# Frobenius method non-polynomial coeficients.

1. Aug 4, 2010

### bp_psy

1. The problem statement, all variables and given/known data
Find the first 3 terms of two independent series solution for the DE.

xy''+2xy'+(6e^x)y=0

2. Relevant equations
Frobenius method. case r1-r2=integer

3. The attempt at a solution
I found that 0 is a regular singular point. I found the indicial equation and found the roots.I then differentiate the Frobenius series and plug everything in the DE. After changing the coefficients I get

$$r(r-1)a_{0}+\sum(n+r)(n+r-1)a_{n}x^{n+r-1}+\sum((2(n+r-1))a_{n-1}x^{n+r-1}+6e^{x}(\sum(a_{n-1}x^{n+r-1} )=0$$
or
$$r(r-1)a_{0}+\sum(n+r)(n+r-1)a_{n}x^{n+r-1}+\sum((2(n+r-1))a_{n-1}x^{n+r-1}+6(1+\sum(\frac{x^n}{n!}))(\sum(a_{n-1}x^{n+r-1} )=0$$
with all the summations starting at 1. My problem is that I have no clue how to find a recurrence relations from this.Do I really have to multiply the last two series and hope for some miracle or there is some other way. My books show no examples of series solutions to this kinds of problems only ones with polynomial coefficients.

2. Aug 4, 2010

### vela

Staff Emeritus
Since the problem is only asking for the first three terms of the solutions, you don't need to find the recurrence relation. Just keep enough terms to solve for a0, a1, and a2.

3. Aug 5, 2010

### bp_psy

Yes that works. Thank you.