1. Particular Integral

In finding the general solution, we need the complementary function and particular integral. To find the complementary function, we set the second order derivative to 0 and solve then for the particular integral, we make assumption.

If the second order derivative is equal to $\displaystyle e^{-2x}$, we assume $\displaystyle Ae^{-2x}$, if the second order derivative is equal to $\displaystyle \sin 2x$, we assume $\displaystyle A\cos 2x + B\sin 2x$, but if the second order derivative is equal to the conbination, ie $\displaystyle e^{-2x}\sin 2x$, what do we assume?

2. Re: Particular Integral

I am puzzled by your use of the term "second order derivative". Second order derivative of what?

In order to solve the non-homogeneous, linear equation, L(y)= f(x), where "L(y)" is the "left side of the equation"- all terms that involve y or a derivative of y, up to any order, not necessarily just two, we first solve the corresponding homogenous equation: L(y)= 0. I think "L(y)= 0" is what you mean but that is not necessarily "setting the second order derivative to 0".

In any case, if the "right side of the equation", the terms that do not involve y or any of its derivatives, is of the form $\displaystyle e^{-x} sin(x)$, then you should try a particular solution of the form $\displaystyle e^{-x}(Acos(x)+ Bsin(x))$ where A and B are to be determined.

(Of course, if either $\displaystyle e^{-x}cos(x)$ or $\displaystyle e^{-x}sin(x)$ is a solution to the associated homogenous equation (if the coefficients of the equation are real numbers, then if one is a solution, the other must be also), then you should try a solution of the form $\displaystyle te^{-x}(Acos(x)+ Bsin(x))$.)