How to Solve a 2nd Order Differential Equation with Singular Points?

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
4 replies · 3K views
jhon
Messages
21
Reaction score
0
I can not how define way to solve these equation

x(l-x)y''+4y'+2y=o

If anyone can help
 
Physics news on Phys.org
That looks like a pretty standard kind of equation. If you are given an initial value problem with y(a), y'(a) given and a is neither 0 nor 1, a standard series solution will work:
[tex]y= \sum_{n=0}^\infty a_nx^n[/tex].

Differentiate term by term and put into the equation to get a recursive equation for [itex]a_n[/itex].

If you are given y(0) and y'(0) you will need to use Frobenius' method with
[tex]y= \sum_{n=0}^\infty a_n x^{n+ c}[/tex]
for some number c (not necessarily a positive integer). Put that into the d.e. and look at the n=0 term to determine c.

If you are given y(l) and y'(l), similarly you will need to use Frobenius' method with
[tex]y= \sum_{n=0}^\infty a_n (x- l)^{n+ c}[/itex][/tex]
 
thnks HallsofIvy

but the question doesn't give me the initial value
 
I could be wrong (Still just a student) but even w/o initial conditions the series method should work. Just you won't know a0 and a1
 
Last edited:
Yes, but x= 0 and x= l are "singular" points. You may find that for some [itex]a_0[/itex] and [itex]a_1[/itex] your solution cannot be extended to x= 0 or x= 1.