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Differential Equations

JEE Main 2026 (Online) 4th April Evening ShiftNumericalMedium📈 Topic in 67% papers+4
Ques. 1JEE Main 2026, 4 April (Evening Shift)

Let f be a twice differentiable function such that f(x)= 0x (t-x) dt- 0x f(t) t dt, x (- 2, 2 ). Then f'' ( 6 )+12f' (- 6 )+f ( 6 ) is equal to .

Answer

5

Solution

Putting u = t - x, 0x (t - x) dt = -x0 u du = [- e u ]-x0 = e x

so f(x) = e x - 0x f(t) t dt, and f(0) = 0. Differentiating,

f'(x) = - x - f(x) x

f'(x) + f(x) x = - x

This is linear with I.F. = e x dx = x, so

f(x) x = - x x dx = - x + C

Since f(0) = 0, 0 = -1 + C, i.e. C = 1. Hence f(x) = x - 1, f'(x) = - x and f''(x) = - x, and

f'' ( 6 ) + 12f' (- 6 ) + f ( 6 ) = - 32 + 12 12 + 32 - 1

∴ The value is 5

Answer: 5

Keep practising — Differential Equations

Ques. 2JEE Main 2026, 2 April (Morning Shift)

If the curve y=f(x) passes through the point (1, e) and satisfies the differential equation dy=y (2+ e x )dx, x>0, then f(e) is equal to :

  1. ee
  2. ee2
  3. e2e
  4. e2e
Ques. 3JEE Main 2026, 2 April (Morning Shift)

Let y=y(x) be the solution curve of the differential equation (1+ x) dydx+(y+1) x=0, y(0)=0. If the curve y=y(x) passes through the point ( , -12 ), then a value of is :

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

Let x=x(y) be the solution of the differential equation 2y2 dxdy-2xy+x2=0, y>1, x(e)=e. Then x (e2 ) is equal to:

  1. 32e2
  2. 23e2
  3. e2
  4. 2e2
Ques. 5JEE 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. 6JEE 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. 7JEE 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 .

Ques. 8JEE 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. 9JEE 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. 10JEE 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. 11JEE 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

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