Rewriting $e^{2xi} + e^{-2xi}$

Question

I'm trying to solve the differential equation $y^{''} + 2y^{'} + 5y = 0$

I get that the general solution is: $$y = Ae^{-x + 2xi} + Be^{-x-2xi} $$ $$y = e^{-x}(Ae^{2xi} + Be^{-2xi}) $$ $$ y = e^{-x}(A\cos(2x) + Ai\sin(2x) + B\cos(-2x) + Bi\sin(-2x))$$

In the book the answer is written as: $$y = e^{-x}(A\cos(2x) + B\sin(2x))$$

How can my answer be rewritten to equal the book's? Or did I do wrong somewhere?

Answer 1

Using $\sin(-2x)=-\sin(2x)$ and $\cos(-2x)=\cos(2x),$ your result is $$y=e^{-x}\left((A+B)\cos(2x) + i(A-B)\sin(2x)\right)$$

So, letting $C=A+B$ and $D=i(A-B)$, you've got $$y=e^{-x}\left(C\cos(2x)+D\sin(2x)\right)$$

Answer 2

if the solution to the charachteristic equation is $a + ib$, then the solution of the differential equation are in the form

$$e^{ax}(\cos(bx) + i\sin(bx))$$

Then of course $e^{ix} = \cos x + i\sin x$, so the two formulas are equivalent.

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You are writing the solution like $$Ae^{(a+ib)x} + Be^{(a-ib)x}$$

But This is equivalent to $$e^{ax}(Ae^{ibx} + Be^{-ibx}) = e^{ax}(A\cos bx + iA\sin bx + B\cos bx - iB\sin bx) = e^{ax}((A+B)\cos x + i(A-B)\sin bx)$$

Thanks to the fact that $\cos x$ is even and $\sin x$ is odd. Now the two constants $A+B$ and $A-B$ are independent, therefore you may just call $A+B = C$ and $A-B=D$ and you get the same solution as written in the book