General solution to the wave equation in one dimension

Question

In physics I am studying the wave equation. We saw that we could use a separable ansatz of the form: $$ u(x,t)=f(x)g(t)$$ to try and solve it. By Rearranging the PDE we get an equation with turns out to be constant: $$ c^2\frac 1f \frac{d^2f}{dx^2}=\frac 1g\frac{d^2g}{dt ^2}=-a^2 $$. We then get two equations that resemble the SHM equation. However, my teacher skipped the solution and wrote down that the solutions where: $$ f(x)=Ae^{\frac{iax}{c}}+Be^{\frac{-iax}{c}}, g(t)=Ce^{iat}+De^{-iat}$$ I have two questions: 1) how de we know that the constant term is $-a^2$, IE why the minus and the square?. 2) Why are the solutions not like the ones to the SHM equation: $$g(t)=C\cos(at)+ D\sin(at), f(x)=A\cos(ax/c) + B\sin(ax/c)$$ Any help would be great!

Answer 1

1. You are granted to find wave-like solutions only with a negative constant. The square of it is just to be sure it is positive and also for convenience in further calculations.
2. Indeed, they are. You have just to remember the Euler's identity $e^{i\phi}=\cos(\phi)+i\sin(\phi)$ and redefine the constants. To see this, let us consider $g(t)$. You will have $$ g(t)=Ce^{iat}+De^{-iat}=C(\cos(at)+i\sin(at))+D(\cos(at)-i\sin(at)). $$ Collecting identical terms one arrives to $$ g(t)=(C+D)\cos(at)+i(C-D)\sin(at) $$ and you are done.