Question
Verify that the given two-parameter family of functions is the general solution of the non-homogeneous differential equation on the indicated interval.
$$ y''-4y'+4y = 2e^{2x}+4x-12 $$ $$ y=C_1e^{2x}+C_2xe^{2x}+x^2e^{2x}+x-2; (-\infty,\infty) $$
I solved it to the point where, after plugging in my $y''$ I get this on the left side of the top equation:
$$ -4C_1e^{2x}-4C_2xe^{2x}-4x^2e^{2x}+2e^{2x}-4 $$
I'm not 100% positive if the above is correct but my curiosity compulsed me to come here and ask this question: Could I combine the $C_1$ and $C_2$ if they happened to have the same $x$ terms? For example, let's say they read $-4C_1xe^{2x}$ and $-4C_2xe^{2x}$. Would this be equal to $-8Cxe^{2x}$? When do I have to respect the difference in $C$'s and when not?
The general solution of a linear ODE can be written as $$ y=y_{h}+y{c} $$ where $y_{h}$ stands for the homogeneous solution and $y_{c}$ is the complementary particular solution. In your case, the ODE $$ y''-4y'+4y=0 $$ has homogeneous solutions $c_{1}e^{2x}$ and $c_{2}xe^{2x}$.
You can easily verify that $x^2e^{2x}+x-2$ is particular solution for $$ y''-4y'+4y = 2e^{2x}+4x-12 $$ With that you are done. $c_{1}$ and $c_{2}$ will be determined from the appropriate initial conditions. I hope that helps.