Can anyone help me with this proof, I've attached my working out so far:
Show that: $\phi =Ae^{-\frac{kt}{2}}sin(pt)cos(qx)$ satisfies the equation
$$\frac{\partial \phi^{2}}{\partial x^2}=\frac{1}{c^2}\left ( \frac{\partial^2\phi}{\partial t^2} + k \frac{\partial \phi}{\partial t}\right )$$
provided that $p^{2}=c^{2}q^{2}-\frac{k^{2}}{4}$.
Working out so far (click to magnify):
On
By chain rule, $\frac{\partial\phi}{\partial{t}} = A\cos(qx)\frac{\partial}{\partial{t}}(e^{\frac{-kt}{2}}\sin(pt))=A\cos(qx)\big[\frac{\partial}{\partial{t}}(e^{\frac{-kt}{2}})\sin(pt)+e^{\frac{-kt}{2}}\frac{\partial}{\partial{t}}(\sin(pt)\big]$.
So, $\frac{\partial\phi}{\partial t} = A\cos{(qx)}e^{\frac{-kt}{2}}\Big[-\frac{k}{2}\sin(pt)+p\cos(pt)\Big]$.
By a similar application of chain rule to this result, you have $\frac{\partial^2\phi}{\partial{t^2}}$.
Retry, using this facts.
You need to use the product rule when computing $\frac{\partial}{\partial t}$, you've only taken the deriviative of one term containing $t$.