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Differential Equations · Question 7

JEE Main 2026 (Online) 4th April Morning ShiftMCQMedium📈 Topic in 23% papers+4−1
1

Let y=y(x) be the solution of the differential equation dydx= (1+x+x2 ) (1-y+y2 ), y(0)= 12. Then (2y(1)-1) is equal to :

  1. 3 ( 11 36 )
  2. 32 ( 11 312 )
  3. 3 ( 11 312 )
  4. 32 ( 11 36 )

Answer

(3) 3 ( 11 312 )

Solution

Separating the variables: dyy2-y+1=(1+x+x2) dx

y2-y+1= (y- 12 )2+ ( 32 )2

2 3 -1 ( 2y-1 3 )=x+ x22+ x33+C

y(0)= 12 -10=0 C=0

At x=1: 2 3 -1 ( 2y(1)-1 3 )=1+ 12+ 13= 116

-1 ( 2y(1)-1 3 )= 11 312

2y(1)-1= 3 ( 11 312 )

Answer: (3)

JEE Main 2026 (Online) 4th April Morning Shift

Keep practising — Differential Equations

2

Let y=y(x) be the solution of the differential equation : dydx+ ( 6x2+ (3x2+2x3+4 )e-2x (x3+2 ) (2+e-2x ) )y=2+e-2x, x (-1, 2), satisfying y(0)= 32. If y(1)= (2+e-2 ), then is equal to :

  1. 138
  2. 613
  3. 1213
  4. 1312
JEE Main 2026 (Online) 4th April Evening Shift
3

Let f be a twice differentiable function such that f(x)= 0x (t-x) dt- 0x f(t) t dt, x (- 2, 2 ). Then f'' ( 6 )+12f' (- 6 )+f ( 6 ) is equal to .

JEE Main 2026 (Online) 4th April Evening Shift
4

If the curve y=f(x) passes through the point (1, e) and satisfies the differential equation dy=y (2+ e x )dx, x>0, then f(e) is equal to :

  1. ee
  2. ee2
  3. e2e
  4. e2e
JEE Main 2026 (Online) 2nd April Morning Shift
5

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
JEE Main 2026 (Online) 2nd April Morning Shift
6

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
7

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
8

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
9

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
10

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
11

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

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