# Finding the solution to Laplace's equation in spherical coordinates

Consider Laplace's equation in spherical coordinates $$\Delta u = \frac{1}{r^2}\frac{\partial }{\partial r}\left(r^2 \frac{\partial u}{\partial r}\right)+\frac{1}{r^2\sin\theta}\frac{\partial}{\partial \theta}\left(\sin\theta\frac{\partial u}{\partial\theta}\right)=0$$

I have let $u=R(r)\Theta(\theta)$ and found the ODE for $R(r)$ to be given by $$r^2R(r)+2rR'(r)-kR(r)=0$$

and the ODE for $\Theta$ to be given by $$\sin\theta\Theta''(\theta)+\cos\theta \Theta(\theta)+k\sin\theta\Theta(\theta)=0$$

In order to solve the ODE for $R(r)$ I am given that $k=n(n+1)$ for $n\in \mathbb{Z}_+$ and I'm told to look for functions of the form $R(r)=Cr^{z}$.

Using the ansatz given above I find that $z(z+1)=n(n+1)$. What do I do from this point though?

The structure of the ODE for $R$ does suggest looking at powers of $r$ - it's a Cauchy–Euler equation. Plugging $R=r^z$ yields two solutions: $z=n$ or $z=-n-1$, with $k=n(n+1)$.
Unfortunately, the ODE for $\Theta$ is not that simple. It's the Legendre differential equation, whose solutions are Legendre polynomials of $\cos\theta$. It's unclear how much of the theory of Legendre polynomials you are supposed to discover on your own. You can find a derivation of some of them here.