Differential Equations

JEE Main 2026 (Online) 5th April Morning ShiftNumericalHard📈 Topic in 14% papers+4

Ques. 1JEE 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 .

Answer

Solution

The equation is homogeneous, so 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 means v = 2 at x = 1, so C = 0 and ( yx ) = e x. Hence

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

The circle is then x2 + y2 - 2px + 2py + 14 = 0, with centre (p, -p) and

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

For a real circle r2 > 0, so 2p2 - 14 > 0 p2 > 7; and r 6 gives 2p2 - 14 36 p2 25.

Hence 7 < p2 25, i.e. p = 3, 4, 5.

∴ Number of integral values of p = 6

Answer: 6

Keep practising — Differential Equations

Ques. 2JEE 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 :

Ques. 3JEE 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. 4JEE 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 :

3 ( 11 36 )
32 ( 11 312 )
3 ( 11 312 )
32 ( 11 36 )

Ques. 5JEE 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 :

138
613
1213
1312

Ques. 6JEE 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. 7JEE 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 :

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

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 :

Ques. 9JEE Main 2026, 2 April (Evening Shift)

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: