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Differential Equations

JEE Main 2025 (Online) 22nd January Evening ShiftNumericalMedium📈 Topic in 71% papers+4
Ques. 1JEE Main 2025, 22 January (Evening Shift)

Let y=f(x) be the solution of the differential equation dydx+ xyx2-1= x6+4x 1-x2, -1<x<1 such that f(0)=0. If 6 -1/21/2f(x) dx=2 - then 2 is equal to .

Answer

27

Solution

I.F. =e xx2-1dx= 1-x2

y 1-x2= x77+2x2+c

f(0)=0 c=0

On [- 12, 12 ] the odd part x77 1-x2 gives 0

6 -1/21/2 f dx=24 01/2 x2 1-x2dx

Put x= : =12 0 /6(1- 2 ) d

=12 ( 6- 34 )=2 -3 3

=3 3 2=27

Answer: 27

Keep practising — Differential Equations

Ques. 2JEE Main 2026, 8 April (Evening Shift)

Let y=y(x) be the solution of the differential equation x 1-x2 dy+ (y 1-x2-x -1x )dx=0, x (0, 1), x 1-y(x)=1. Then y ( 12 ) equals :

  1. 3- 3
  2. 4- 3
  3. 4- 2 3
  4. 3- 2 3
Ques. 3JEE Main 2026, 6 April (Morning Shift)

Let y=y(x) be the solution of the differential equation (x2-x x2-1 )dy + (y (x- x2-1 )-x )dx=0, x 1. If y(1)=1, then the greatest integer less than y ( 5 ) is .

Ques. 4JEE Main 2026, 6 April (Evening Shift)

Let f: R R be such that f(xy)=f(x)f(y), for all x, y R and f(0) 0. Let g: [1, ) R be a differentiable function such that x2g(x)= 1x (t2f(t)-tg(t) )dt. Then g(2) is equal to :

  1. 138
  2. 1116
  3. 1532
  4. 1764
Ques. 5JEE Main 2026, 5 April (Morning Shift)

Let y=y(x) be the solution of the differential equation x ( yx )dy= (y ( yx )-x )dx, y(1)= 2 and let = ( y (e12 )e12 ). Then the number of integral values of p, for which the equation x2+y2-2px+2py+ +2=0 represents a circle of radius r 6, is .

Ques. 6JEE Main 2026, 5 April (Evening Shift)

Let f(x) and g(x) be twice differentiable functions satisfying f''(x)=g''(x) for all x R, f'(1)=2g'(1)=4 and g(2)=3f(2)=9. Then f(25)-g(25) is equal to :

  1. 20
  2. 40
  3. -20
  4. -40
Ques. 7JEE Main 2026, 5 April (Evening Shift)

Let y=y(x) be the solution of the differential equation ( x)1/2 dy= ( 3x-( x)3/2y )dx, 0<x< 2, y ( 4 )= 6 25. If y ( 3 )= 45 , then 4 equals .

Ques. 8JEE Main 2026, 4 April (Morning Shift)

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 )
Ques. 9JEE Main 2026, 4 April (Evening Shift)

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
Ques. 10JEE Main 2026, 4 April (Evening Shift)

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 .

Ques. 11JEE Main 2026, 2 April (Morning Shift)

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

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