what is asking to do to show that u(x,t)=$\phi(x)\theta(t)$ is a solution to the wave equation?

Question

I been asked the following if $\phi^{''}(x)+a^2 \phi(x) =0$ and $\theta^{''}(t)+c^2a^2 \theta (t) =0$ show that $u(x,t)=\phi(x)\theta(t)$ is a solution to the wave equation

My question is do I have to arrive to the wave equation from the above equation, that I shown?or on the wave equation I sub the partials of u and them sub the first equation witch prove 0=0 ?

Edit: by other words what is asking to do or show? I have prove both ways but I don't know what is the right anwser

Answer 1

You just need to show that the function $u(x,t) = \phi(x)\theta(t)$ satisfies the wave equation. That is, show that $$\frac{\partial^2 u(x,t)}{\partial t^2} = c^2 \frac{\partial^2 u(x,t)}{\partial x^2}.$$ Since you have $u(x,t) = \phi(x)\theta(t)$, this reduces to showing that $$ \phi(x)\frac{\partial^2 \theta(t)}{\partial t^2} = c^2 \frac{\partial^2 \phi(x)}{\partial x^2} \theta(t). $$

Answer 2

@Kelly: You need to complete what Tom has left:

$$\phi(x)\frac{\partial^2 \theta(t)}{\partial t^2} = c^2 \frac{\partial^2 \phi(x)}{\partial x^2} \theta(t).\tag{1}$$

Divide both sides by $\phi(x)\theta(t)$, you will get

$$\frac{1}{\theta(t)}\frac{\partial^2 \theta(t)}{\partial t^2} = c^2 \frac{1}{\phi(x)} \frac{\partial^2 \phi(x)}{\partial x^2} .\tag{2}$$

Since LHS is a function of $t$ and RHS is a function of $x$, (2) must be equal to a constant, called $h$

$$\frac{1}{\theta(t)}\frac{\partial^2 \theta(t)}{\partial t^2} = h \tag{3}$$

$$ c^2 \frac{1}{\phi(x)} \frac{\partial^2 \phi(x)}{\partial x^2}= h \tag{4}$$

Now you can say that any function $u(x,t)=\theta(t)\phi(x)$ with $\theta(t)$ and $\phi(x)$ satisfying (3) and (4) is a solution to the original wave equation.