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Differential Equations · Question 16

JEE Main 2026 (Online) 24th January Evening ShiftMCQMedium📈 Topic in 71% papers+4−1
1

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

Answer

(3) 23

Solution

t2y(x) - x2y(t) = y(x)(t2 - x2) + x2 (y(x) - y(t) )

⇒ t2y(x) - x2y(t)x - t = -(t + x) y(x) + x2 y(x) - y(t)x - t

As t x: -2x y + x2 dydx = 3

⇒ dydx - 2xy = 3x2 (linear)

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

⇒ yx2 = 3x4dx = - 1x3 + C

y(1) = 2 ⇒ 2 = -1 + C ⇒ C = 3

⇒ y = 3x2 - 1x

∴ y(2) = 12 - 12 = 232 ⇒ 2y(2) = 23

Answer: (3)

JEE Main 2026 (Online) 24th January Evening Shift

Keep practising — Differential Equations

2

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
3

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
4

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
5

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
6

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
7

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
8

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
9

Let y = y(x) be the solution of the differential equation x dydx - 2y = 2 + 3 x, x (- 2, 2 ), y(0) = - 74. Then y ( 6 ) is equal to :

  1. - 52
  2. -3 2 - 7
  3. - 54
  4. -3 3 - 7
JEE Main 2026 (Online) 21st January Evening Shift
10

Let f(x) = x - 1 and g(x) = ex for x R. If dydx = (e-2 x g(f(f(x))) - y x ), y(0) = 0, then y(1) is

  1. 1-e3e4
  2. e-1e4
  3. 1-e2e4
  4. 2e-1e3
JEE Main 2025 (Online) 8th April Evening Shift
11

Let y = y(x) be the solution curve of the differential equation x (x2 + ex )dy + (ex(x - 2)y - x3 )dx = 0, x > 0, passing through the point (1, 0). Then y(2) is equal to :

  1. 22 + e2
  2. 44 - e2
  3. 44 + e2
  4. 22 - e2
JEE Main 2025 (Online) 7th April Morning Shift

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