2) is equal to :

  1. 138
  2. 1116
  3. 1532
  4. 1764
Ques. 3JEE Main 2026, 4 April (Evening Shift)

Let y=y(x) be the solution of the differential equation : dydx+ ( 6x2+ (3x2+2x3+4 )e-2x (x3+2 ) (2+e-2x ) )y=2+e-2x, x (-1, 2), satisfying y(0)= 32. If y(1)= (2+e-2 ), then is equal to :

  1. 138
  2. 613
  3. 1213
  4. 1312
Ques. 4JEE Main 2026, 28 January (Morning Shift)

Let y = y(x) be the solution of the differential equation x dydx - 2y = x3 (2 - x3 ) 2 y, x 0. If y(2) = 0, then (y(1)) is equal to

  1. - 74
  2. - 34
  3. 34
  4. 74
Ques. 5JEE Main 2026, 28 January (Evening Shift)

Let y = y(x) be the solution of the differential equation x dydx - y = x2 x, x (0, ). If y ( 2 ) = 2, then 6y ( 6 ) - 8y ( 4 ) is equal to :

  1. -3
  2. 3
  3. -
Ques. 6JEE Main 2026, 24 January (Evening Shift)

Let y = y(x) be a differentiable function in the interval (0, ) such that y(1) = 2, and t x ( t2y(x) - x2y(t)x - t ) = 3 for each x > 0. Then 2y(2) is equal to :

  1. 27
  2. 18
  3. 23
  4. 12
Ques. 7JEE Main 2026, 23 January (Morning Shift)

Let y = y(x) be the solution of the differential equation x4 dy + (4x3y + 2 x) dx = 0, x > 0, y ( 2 ) = 0. Then 4 y ( 3 ) is equal to :

  1. 92
  2. 72
  3. 64
  4. 81
Ques. 8JEE Main 2026, 22 January (Morning Shift)

Let f : [1, ) R be a differentiable function. If 6 1x f(t) dt = 3xf(x) + x3 - 4 for all x 1, then the value of f(2) - f(3) is :

  1. 4
  2. 3
  3. -4
  4. -3
Ques. 9JEE Main 2026, 21 January (Morning Shift)

Let y = y(x) be the solution curve of the differential equation (1 + x2)dy + (y - -1 x)dx = 0, y(0) = 1. Then the value of y(1) is :

  1. 4e /4 - 2 - 1
  2. 2e /4 + 4 - 1
  3. 4e /4 + 2 - 1
  4. 2e /4 - 4 - 1
Ques. 10JEE Main 2026, 21 January (Evening Shift)

Let y = y(x) be the solution of the differential equation x dydx - 2y = 2 + 3 x, x (- 2, 2 ), y(0) = - 74. Then y ( 6 ) is equal to :

  1. - 52
  2. -3 2 - 7
  3. - 54
  4. -3 3 - 7
Ques. 11JEE Main 2025, 8 April (Evening Shift)

Let f(x) = x - 1 and g(x) = ex for x R. If dydx = (e-2 x g(f(f(x))) - y x ), y(0) = 0, then y(1) is

  1. 1-e3e4
  2. e-1e4
  3. 1-e2e4
  4. 2e-1e3
Ques. 12JEE Main 2025, 7 April (Evening Shift)

Let y = y(x) be the solution of the differential equation (x2 + 1)y' - 2xy = (x4 + 2x2 + 1) x, y(0) = 1. Then -33 y(x) dx is :

  1. 36
  2. 24
  3. 18
  4. 30
Ques. 13JEE Main 2025, 4 April (Morning Shift)

Let f : [0, ) R be a differentiable function such that f(x) = 1 - 2x + 0x ex-t f(t) dt for all x [0, ). Then the area of the region bounded by y = f(x) and the coordinate axes is

  1. 5
  2. 2
  3. 2
  4. 12
Ques. 14JEE Main 2025, 4 April (Evening Shift)

If a curve y = y(x) passes through the point (1, 2 ) and satisfies the differential equation (7x4 y - ex cosec y ) dxdy = x5, x 1, then at x = 2, the value of y is :

  1. 2e2 + e64
  2. 2e2 - e64
  3. 2e2 - e128
  4. 2e2 + e128
Ques. 15JEE Main 2025, 3 April (Morning Shift)

Let g be a differentiable function such that 0x g(t) dt = x - 0x t g(t) dt, x 0 and let y = y(x) satisfy the differential equation dydx - y x = 2(x+1) x g(x), x [0, 2 ). If y(0) = 0, then y ( 3 ) is equal to

  1. 4 3
  2. 2 3
  3. 2 3 3
  4. 4 3 3
Ques. 16JEE Main 2025, 3 April (Evening Shift)

Let y = y(x) be the solution of the differential equation dydx + 3( 2 x)y + 3y = 2 x, y(0) = 13 + e3. Then y ( 4 ) is equal to :

  1. 43
  2. 23 + e3
  3. 43 + e3
  4. 23
Ques. 17JEE Main 2025, 2 April (Evening Shift)

Let f : [1, ) [2, ) be a differentiable function. If 10 1x f(t) dt = 5xf(x) - x5 - 9 for all x 1, then the value of f(3) is :

  1. 22
  2. 26
  3. 32
  4. 18
Ques. 18JEE Main 2025, 29 January (Morning Shift)

Let y = y(x) be the solution of the differential equation : x( e( x))2 dy + ( x - 3y x e( x))dx = 0, x (0, 2 ). If y ( 4 ) = - 1 e 2, then y ( 6 ) is equal to :

  1. 2 e(3) - e(4)
  2. - 1 e(4)
  3. 1 e(4) - e(3)
  4. 1 e(3) - e(4)
Ques. 19JEE Main 2025, 29 January (Evening Shift)

If for the solution curve y = f(x) of the differential equation dydx + ( x)y = 2 + x(1 + 2 x)2, x ( - 2, 2 ), f ( 3 ) = 310, then f ( 4 ) is equal to:

  1. 5 - 32 2
  2. 4 - 214
  3. 9 3 + 310(4 + 3)
  4. 3 + 110(4 + 3)
Ques. 20JEE Main 2025, 24 January (Morning Shift)

Let y=y(x) be the solution of the differential equation (xy-5x2 1+x2 )dx+ (1+x2 )dy=0, y(0)=0. Then y( 3) is equal to

  1. 5 32
  2. 152
  3. 143
  4. 2 2
Ques. 21JEE Main 2025, 24 January (Evening Shift)

Let f:(0, ) be a function which is differentiable at all points of its domain and satisfies the condition x2f'(x)=2xf(x)+3, with f(1)=4. Then 2f(2) is equal to :

  1. 19
  2. 23
  3. 29
  4. 39
Ques. 22JEE Main 2025, 22 January (Morning Shift)

Let x=x(y) be the solution of the differential equation y2 dx+ (x- 1y )dy=0. If x(1)=1, then x ( 12 ) is :

  1. 32+e
  2. 12+e
  3. 3+e
  4. 3-e
Ques. 23JEE Main 2025, 22 January (Evening Shift)

If x=f(y) is the solution of the differential equation (1+y2 )+ (x-2e -1y ) dydx=0, y (- 2, 2 ) with f(0)=1, then f ( 1 3 ) is equal to :

  1. e /4
  2. e /12
  3. e /6
  4. e /3
Ques. 24JEE Main 2024, 8 April (Evening Shift)

Let y = y(x) be the solution curve of the differential equation y dydx + 2x y = x3 y, y(1) = 0. Then y( 3) is equal to:

  1. 6
  2. 12
  3. 3
  4. 4
Ques. 25JEE Main 2024, 6 April (Morning Shift)

Let y = y(x) be the solution of the differential equation (2x e x) dydx + 2y = 3x e x, x > 0 and y (e-1 ) = 0. Then, y(e) is equal to

  1. - 3e
  2. - 32e
  3. - 23e
  4. - 2e
Ques. 26JEE Main 2024, 6 April (Morning Shift)

Let y = y(x) be the solution of the differential equation (1 + x2 ) dydx + y = e -1x, y(1) = 0. Then y(0) is

  1. 14 (e /2 - 1 )
  2. 12 (1 - e /2 )
  3. 14 (1 - e /2 )
  4. 12 (e /2 - 1 )
Ques. 27JEE Main 2024, 5 April (Morning Shift)

If y = y(x) is the solution of the differential equation dydx + 2y = (2x), y(0) = 34, then y ( 8 ) is equal to :

  1. e- /8
  2. e /4
  3. e- /4
  4. e /8
Ques. 28JEE Main 2024, 4 April (Evening Shift)

Let y = y(x) be the solution of the differential equation (x2 + 4)2dy + (2x3y + 8xy - 2)dx = 0. If y(0) = 0, then y(2) is equal to

  1. 2
  2. 8
  3. 16
  4. 32
Ques. 29JEE Main 2024, 1 February (Evening Shift)

Let be a non-zero real number. Suppose f: R R is a differentiable function such that f(0)=2 and x - f(x)=1. If f (x)= f(x)+3, for all x R, then f (- e 2 ) is equal to :

  1. 7
  2. 9
  3. 3
  4. 5
Ques. 30JEE Main 2024, 31 January (Morning Shift)

Let y=y(x) be the solution of the differential equation dydx= ( x)+y x( x- x x), x (0, 2 ) satisfying the condition y ( 4 )=2. Then, y ( 3 ) is

  1. 3 (2+ e 3 )
  2. 3 (1+2 e 3 )
  3. 3 (2+ e 3 )
  4. 32 (2+ e 3 )