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Differential Equations

JEE Main 2025 (Online) 3rd April Morning ShiftMCQMedium📈 Topic in 71% papers+4−1
Ques. 1JEE 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

Answer

(1) 4 3

Solution

Differentiating 0x g(t) dt = x - 0x t g(t) dt w.r.t. x:

g(x) = 1 - x g(x)

g(x) = 11+x

the differential equation becomes

dydx - y x = 2(x+1) x 11+x = 2 x

It is linear in y; I.F. = e- x dx = e x = x

ddx(y x) = 2 x x = 2

y x = 2x + c

y(0) = 0 c = 0

y = 2x x

y ( 3 ) = 2 3 2 = 4 3

Answer: (1)

Keep practising — Differential Equations

Ques. 2JEE 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. 3JEE Main 2025, 2 April (Morning Shift)

Let f : R R be a thrice differentiable odd function satisfying f'(x) 0, f''(x) = f(x), f(0) = 0, f'(0) = 3. Then 9f( e 3) is equal to .

Ques. 4JEE 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. 5JEE Main 2025, 2 April (Evening Shift)

Let y = y(x) be the solution of the differential equation dydx + 2y 2 x = 2 2 x + 3 x 2 x such that y(0) = 54. Then 12 (y ( 4 ) - e-2 ) is equal to .

Ques. 6JEE 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. 7JEE Main 2025, 29 January (Morning Shift)

Let f : (0, ) R be a twice differentiable function. If for some a 0, 01 f( x) d = af(x), f(1) = 1 and f(16) = 18, then 16 - f' ( 116 ) is equal to .

Ques. 8JEE 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. 9JEE Main 2025, 28 January (Morning Shift)

Let for some function y = f(x), 0x t f(t) dt = x2 f(x), x > 0 and f(2) = 3. Then f(6) is equal to

  1. 1
  2. 6
  3. 2
  4. 3
Ques. 10JEE Main 2025, 28 January (Evening Shift)

If y = y(x) is the solution of the differential equation, 4 - x2 dydx = ( ( -1 ( x2 ) )2 - y ) -1 ( x2 ), -2 x 2, y(2) = 2 - 84, then y2(0) is equal to .

Ques. 11JEE 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

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