$\frac{{\partial }^{2}u}{\partial t^2}=c^2\frac{{\partial }^{2}u}{\partial x^2}$

Step 1 ：First, we rewrite the given partial differential equation as $\frac{{\partial }^{2}u}{\partial t^2}-c^2\frac{{\partial }^{2}u}{\partial x^2}=0$

Step 2 ：Now, we assume a solution of the form $u(x,t)=f(x)g(t)$

Step 3 ：Substitute this assumed solution into the given equation: $f(x)g^{″}(t)-c^2{f}^{″}(x)g(t)=0$

Step 4 ：Divide both sides by $f(x)g(t)$: $\frac{g^{″}(t)}{g(t)}=c^2\frac{f^{″}(x)}{f(x)}$

Step 5 ：Since the left side depends only on $t$ and the right side depends only on $x$, both sides must be equal to a constant, say $-\lambda$: $\frac{g^{″}(t)}{g(t)}=c^2\frac{f^{″}(x)}{f(x)}=-\lambda$

Step 6 ：Now, we have two ordinary differential equations: $g^{″}(t)+\lambda g(t)=0$ and $f^{″}(x)-\frac{\lambda }{c^2}f(x)=0$

Step 7 ：Solve these two ordinary differential equations to find the general solution for $u(x,t)$

Step 8 ：Finally, apply any given initial or boundary conditions to find the specific solution for the problem