Find the general solution..

y'' + 4y' = sin3x

The general solution of a differential equation of the form ay'' + by' + cy is given by Ae^mx + Be^nx; where m and n are the root of the quadratic equation ax^2 + bx + c
The general solution of y'' + 4y' is given by Ae^mx + Be^nx; where m and n are the root of the quadratic equation x^2 + 4x = 0
x(x + 4) = 0
x = 0 or x = -4

Therefore, the general solution is Ae^(-4x) + Be^0x = Ae^(-4x) + B