$\displaystyle y'' - 2 y' + 1 y = -18 e^{1 t}$

Find a particular solution to this equation:

How exactly am I supposed to answer this?

My homologous equation renders repeated roots:

[Math] r_1=e^x, r_2=xe^x[/tex]

And when solve this with undetermined coefficients, I get:

$\displaystyle

Y_p=Ae^t

$

but since that e^x and xe^x already exist in the homologous solution, I should get:

$\displaystyle

Y_p=\alpha T^2e^t$

Is this the correct answer? I cant solve for alpha, because I have no initial values missing. My assignment wont accept this format. What am I missing?