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Differential Equations · Question 4

JEE Main 2026 (Online) 5th April Morning ShiftNumericalHard📈 Topic in 10% papers+4
1

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 .

Answer

11

Solution

Put y=vx, dy=v dx+x dv:

x v (v dx+x dv)=(vx v-x) dx

x2 v dv=-x dx v dv=- dxx

- v=- e x+C

y(1)= 2 v= 2 at x=1 C=0

( yx )= e x

= ( y (e12 )e12 )= e e12=12

Circle: x2+y2-2px+2py+14=0, centre (p, -p)

r2=p2+p2-14=2p2-14

r 6 2p2-14 36 p2 25 -5 p 5

Number of integral values of p=11

Answer: 11

JEE Main 2026 (Online) 5th April Morning Shift

Keep practising — Differential Equations

2

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
JEE Main 2026 (Online) 5th April Evening Shift
3

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 .

JEE Main 2026 (Online) 5th April Evening Shift
4

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 )
JEE Main 2026 (Online) 4th April Morning Shift
5

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
6

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
7

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
8

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
9

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
10

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
11

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

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