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Differential Equations

JEE Main 2026 (Online) 24th January Morning ShiftNumericalMedium📈 Topic in 33% papers+4
Ques. 1JEE Main 2026, 24 January (Morning Shift)

Let a differentiable function f satisfy the equation 036 f ( tx36 )dt = 4 f(x). If y = f(x) is a standard parabola passing through the points (2, 1) and (-4, ), then is equal to .

Answer

64

Solution

Put tx36 = u, so dt = 36xdu and the limits t: 0 36 become u: 0 x. The equation becomes

36x 0x f(u) du = 4 f(x)

9 0x f(u) du = x f(x)

Differentiating w.r.t. x:

9f(x) = f(x) + x f'(x)

f'(x)f(x) = 9 - 1x

f(x) = cx 9 -

Since it is a standard parabola y = cx2, 9 - = 2, so = 3.

It passes through (2, 1), so 1 = 4c, i.e. c = 14 and the parabola is x2 = 4y; through (-4, ) this gives = 164 = 4.

= 43 = 64

Answer: 64

Keep practising — Differential Equations

Ques. 2JEE 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. 3JEE 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. 4JEE Main 2026, 23 January (Morning Shift)

Let f be a twice differentiable non-negative function such that (f(x))2 = 25 + 0x ((f(t))2 + (f'(t))2 )dt. Then the mean of f( e(1)), f( e(2)), , f( e(625)) is equal to .

Ques. 5JEE Main 2026, 23 January (Evening Shift)

If the solution curve y = f(x) of the differential equation (x2 - 4)y' - 2xy + 2x(4 - x2)2 = 0, x > 2, passes through the point (3, 15), then the local maximum value of f is

Ques. 6JEE Main 2026, 22 January (Morning Shift)

Let the solution curve of the differential equation x dy - y dx = x2 + y2 dx, x > 0, y(1) = 0, be y = y(x). Then y(3) is equal to

  1. 6
  2. 4
  3. 1
  4. 2
Ques. 7JEE 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. 8JEE Main 2026, 22 January (Evening Shift)

If y = y(x) satisfies the differential equation 16 ( 9x + x x ) (4 + 9 + x ) y dy = (1 + 2 y) dx, x > 0 and y(256) = 2, y(49) = , then 2 is equal to :

  1. 2 2 - 1
  2. 2 - 1
  3. 2( 2 - 1)
  4. 3( 2 - 1)
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

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