So I have the wave equation
$$\frac{\partial^2u}{\partial t^2} = c^2 \frac{\partial^2u}{\partial x^2} $$
and I know the process to split it into the two ODE's which are
$F''(x)-nF(X) = 0$ (Used $n$ because I couldn't find $\lambda$, where $n$ is the separation constant)
$$G''(t) - c^2nG(t)=0$$
I know how to go on from there when the separation constant is $<0$ However I am unsure what happens when the separation constant is $= 0$ or $> 0$
Your ode equation for $F(x)$ will become $$F''(x)=0$$ which gives $$F(x)=Ax+B$$ Now we do not know your boundary conditions but let say $F(0)=0=F(L)$, then $F(x)=0$, a trivial solution.
The ode for $F(x)$ will have the characteristics equation of the form $r^2-n=0$ with the roots $r=\pm n$ and thus the solution is
$$F(x)=C\exp{(\sqrt{n}x)}+D\exp{(-\sqrt{n}x)}$$
let say with homogeneous boundary conditions of the first type, again we will endup with trivial solution.
The characteristic equation in this case is $r^2+\lambda^2=0$ with imaginary roots $r=\pm \iota \lambda$ then the solution to the ode is $$F(x)=A_1\cos{(\lambda\,x)}+B_{1}\sin{(\lambda x)}$$ Now utilizing one the the three types of boundary conditions (say first type), we will have $A_1=0$ and for nontrivial solution we will take $B_1\neq0$, thus \begin{align} \sin{(\lambda L)}=0\\ \Rightarrow \lambda L = \sin^{-1}{(0)}\\ \Rightarrow \lambda = \frac{n_1\pi}{L},\,\, n_1=1,2,3... \end{align} Finally, $F(x)=B_{n_1}\sin{(\frac{n_1\pi}{L} x)}$. Subbing the eignvalues into the temporal equation $G(t)$, we get $$G''(t)-c^2\frac{n_1\pi}{L}G(t)=0$$ I think you can solve for $G(t)$.