y''-4y+4y=xe^2x
what is the general private solution
usuallu
its
y_p=(ax+b)e^2x
but here we have
a double root of 2
I think you mean particular solution.
From just looking at the right hand side we would "guess" that
$\displaystyle y_p=(ax+b)^{2x}$
This is correct, but since the complimentary solution is $\displaystyle c_1e^{2x}+c_2xe^{2x}$ as you said a double root we have to multiply our guess by $\displaystyle x^2$ This is justified by using reduction of order. so we get
$\displaystyle y_p=(ax^3+bx^2)e^{-2x}$
Now just plug in to find a and b.