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I'm trying to find the general solution of $$u_{tt}=c^2\frac{1}{x}\frac{\partial}{\partial x}\left( x \frac{\partial u}{\partial x} \right)$$ by using the ansatz $u(x,t)=f(x)g(t)$, assuming the seperation constant is zero.

So far I've done the following:

$$u_{tt}=f(x)g''(t)$$

$$u_x=f'(x)g(t)$$

and therefore I have $$f(x)g''(t)=\frac{c^2g(t)f'(x)}{x}+c^2g(t)f''(x)$$

I'm not sure how to progress from here though.

EDIT

Continuing from the advice given in the comments, I've found that $$f''(x)+f'(x)=0$$ which gives me $f(x)=A+Be^{-x}$. Similarly I have found that $g(t)=at+b$. Therefore, the general solution is $$u(x,t)=(A+Be^{-x})(at+b)$$ Is this now correct?

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    divide by $g(t)f(x)$ you can then conclude that the r.h.s as well as the l.h.s have to equal the same constant – 2017-02-08
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    @tired Thank you. I've made an edit, could you have a look at this too? – 2017-02-08

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