I'm trying to solve Laplace's equation in the disc $r\leq a$ using separation of variables (so relatively simple stuff, compared to what's often on here), and I've proceeded like so thus far:
Let $T(r,\theta)$ be a solution to $\frac{\partial^2 T}{\partial r^2}+\frac{1}{r}\frac{\partial T}{\partial r}+\frac{1}{r^2}\frac{\partial^2 T}{\partial \theta^2}=0$. Substituting $T(r,\theta)=F(r)G(\theta)$ into this equation, and equating both functions to a constant, say, $\lambda$, gives:
(1) $r^2F''(r)+rF'(r)=\lambda F(r)$
(2) $G''(\theta)=-\lambda G(\theta)$
First considering (2), we know that $G(\theta)$ must be $2\pi$ periodic, which is only possible if either $\lambda=n^2$ for $n\in\mathbb{N}$, in which case $G(\theta)=A\sin(n\theta)+B\cos(n\theta)$, or $\lambda=0$, giving $G(\theta)=c$ for some constant $c$.
Now looking back at (1), $\lambda=0$ gives $F(r)=A + B\log{r}$, but what about when $\lambda=n^2$?
I have in my online notes that the solution is a linear combination of $r^n$ and $r^{-n}$, and I can see that these fit, but apart from getting that answer by a lucky guess, I don't see how one would come about it. The notes say that this equation is of 'Euler's type', but I'm not too sure what that means either, and if I search for it on the internet I just come across a lot of other unrelated equations that Euler had something to do with.
Thanks in advance, and this is my first question, so sorry if I've done anything wrong!
Too late to help I imagine but the $\lambda=n^2$ eigenvalues arise because we want a single-valued solution, so $G(\theta+2\pi)=G(\theta)$. Having found the possible values for $\lambda$, the equation for $F$ is equidimensional (each time we differentiate we multiply by $r$, and $r \frac{d}{dr} r^k=kr^k$), so we try $F=r^k$ and find the values of $k$ that work.