# Solving an ODE with variable coefficients

1. Dec 5, 2014

### c0der

1. The problem statement, all variables and given/known data
Solve the following:

y'' = c2 / (x2 + c1*x) * y
c1, c2 are constants, x is variable

2. Relevant equations
As above

3. The attempt at a solution
I have used the method of Frobenius and regular power series and obtained an infinite series on top of an infinite series, which is indefinite. What are the normal methods to solve that ODE? It's not quite in the Frobenius or Bessel forms

2. Dec 5, 2014

### ShayanJ

I used a regular power series and there was no problem. Just multiply the equation by $x^2+c_1 x$ to get $x^2 y''+c_1 x y''-c_2 y=0$.
The regular series gives only one of the solutions.So you should also consider a solution of the form $y=\sum_{n=0}^\infty a_n x^{n+m}$.

Last edited: Dec 5, 2014
3. Dec 5, 2014

### Ray Vickson

Maple gets a solution in terms of hypergeometric functions with (variable) argument $-c_1/x$. Thus, the power-series expansion of Maple's solution will be in powers of $1/x$, not of $x$ itself.

4. Dec 13, 2014

### c0der

Thank you, I have done this however as follows:

Equating coefficients of x0 gives:
a0=0

For x1:

a2 = c2/2c1

Equating coefficients of xm:

am+1 = [ c2 - (m-1)m ] / [ c1m(m+1) ] am for m>=2

Then:

a3 = [ (c2 - 2) / 3!2!c12 ] a1

a4 = [ (c2 - 6)(c2 - 2)c2 / 4!3!c13 ] a1

a5 = [ (c2 - 12)(c2 - 6)(c2 - 2)c2 / 5!4!c14 ] a1

a6 = [ (c2 - 20)(c2 - 12)(c2 - 6)(c2 - 2)c2 / 6!5!c15 ] a1

How can I correlate this with the hypergeometric series or any other recognizable series? The numerator terms in the brackets have a pattern to it 2+4=6 6+6=12 12+8=20 etc but getting the terms in powers of m is difficult

5. Dec 13, 2014

### haruspex

Is that (c2 / (x2 + c1*x)) * y or c2 / ((x2 + c1*x) * y)?

6. Dec 14, 2014

### c0der

(c2 / (x2 + c1x)) * y