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Differential Equations · Question 13

JEE Main 2026 (Online) 28th January Morning ShiftMCQMedium📈 Topic in 71% papers+4−1
1

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

Answer

(4) 74

Solution

Divide by 2 y:

x 2 y dydx - 2 y = x3(2 - x3)

Put y = t 2 y dydx = dtdx

dtdx - 2xt = x2(2 - x3) (linear)

I.F. = e- 2xdx = 1x2

ddx ( tx2 ) = 2 - x3

tx2 = 2x - x44 + C

y(2) = 0 t = 0 at x = 2: 0 = 4 - 4 + C C = 0

y = x2 (2x - x44 )

At x = 1: (y(1)) = 2 - 14 = 74

Answer: (4)

JEE Main 2026 (Online) 28th January Morning Shift

Keep practising — Differential Equations

2

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. -
JEE Main 2026 (Online) 28th January Evening Shift
3

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 .

JEE Main 2026 (Online) 24th January Morning Shift
4

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
JEE Main 2026 (Online) 24th January Evening Shift
5

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
JEE Main 2026 (Online) 23rd January Morning Shift
6

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 .

JEE Main 2026 (Online) 23rd January Morning Shift
7

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

JEE Main 2026 (Online) 23rd January Evening Shift
8

Let the solution curve of the differential equation x dy - y dx = x2 + y2 dx, x > 0, y(1) = 0, be y = y(x). Then y(3) is equal to

  1. 6
  2. 4
  3. 1
  4. 2
JEE Main 2026 (Online) 22nd January Morning Shift
9

Let f : [1, ) R be a differentiable function. If 6 1x f(t) dt = 3xf(x) + x3 - 4 for all x 1, then the value of f(2) - f(3) is :

  1. 4
  2. 3
  3. -4
  4. -3
JEE Main 2026 (Online) 22nd January Morning Shift
10

If y = y(x) satisfies the differential equation 16 ( 9x + x x ) (4 + 9 + x ) y dy = (1 + 2 y) dx, x > 0 and y(256) = 2, y(49) = , then 2 is equal to :

  1. 2 2 - 1
  2. 2 - 1
  3. 2( 2 - 1)
  4. 3( 2 - 1)
JEE Main 2026 (Online) 22nd January Evening Shift
11

Let y = y(x) be the solution curve of the differential equation (1 + x2)dy + (y - -1 x)dx = 0, y(0) = 1. Then the value of y(1) is :

  1. 4e /4 - 2 - 1
  2. 2e /4 + 4 - 1
  3. 4e /4 + 2 - 1
  4. 2e /4 - 4 - 1
JEE Main 2026 (Online) 21st January Morning Shift

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