calculus - Differential equation with complex roots and undetermined coefficients

Hey guys i need help with this problem:

$y''-4y'+5y=6e^{2x}sinx$

I get the roots from the characteristic equation. $r^2-4r+5=0$ $\implies r=2\pm i$

Then i set up:

$y_{h}=c_{1}e^{2x}sinx+c_{2}e^{2x}cosx$

And i know that:

$y_p=e^{2x}(Asinx+Bcosx)x$

But the problem arrives when i differentiate. The problem is the x on the end i dont know how to work with it, i can solve the other differential equation types. For this problem i end up with:

$$-xBcosx-2Bcosx+4Acosx-2xBsinx-xAsinx+2Bsinx$$

After differentiating and putting it back into the equation. But i dont know what to do with the x's.

Answer

If I may suggest.

You can make your life much easier if you start from the very beginning using $$y=e^{2x}z \implies y'=e^{2 x} \left(z'+2 z\right)\implies y''=e^{2 x} \left(z''+4 z'+4z\right)$$ This makes the differential equation to be $$z''+z=6\sin(x)$$ and then, almost as you wrote, the particular solution will be $$z_p=x(A \sin(x)+B \cos(x))$$ $$z'_p=\sin (x) (A-B x)+\cos (x) (A x+B)$$ $$z''_p=\cos (x) (2 A-B x)-\sin (x) (A x+2 B)$$ $$z_p''+z_p=2 A \cos (x)-2 B \sin (x)=6\sin(x)$$ which is easy to solve.