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Differential Equations · Question 27

JEE Main 2025 (Online) 7th April Evening ShiftMCQMedium📈 Topic in 71% papers+4−1
1

Let y = y(x) be the solution of the differential equation (x2 + 1)y' - 2xy = (x4 + 2x2 + 1) x, y(0) = 1. Then -33 y(x) dx is :

  1. 36
  2. 24
  3. 18
  4. 30

Answer

(2) 24

Solution

Divide by (x2 + 1); note x4 + 2x2 + 1 = (x2 + 1)2:

dydx - 2xx2 + 1y = (x2 + 1) x

I.F. = e- 2xx2 + 1dx = e- (x2 + 1) = 1x2 + 1

yx2 + 1 = x dx = x + C

y(0) = 1 C = 1

y = (x2 + 1)( x + 1) = (x2 + 1) x + (x2 + 1)

(x2 + 1) x is odd, so its integral over [-3, 3] is zero:

-33 y dx = -33 (x2 + 1) dx = 2 [ x33 + x ]03

= 2(9 + 3) = 24

Answer: (2)

JEE Main 2025 (Online) 7th April Evening Shift

Keep practising — Differential Equations

2

Let f : [0, ) R be a differentiable function such that f(x) = 1 - 2x + 0x ex-t f(t) dt for all x [0, ). Then the area of the region bounded by y = f(x) and the coordinate axes is

  1. 5
  2. 2
  3. 2
  4. 12
JEE Main 2025 (Online) 4th April Morning Shift
3

If a curve y = y(x) passes through the point (1, 2 ) and satisfies the differential equation (7x4 y - ex cosec y ) dxdy = x5, x 1, then at x = 2, the value of y is :

  1. 2e2 + e64
  2. 2e2 - e64
  3. 2e2 - e128
  4. 2e2 + e128
JEE Main 2025 (Online) 4th April Evening Shift
4

Let g be a differentiable function such that 0x g(t) dt = x - 0x t g(t) dt, x 0 and let y = y(x) satisfy the differential equation dydx - y x = 2(x+1) x g(x), x [0, 2 ). If y(0) = 0, then y ( 3 ) is equal to

  1. 4 3
  2. 2 3
  3. 2 3 3
  4. 4 3 3
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5

Let y = y(x) be the solution of the differential equation dydx + 3( 2 x)y + 3y = 2 x, y(0) = 13 + e3. Then y ( 4 ) is equal to :

  1. 43
  2. 23 + e3
  3. 43 + e3
  4. 23
JEE Main 2025 (Online) 3rd April Evening Shift
6

Let f : R R be a thrice differentiable odd function satisfying f'(x) 0, f''(x) = f(x), f(0) = 0, f'(0) = 3. Then 9f( e 3) is equal to .

JEE Main 2025 (Online) 2nd April Morning Shift
7

Let f : [1, ) [2, ) be a differentiable function. If 10 1x f(t) dt = 5xf(x) - x5 - 9 for all x 1, then the value of f(3) is :

  1. 22
  2. 26
  3. 32
  4. 18
JEE Main 2025 (Online) 2nd April Evening Shift
8

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 .

JEE Main 2025 (Online) 2nd April Evening Shift
9

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)
JEE Main 2025 (Online) 29th January Morning Shift
10

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 .

JEE Main 2025 (Online) 29th January Morning Shift
11

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
JEE Main 2025 (Online) 28th January Morning Shift

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