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Differential Equations

JEE Main 2025 (Online) 2nd April Evening ShiftNumericalMedium📈 Topic in 67% papers+4
Ques. 1JEE 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 .

Answer

21

Solution

This is a linear equation with P = 2 2 x, so the integrating factor is e 2 2 x dx = e2 x and

y e2 x = (2 + 3 x) 2 x e2 x dx

Putting x = t, 2 x dx = dt:

y e2t = (2 + 3t)e2t dt = e2t + 3 ( t2 - 14 )e2t + C

y = 14 + 32 x + C e-2 x

Since y(0) = 54, 14 + C = 54, so C = 1 and

y ( 4 ) = 14 + 32 + e-2 = 74 + e-2

12 (y ( 4 ) - e-2 ) = 12 74 = 21

Answer: 21

Keep practising — Differential Equations

Ques. 2JEE 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. 3JEE 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. 4JEE 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. 5JEE 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. 6JEE 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. 7JEE 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. 8JEE 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. 9JEE 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. 10JEE 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. 11JEE 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

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