Solving 2nd order PDE of single variable

[nigels]

I've been getting pretty rusty in terms of derivation in recent years. Encountered this problem which I can't derive the steps despite knowing the solution.

\frac{\partial^2 u}{\partial r^2} + \frac{\partial u}{\partial r}\left(\beta + \frac{1}{r}\right)+\frac{\beta}{r}u=0

Known solution:

u(r) = \beta \cdot \exp(-\beta r)

Thank you very much for your help!

 

[Jufro]

Since the differential equation has the form of:

(\frac{∂}{∂r})² u(r) + a(r) \frac{∂}{∂r} u(r) + b(r) u(r) =0​

With a(r) having a simple pole at r=0, and b(r) having no more than a pole of order 2 at r=0, it is sufficient to continue with the Frobenius Method:

u(r) = r^σ \sum k_nrⁿ​

You can then substitute this into the equation to get something like:

\sum k_n ((n+σ)(n+σ-1)+(n+σ)+(n+σ+1)βr) r^n+σ-1 = 0​

This means that the indicial equation becomes:

σ²-σ+σ=0
σ=0, multiplicity 2.​

Using σ=0 the next step is to determine the recursion relationship which will give you k_n. This get's one independent solution that you can show is the same as the series for βe^-βr.
And then you're done.