My problem:

$\displaystyle y'' + 4y' + 5y = 5t + e^{-t}$

I found the general solution of the homogenous equation

$\displaystyle r^2 + 4r + 5 = 0$

to be

$\displaystyle Y_c = C_1e^{-2t}cos(t) + C_2e^{-2t}sin(t)$

since

$\displaystyle r = -2 + i, -2 - i$ (complex)

Now, I have tried all of the following as particular solutions and all have come up 'no solution'

$\displaystyle Be^{-t}$

$\displaystyle Bte^{-t}$

$\displaystyle A + Be^{-t}$

$\displaystyle A + Bte^{-t}$

$\displaystyle At + Be^{-t}$

$\displaystyle At + Bte^{-t}$

$\displaystyle At^2 + Be^{-t}$

$\displaystyle At^2 + Bte^{-t}$

$\displaystyle At^2 + Bt^2e^{-t}$

$\displaystyle At^3 + Be^{-t}$

$\displaystyle At^3 + Bte^{-t}$

$\displaystyle At^3 + Bte^{-t}$

$\displaystyle At^3 + Ct^2 + Dt + Be^{-t}$

All of them come up with conflicting values (A will equal 0 in one spot and need to be a value in another).

Can anyone tell what I am doing wrong?

thanks