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Differential Equations · Question 11

JEE Main 2026 (Online) 2nd April Morning ShiftMCQEasy📈 Topic in 23% papers+4−1
1

Let y=y(x) be the solution curve of the differential equation (1+ x) dydx+(y+1) x=0, y(0)=0. If the curve y=y(x) passes through the point ( , -12 ), then a value of is :

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

Answer

(4) 2

Solution

(1+ x) dydx+(y+1) x= ddx [(1+ x)(y+1) ]=0

(1+ x)(y+1)=C

y(0)=0 (1)(1)=C C=1

Point ( ,- 12 ): (1+ ) 12=1

=1 = 2 (one value)

Answer: (4)

JEE Main 2026 (Online) 2nd April Morning Shift

Keep practising — Differential Equations

2

Let x=x(y) be the solution of the differential equation 2y2 dxdy-2xy+x2=0, y>1, x(e)=e. Then x (e2 ) is equal to:

  1. 32e2
  2. 23e2
  3. e2
  4. 2e2
JEE Main 2026 (Online) 2nd April Evening Shift
3

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
JEE Main 2026 (Online) 28th January Morning Shift
4

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
5

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
6

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
7

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
8

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
9

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
10

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
11

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

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