Is there a rule for simplifying complex powers of e into sin and cos terms?

Question

Part of the algebraic simplifications involved in a solution to the differential equation $$ y''+ y' - 6y = 4 \cos(2x) $$ make the following jump: $$ -\frac{1}{13}(5+i) e^{2ix} = -\frac{1}{13}(5+i)(\cos 2x + i \sin 2x) $$

I have understood the other examples, so i think this is a more simple algebra issue. Is there some rule for this kind of complex power of e simplification? For context, the surrounding algebra is as follows:

I understand that we want to take the real part of the particular solution $z_p$ to have the general solution, but that algebra jump is defying me.

Answer 1

Use that $$e^{ix}=\cos(x)+i\sin(x)$$

Answer 2

You need$$\cos kx=\frac{e^{ikx}+e^{-ikx}}{2},\,\sin kx=\frac{e^{ikx}-e^{-ikx}}{2i}.$$ In your example $(D+3)(D-2)y=(D^2+D-6)y=2e^{2ix}+2e^{-2ix}$ with $D:=\dfrac{d}{dx}$, so constants $A,\,B$ exist for which $y=Ae^{2ix}+Be^{-2ix}$ is one solution. Equating $e^{2ix}$ coefficients gives $A$ by substitution; you can get $B$ the same way from the $e^{-2ix}$ coefficient. To the resulting solution you must add $Ce^{-3x}+De^{2x}$ to get the general solution.