What is the particular solution for $y''-y'=e^{2x} + \sin x$ I calculated homogenous already, I'm just struggling a bit with the right side. Would $y_p$ be $= ++e^x$ or $= ++e^{2x}$?
Would the power in front of the root be the roots found from the homogenous part?
Sorry, it's been a while since I did ODE's. All help is appreciated.
 A: It would be neither. In either case, no amount of differentiating $A + Bx + Ce^x$ or $A + Bx + Ce^{2x}$ will produce the $\sin x$ on the right hand side. When a $\sin$ or $\cos$ term appears on the right hand side, it is best to include an $A\sin x + B\cos x$ type term, and yes, both the $\sin$ and the $\cos$ terms are necessary, as $\cos$ can appear through differentiation.
You also won't need the $A + Bx$ terms. These terms are for when a degree one polynomial appears on the right hand side, e.g. if the right hand side were $e^{2x} + x$ or $e^{2x} - 3x - 2$. Once again, even if there were no constant term (such as in the former example), you need the constant term $A$ in $A + Bx$, as a constant can appear through differentiation.
The exponential term should be $Ae^{2x}$ as in your second guess, because the $e^{2x}$ term on the right hand side can never appear through differentiating $e^x$.
So, the form you should use for finding the particular solution would be $Ae^{2x} + B\sin x + C \cos x$.
You should also verify that none of the terms in your guess for the particular solution belong to the homogeneous solution. For any such terms, you should try multiplying the term by $x$. But, this is not a problem here, as hopefully you can see from your own working.
A: $$y''_p-y'_p=e^{2x} + \sin x$$
$$(y'_pe^{-x})'=e^{x} + e^{-x}\sin x$$
And integrate. No need for the constants of integration.
