# Euler-Cauchy equation

While trying to solve for the Laplacian in polar coordinates, one encounters the famous Euler-Cauchy differential equation.

$x^{2}{\frac {d^{2}y}{dx^{2}}}+2x{\frac {dy}{dx}}-l(l+1)y=0$

or

${\frac {d^{2}y}{dx^{2}}}+ \frac{2}{x}{\frac {dy}{dx}}- \frac{l(l+1)}{x^2}y=0$

How does one find a solution to this differential equation ? Well, most places that I have read simply dictate that the solutions to this differential equation as $y = x^l, x^{-(l+1)}$ without explaining why only these two are the only solutions. The purpose of this post is to explain why!

We can clearly see that x = 0 is a singular point. Therefore a simple power series solution won’t work. Hence we use the frobenius method to get rid of the singularity i.e

$y= \sum\limits_{n=0}^{\infty} a_{n}x^{n + \lambda}$

Let’s now compute its derivatives wrt x.

$\frac{dy}{dx}= \sum\limits_{n=0}^{\infty} (n+\lambda)a_{n}x^{n +\lambda -1}$

$\frac{d^{2}y}{dx^{2}}= \sum\limits_{n=0}^{\infty} (n + \lambda)(n + \lambda -1)a_{n}x^{n + \lambda -2}$

Putting the values for $y,\frac{dy}{dx}$ and $\frac{d^{2}y}{dx^{2}}$ back into the differential equation, we get the following form.

$x^{2}{\sum\limits_{n=0}^{\infty} (n + \lambda)(n + \lambda -1)a_{n}x^{n-2}}$

$+2x{\sum\limits_{n=0}^{\infty}(n+\lambda)a_{n}x^{n-1}}$

$- l(l+1){\sum\limits_{n=0}^{\infty} a_{n}x^{n}}=0$

Bringing the $x,x^{2}$ terms inside the summation, we obtain the following form:

$\sum\limits_{n=0}^{\infty}x^n \left({(n+\lambda)(n+\lambda -1)a_n + 2(n+\lambda)a_n -l(l+1)a_n}\right)=0$

$\sum\limits_{n=0}^{\infty}a_n x^n \left({(n+\lambda)(n+\lambda +1) -l(l+1)}\right) = 0$

Obviously, the coefficients of the summation cannot be zero and x=0 is a trivial solution. Therefore, we get the indicial equation.

$(n+\lambda)(n+\lambda+1) - l(l+1)= 0$ or

$(n+\lambda)(n+\lambda+1) = l(l+1)$

(i) When n = 0, the indicial equation becomes

$\lambda(\lambda+1) = l(l+1)$

The values of the $\lambda$ that solve for this equation are $l,-(l+1)$

$\lambda = l , -(l+1)$

(ii) When n = 1, the indicial equation becomes

$(\lambda + 1)(\lambda+2) = l(l+1)$

The values of the $\lambda$ that solve for this equation are $(l-1),-(l+2)$

$\lambda = (l-1) , -(l+2)$

$\vdots$

And in general:

$\lambda_n = (l-n), -(l+n+1)$

$\lambda_{n_1} = (l-n)$, $\lambda_{n_2} = -(l+n+1)$

Here is the crux of it all.
Lets now write down the solution for this differential equation in its glory:

$y= \sum\limits_{n=0}^{\infty} \left(a_{n}x^{n + \lambda_{n_1}} + b_n x^{n+ \lambda_{n_{2}}}\right)$

$y= \sum\limits_{n=0}^{\infty} \left(a_{n}x^{n + (l-n)} + b_n x^{n-(l+n+1)}\right)$

$y= \sum\limits_{n=0}^{\infty} \left(a_{n}x^{l} + b_n x^{-(l+1)}\right)$

$y = \left(a_0 + a_1 + a_2 + \hdots \right) x^{l} + \left(b_0 + b_1 + b_2 + \hdots \right) x^{-(l+1)}$

$y = A x^{l} + B x^{-(l+1)}$

where A and B are constants.

This is the general solution to the Euler-Cauchy differential equation. 😀