# Series solution about a singular point--help with y(2)

• Sep 7th 2012, 08:57 PM
phys251
Series solution about a singular point--help with y(2)
The given equation is x(x-1)y'' + 3y' - 2y = 0.

I was able to discern that a(k+1) = a(k) * [(k+r)(k+r-1)-2]/[(k+r+1)(k+r-3)], and the roots are 0 and 4. That recursive equation gives me the correct coefficients for a(0) through a(3), and of course, a(4) is undefined. But I can't get it right from a(5) on. I used the root 0 to get those first few a's; was I supposed to switch over to 4?
• Sep 8th 2012, 04:54 AM
BobP
Re: Series solution about a singular point--help with y(2)
Regard the \$\displaystyle a_{4}\$ coefficient from the \$\displaystyle r=0\$ case as being arbitrary and then calculate the remaining coefficients in terms of \$\displaystyle a_{4}.\$
That gets you a solution containing two arbitrary constants, \$\displaystyle a_{0}\$ and \$\displaystyle a_{4},\$ in which case you have the general solution of the differential equation.
If you generate the series based on \$\displaystyle r=4,\$ you will find that it duplicates the second part of the \$\displaystyle r=0\$ solution
• Sep 8th 2012, 11:59 AM
phys251
Re: Series solution about a singular point--help with y(2)
Oh goodness--I made the remaining terms in terms of a(5), not a(4). Changing that fixed the problem.