What are the general expression for the general solutions satisfying the boundary conditions $u_x(0, t)$ = 0 = $u_x(L, t)$, for all t > 0.

Question

So I have this question here, and I'm given that

$u_{xx}$ $-$ $\frac {1}{v^2}$$u_{tt}$ = $0$, where 0 < x < L, v is a constant, and where the subscripts indicate differentiation.

So what I've done is say that $u(x,t)$ = $X(x)$$Y(t)$

but I am really unsure what to do from here. Any hints would be great. Thanks

So carrying on from here what I think is that $\frac {1}{X(x)}$$X(x)_{xx}$ = $\frac {1}{Y}$$\frac {1}{v^2}$$Y(t)_{tt}$

So setting $\frac {1}{X(x)}$$X(x)_{xx}$ = $-a^2$ where a is a constant and doing the same for $Y(t)$