Help with Method of Undetermined Coefficients 
Solve the following ODE using the method of undetermined coefficients:
$$y''-4y'+4y=(x+1)e^{2x}$$

My work. I have the homogeneous solution:
$$y_h=C_1e^{2t}+C_2te^{2t}$$
I use this function to get the particular solution of the ODE:
$$y=(Ax+B)xe^{2x}$$
And after substitution on the ODE, I get:
$$2A=(x+1)e^{2x}$$
What about $B$? I don't know what I am missing. Could someone give me a hint? Thank you.
 A: Hint. Since the characteristic polynomial is $(z-2)^2$ then the multiplicity $m$ of the root $2$ is $2$. It follows that for $f(x)=(x+1)e^{2x}$ you should try a particular solution of the form
$$y(x)=(Ax+B)x^me^{2x}=(Ax+B)x^2e^{2x}.$$
P.S. If you use the form $y(x)=(Ax+B)2e^{2x}$ then
after the substitution you should get $2Ae^{2x}=xe^{2x}+e^{2x}$ which is impossible to solve.
A: For the particular solution, Robert's suggestion is the best
$$y_p=(ax^3+bx^2)e^{2x}$$
Since you already have terms in $e^{2x},xe^{2x}$
Another solution
$$y''-4y'+4y=(x+1)e^{2x}$$
$$y''e^{-2x}-4y'e^{-2x}+4ye^{-2x}=(x+1)$$
$$y''e^{-2x}-2y'e^{-2x}-2y'e^{-2x}+4ye^{-2x}=(x+1)$$
$$(y'e^{-2x})'-2(ye^{-2x})'=(x+1)$$
Simply integrate
$$y'e^{-2x}-2ye^{-2x}=\frac {x^2}2+x+K_1$$
$$(ye^{-2x})'=\frac {x^2}2+x+K_1$$
Integrate again
$$ye^{-2x}=\frac {x^3}6+\frac {x^2}2+K_1x+K_2$$
Therefore
$$\boxed{y(x)=e^{2x}\left(\frac {x^3}6+\frac {x^2}2+K_1x+K_2\right )}$$
You can see that the particular solution $y_p$ is
$$y_p=e^{2x}(\frac {x^3}6+\frac {x^2}2)$$
A: You've probably made a mistake substituting it back in. After simplifying, you should arrive at
$$
6Axe^{2x}+2Be^{2x}=xe^{2x}+e^{2x}
$$
