Solving ODEs w/ Frobenius Method: Q on b, c as Funcs of x

[Master J]

I have been looking at the Frobenius method for solving ODEs of the form. I have a few questions on it.


(x^2)y'' + xby' + cy = 0

If b and c are functions of x, does one use the Frobenius method, where as if they are constants, it is an Euler Cauchy equation and you use y = x^r ??

Thats the first Q. anyway.

Thanks folks!

 

[matematikawan]

y=x^r only works for the Euler-Cauchy equation, i.e. b and c are constants.

If b and c are functions of x, you cannot use the try function y=x^r.

 

[Master J]

Thanks for cledaring that up.

I'm having trouble applying the method. My textbook, (which I won't name but it's approach and exlanation in this section is absolutely terrible) isn't helping me much.

I have been trying to solve, for example,

xy'' + 5y' + xy = 0

So I get

SUM(n + r)(N + r -1)(a_n)x^(n + r -2) + SUM(5)(n + r)(a_n)x^(n + R -2) + SUM(a_n)x^(n + r +1) = 0

where SUM is the sum to infinity from n = 0.

and the general solution is of form y=(x^r)SUM(a_n)(x^n)


I don't know what to do know. The book's next steps are done without explanation really.
Can someone help me??

 

[matematikawan]

x=0 is a regular singular point. There is at least one solution of the form y=(x^r)SUM(a_n)(x^n)
where r satisfy the indicial equation r²+4r=0.
r₁=0 , r₂=-4 and r₁-r₂ is an integer in this case.

The Frobenius method only guarantee for r=0 (the larger root) but not for r=-4 (but there is no harm for trying)

If you know anything about Bessel equation, I would suggest you solve the equation using the substitution z=x²y.