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Differential Equations

JEE Main 2025 (Online) 29th January Morning ShiftNumericalMedium📈 Topic in 33% papers+4
Ques. 1JEE 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 .

Answer

112

Solution

Put x = t, so d = dtx and the given relation becomes

1x 0x f(t) dt = af(x) 0x f(t) dt = a x f(x)

Differentiating w.r.t. x:

f(x) = af(x) + a x f'(x) f'(x)f(x) = 1 - aa 1x

Integrating, f(x) = k x + c, where k = 1 - aa, so f(x) = Cxk.

f(1) = 1 gives C = 1, and f(16) = 18 gives

24k = 2-3 k = - 34 (then a = 4 0)

Hence f(x) = x-3/4 and f'(x) = - 34x-7/4, so

f' ( 116 ) = - 34 167/4 = - 34 27 = -96

16 - f' ( 116 ) = 16 + 96 = 112

Answer: 112

Keep practising — Differential Equations

Ques. 2JEE 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. 3JEE 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. 4JEE 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. 5JEE 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. 6JEE Main 2025, 24 January (Morning Shift)

Let f be a differentiable function such that 2(x+2)2f(x)-3(x+2)2=10 0x(t+2)f(t) dt, x 0. Then f(2) is equal to .

Ques. 7JEE 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. 8JEE Main 2025, 24 January (Evening Shift)

Let y=y(x) be the solution of the differential equation 2 x dydx= 2x-4y x, x (0, 2 ). If y ( 3 )=0, then y' ( 4 )+y ( 4 ) is equal to .

Ques. 9JEE Main 2025, 23 January (Morning Shift)

Let a curve y=f(x) pass through the points (0,5) and ( e 2,k). If the curve satisfies the differential equation 2(3+y)e2xdx-(7+e2x)dy=0, then k is equal to

  1. 32
  2. 8
  3. 4
  4. 16
Ques. 10JEE Main 2025, 23 January (Evening Shift)

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

  1. 2 ( e 2 )-1
  2. 1-2 ( e 2 )2
  3. 1-2 ( e 2 )
  4. 2 ( e 2 )2-1
Ques. 11JEE 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

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