Hi. I have to solve this:

$\displaystyle y''-3y'+2y=e^x$,
Using the replacement $\displaystyle y=\phi y_1$ being $\displaystyle y_1$ a solution of the homogeneous differential equation. I can't do it "traditionally", I have to use this method.

So I have to solve that $\displaystyle \phi ''y_1+ \phi ' \left[2y_1'+Py_1] \right]=e^x $

So, this is what I did:

$\displaystyle y_1=C_1 e^x+C_2e^{2x}\rightarrow y_1'=C_1 e^x+2C_2e^{2x}$
$\displaystyle \phi ''C_1 e^x+\phi '' C_2 e^{2x}+ \phi ' \left[C_2e^{2x}-C_1e^x \right]=e^x $
What should I do from here? I don't know how to handle $\displaystyle \phi$

Bye there.