Regular singular points of 2nd order ODE

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
1 reply · 2K views
Jerbearrrrrr
Messages
124
Reaction score
0

Homework Statement


[PLAIN]http://img265.imageshack.us/img265/6778/complex.png

I did the coefficient of the w' term. What about the w term?

This seems like a fairly standard thing, but I can't seem to find it anywhere.
What ansatz should I use for q, if the eqn is written w''+pw'+qw?
C/(z-a)²+ D/(z-b)²?
Any conditions, except for the one generated by z->1/t substitution?
Or should I use C+cz on the top, etc?
 
Last edited by a moderator:
Physics news on Phys.org
Jerbearrrrrr said:

Homework Statement


[PLAIN]http://img265.imageshack.us/img265/6778/complex.png

I did the coefficient of the w' term. What about the w term?

This seems like a fairly standard thing, but I can't seem to find it anywhere.
What ansatz should I use for q, if the eqn is written w''+pw'+qw?
C/(z-a)²+ D/(z-b)²?
Any conditions, except for the one generated by z->1/t substitution?
Or should I use C+cz on the top, etc?

Well, if a 2nd order ODE has regular singular points at [itex]z=a[/itex] and [itex]z=b[/itex], then [itex]q[/itex] has poles up to 2nd order at those points, and the most general form of [itex]q[/itex] is then

[tex]q(x)=\frac{g(z)}{(z-a)^2(z-b)^2}[/tex]

where [itex]g(z)[/itex] is analytic everywhere. You should have used similar reasoning to find

[tex]p(z)=\frac{f(z)}{(z-a)(z-b)}[/itex]<br /> <br /> Then just apply the linearity condition to find [itex]f[/itex] and [itex]g[/itex].[/tex]
 
Last edited by a moderator: