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