If `y(x)` satisfies the differential equation `y^(prime)-ytanx=2xs e c x` and `y(0)=0` , then

Let y(x) be a solution of the differential equation (1+e^x)y^(prime)+y e^x=1. If y(0)=2 , then which of the following statements is (are) true? (a) y(-4)=0 (b) y(-2)=0 (c) y(x) has a critical point in the interval (-1,0) (d) y(x) has no critical point in the interval (-1,0)

If y(t) satisfies the differential equation y'(t)+2y(t)=2e^(-2t),y(0)=2 then y(1) equals

The solution to the differential equation ylogy+x y^(prime)=0, where y(1)=e , is

A function y=f(x) satisfies the differential equation (d y)/(d x)+x^2 y=-2 x, f(1)=1 . The value of |f^( prime prime)(1)| is

Solve the differential equation y'-(y)/(x)+cosec((y)/(x))=0 when y(1)=0.

Solve the differential equation ye^(x/y)dx=(xe^(x/y)+y^(2))dy(yne0).

Verify the differential equation y = Ax : xy' = y (x ne 0)

Differential equation y log y dx - x dy = 0

If the solution of the differential equation (dy)/(dx)-y=1-e^(-x) and y(0)=y_0 has a finite value, when x->oo, then the value of |2/(y_0)| is__

Solve the differential equation: (1+y^2) + ( x - e^(tan^-1 y) ) dy/dx = 0