[SPLIT] y'' - 3y' + 2y = - 6e^{4x}, y(0) = y'(0) = 1, x = 1

Another problem... how do I solve this one?
y'' − 3y' + 2y = − 6e^{4x}
y(0) = 1
y'(0) = 1
x = 1

I think I have to do the left part first (y'' − 3y' + 2y = 0) and then the right part of the equation (I don't now how to solve this one).
Where do I put in y(0) and y'(0)?

Homogeneous solution:
\(\displaystyle \L\\y_{h}=C_{1}e^{x}+C_{2}e^{2x}\)

Particular solution:
\(\displaystyle \L\\y_{1}=Ae^{4x}\)

\(\displaystyle \L\\4Ae^{4x}-6Ae^{4x}+2Ae^{4x}=-6Ae^{4x}\)

\(\displaystyle \L\\6Ae^{4x}=0\)

\(\displaystyle \L\\A=0\)

\(\displaystyle \L\\y_{p}=0\)

Is this correct? I found similar equation solved this way but I'm not sure about numbers 4, 6 and 2.

Particular solution, \(\displaystyle \L\ y = -e^{4x}\)

Hence, general solution, \(\displaystyle \L\ y = u(x) - e^{4x}\), where \(\displaystyle \L\ u" - 3u' + 2u = 0\)

auxiliary: \(\displaystyle \L\ r^2 - 3r + 2 = 0\), so \(\displaystyle \L\ r = 2, 1\)

Therefore, the general solution for u(x) is \(\displaystyle \L\ u = Ae^{2x} + Be^{x}\)

Hence, general solution for y:

\(\displaystyle \L\ y = Ae^{2x} + Be^{x} - e^{4x}\)

Thank you morson!

\(\displaystyle \L\\y=C_{1}e^{2x}+C_{2}e^{x}-e^{4x}\)

\(\displaystyle y(0)=1\) \(\displaystyle \L\\1=C_{1}+C_{2}-1\)

\(\displaystyle y'(0)=1\) \(\displaystyle \L\\1=2C_{1}+C_{2}-4\)

\(\displaystyle \L\\C_{1}=3\)

\(\displaystyle \L\\C_{2}=-1\)

\(\displaystyle \L\\y(1)=3e^{2}-e-e^{4}=-35,14926\)

I hope this is the correct solution

Looks fine, if the comma is supposed to denote a decimal point.