Solve the differential equation

y ′′ − 3 y ′ − 4 y = −3 x+5

I've worked out the independent solution which is

r^2 - 3r - 4 = 0

(r + 1) (r - 4)

{e^(-1*x), e^(4*x)}

(as a set of function)

Now my question is how do I find the solution of the original d.e. is of form

y(x) = u1v1 + u2v2

And the general solution?