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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.
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.
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 $$y_p=e^{2x}(\frac {x^3}6+\frac {x^2}2)$$
You've probably made a mistake substituting it back in. After simplifying, you should arrive at $$ 6Axe^{2x}+2Be^{2x}=xe^{2x}+e^{2x} $$
