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Differential Equations

JEE Main 2025 (Online) 23rd January Evening ShiftMCQHard📈 Topic in 14% papers+4−1
Ques. 1JEE 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

Answer

(4) 2 ( e 2 )2-1

Solution

y dy=(x dy-y dx) xy

dyy=- xy d(x/y) (as x dy-y dx=-y2 d(x/y))

Integrate: y= xy+C; x(1)= 2 C=0

At y=2: x(2)2= 2

x(2)=2 2 x(2)2-1=2( 2)2-1

Answer: (4)

Keep practising — Differential Equations

Ques. 2JEE 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
Ques. 3JEE Main 2025, 22 January (Evening Shift)

If x=f(y) is the solution of the differential equation (1+y2 )+ (x-2e -1y ) dydx=0, y (- 2, 2 ) with f(0)=1, then f ( 1 3 ) is equal to :

  1. e /4
  2. e /12
  3. e /6
  4. e /3
Ques. 4JEE Main 2025, 22 January (Evening Shift)

Let y=f(x) be the solution of the differential equation dydx+ xyx2-1= x6+4x 1-x2, -1<x<1 such that f(0)=0. If 6 -1/21/2f(x) dx=2 - then 2 is equal to .

Ques. 5JEE Main 2024, 9 April (Morning Shift)

The solution of the differential equation (x2 + y2 )dx - 5xy dy = 0, y(1) = 0, is :

  1. |x2 - 4y2 |5 = x2
  2. |x2 - 2y2 |6 = x
  3. |x2 - 2y2 |5 = x2
  4. |x2 - 4y2 |6 = x
Ques. 6JEE Main 2024, 9 April (Morning Shift)

The solution curve, of the differential equation 2y dydx + 3 = 5 dydx, passing through the point (0, 1) is a conic, whose vertex lies on the line :

  1. 2x + 3y = -9
  2. 2x + 3y = -6
  3. 2x + 3y = 9
  4. 2x + 3y = 6
Ques. 7JEE Main 2024, 9 April (Evening Shift)

Let 0x 1 - (y'(t))2 dt = 0x y(t) dt, 0 x 3, y 0, y(0) = 0. Then at x = 2, y'' + y + 1 is equal to

  1. 2
  2. 2
  3. 1/2
  4. 1
Ques. 8JEE Main 2024, 9 April (Evening Shift)

For a differentiable function f : R R, suppose f'(x) = 3f(x) + , where R, f(0) = 1 and x - f(x) = 7. Then 9f(- e 3) is equal to .

Ques. 9JEE Main 2024, 8 April (Morning Shift)

Let f(x) be a positive function such that the area bounded by y = f(x), y = 0 from x = 0 to x = a > 0 is e-a + 4a2 + a - 1. Then the differential equation, whose general solution is y = c1 f(x) + c2, where c1 and c2 are arbitrary constants, is

  1. (8ex + 1 ) d2ydx2 - dydx = 0
  2. (8ex + 1 ) d2ydx2 + dydx = 0
  3. (8ex - 1 ) d2ydx2 - dydx = 0
  4. (8ex - 1 ) d2ydx2 + dydx = 0
Ques. 10JEE Main 2024, 8 April (Morning Shift)

Let y = y(x) be the solution of the differential equation (1 + y2)e xdx + 2 x (1 + e2 x )dy = 0, y(0) = 1. Then y ( 4 ) is equal to

  1. 1e2
  2. 2e2
  3. 2e
  4. 1e
Ques. 11JEE Main 2024, 8 April (Evening Shift)

Let y = y(x) be the solution curve of the differential equation y dydx + 2x y = x3 y, y(1) = 0. Then y( 3) is equal to:

  1. 6
  2. 12
  3. 3
  4. 4

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