Find the general solution of: y'' + 4y' + 13y = sin(x)

First we find the auxilary equation by substituting y with m^0, y' with m^1 and y'' with m^2. We get m^2 + 4m + 13 and find the roots using the differential equation, m = (-4 +- (16-4x1x13)^0.5)/(2x1).

From this we get complex roots m = -2 + 3i and m = -2 - 3i. Now we solve the homogenous form using these roots, y = e^(-2x) (Acos(3x) + Bsin(3x)).

So we have solved the differential equation for when the right hand side is equal to zero but we must solve it for when the RHS is equal to sin(x) so we need to take y = psin(x) + qcos(x) and find y' and y'' to substitute into the LHS. So y' = -psin(x) + qcos(x) and y'' = -pcos(x) - qsin(x).

By comparing coefficients of the substitued LHS and the RHS we get, (-pcos(x) - qsin(x)) + 4(-psin(x) + qcos(x)) + 13(pcos(x) + qsin(x)) == sin(x). After comparing coefficients and solving the resulting simultaneous equation we find, p = -1/40 and q = 3/40.

Now we just put all the parts together to obtain the general solution to the equation, y = e^(-2x) (Acos(3x) + Bsin(3x)) - (1/40)cos(x) + (3/40)sin(x).