`(dy)/(dx) = y + e^(-2x)`

(dy)/(dx) + 2y = e ^(-2x) sinx , given that y=0 when x = 0

If y = f(x) is the solution of the differential equaiton e^(3y) ((dy)/(dx) - 1) = e^(2x) and y(0) = 0 then y(x) = log (Ae^(3x) - Be^(2x))^((1)/(3)) where the value of (A + B) is:

Solve: (dy)/(dx)+3y=e^(-2x)

Solve: (dy)/(dx)+2y=e^(-x)

Solve the following differential equation: (dy)/(dx)=2y=e^(3x)

Assertion: Integrating factor of (dy )/(dx) + y = x^2 is e^x Reason: Integrating factor of ( dy )/(dx) + P( x) y= Q (x) is e^(int p(x) dx)

The general solution of (dy)/(dx) = 2x e^(x^(2)-y) is

(dy) / (dx) = e ^ (x + y)

(dy)/(dx)=(x+e^(2x))/(y)