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 .
Differential Equations
Previous year questions with solutions · 2024 – 2026
Year-wise statistics questions · papers · average per paper · weightage
| Year | Questions | Papers | Avg / paper | Weightage |
|---|---|---|---|---|
| 2026 | 24 | 19 | 1.3 | 5.05% same as last year |
| 2025 | 24 | 19 | 1.3 | 5.05% ▼ 24.2% vs last year |
| 2024 | 40 | 20 | 2 | 6.67% |
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 .
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 .
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
Let y = y(x) be the solution of the differential equation dydx + 2y 2 x = 2 2 x + 3 x 2 x such that y(0) = 54. Then 12 (y ( 4 ) - e-2 ) is equal to .
If y = y(x) is the solution of the differential equation, 4 - x2 dydx = ( ( -1 ( x2 ) )2 - y ) -1 ( x2 ), -2 x 2, y(2) = 2 - 84, then y2(0) is equal to .
Let f be a differentiable function such that 2(x+2)2f(x)-3(x+2)2=10 0x(t+2)f(t) dt, x 0. Then f(2) is equal to .
Let y=y(x) be the solution of the differential equation 2 x dydx= 2x-4y x, x (0, 2 ). If y ( 3 )=0, then y' ( 4 )+y ( 4 ) is equal to .
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 .
For a differentiable function f : R R, suppose f'(x) = 3f(x) + , where R, f(0) = 1 and x - f(x) = 7. Then 9f(- e 3) is equal to .
Let f be a differentiable function in the interval (0, ) such that f(1) = 1 and t x t2 f(x) - x2 f(t)t - x = 1 for each x > 0. Then 2f(2) + 3f(3) is equal to .
Let y = y(x) be the solution of the differential equation dydx + 2x (1 + x2 )2y = xe 1 (1 + x2 ); y(0) = 0. Then the area enclosed by the curve f(x) = y(x)e- 1 (1 + x2 ) and the line y - x = 4 is .
Let the solution y=y(x) of the differential equation dydx-y=1+4 x satisfy y( )=1. Then y ( 2 )+10 is equal to .
If x=x(t) is the solution of the differential equation (t+1)dx= (2x+(t+1)4 )dt, x(0)=2, then, x(1) equals .
Let y=y(x) be the solution of the differential equation (1-x2 )dy= [xy+ (x3+2 ) 3 (1-x2 ) ]dx, -1<x<1, y(0)=0. If y ( 12 )= mn, m and n are co-prime numbers, then m+n is equal to .