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

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