Let x=x(y) be the solution of the differential equation y2 dx+ (x- 1y )dy=0. If x(1)=1, then x ( 12 ) is :
(4) 3-e
With y as the independent variable the equation is linear: dxdy+ xy2= 1y3, I.F. =e-1/y.
x e-1/y= e-1/yy3dy; put t= 1y: =- te-tdt=(t+1)e-t+C
x= 1y+1+Ce1/y; x(1)=1 C=- 1e
x ( 12 )=2+1- 1e e2=3-e
Answer: (4)
If x=f(y) is the solution of the differential equation (1+y2 )+ (x-2e -1y ) dydx=0, y (- 2, 2 ) with f(0)=1, then f ( 1 3 ) 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 .
The solution of the differential equation (x2 + y2 )dx - 5xy dy = 0, y(1) = 0, is :
The solution curve, of the differential equation 2y dydx + 3 = 5 dydx, passing through the point (0, 1) is a conic, whose vertex lies on the line :
Let 0x 1 - (y'(t))2 dt = 0x y(t) dt, 0 x 3, y 0, y(0) = 0. Then at x = 2, y'' + y + 1 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(x) be a positive function such that the area bounded by y = f(x), y = 0 from x = 0 to x = a > 0 is e-a + 4a2 + a - 1. Then the differential equation, whose general solution is y = c1 f(x) + c2, where c1 and c2 are arbitrary constants, is
Let y = y(x) be the solution of the differential equation (1 + y2)e xdx + 2 x (1 + e2 x )dy = 0, y(0) = 1. Then y ( 4 ) is equal to
Let y = y(x) be the solution curve of the differential equation y dydx + 2x y = x3 y, y(1) = 0. Then y( 3) is equal to:
Let |x| = |y|exy - , , N be the solution of the differential equation x dy - y dx + xy(x dy + y dx) = 0, y(1) = 2. Then + is equal to