Frobeniuns Method/Generalized Power Series to DiffEQ solutions

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(Working out of Boas chapter 12, section 11)

[tex]
3xy'' + (3x + 1)y' + y = 0
[/tex]

I'm asked to solve the differential equation using the method of Frobenius but I'm finding the way Boas introduces/explains/exemplifies the method to be incredibly confusing. So, I used some google-fu and was even more confused. Seems like everyone has a different plan of attack for these problems.

What I've done so far is to assume
[tex]
y = \sum_{n=0}^{\infty} c_n x^{n+s}
[/tex]

*I know in a normal expansion it's simpy [itex]x^n[/itex] but from my understanding we're to multiply the summation by another factor of [itex]x^s[/itex], or whatever variable we choose.

...doing like wise for the respective derivatives:
[tex]
y' = \sum_{n=0}^{\infty} c_n (n+s) x^{n+s-1}
[/tex]

[tex]
y'' = \sum_{n=0}^{\infty} c_n (n+s-1) (n+s) x^{n+s-2}
[/tex]

then we substitute these expansions into the respective places within the original equation.

Now this is where I'm getting REALLY REALLY confused. Boas says to make a table and then from there find the indicial equation. From various pdf's and youtube videos I'm getting different information. Can anyone please point me onto the right path?

Thanks.
 

Answers and Replies

  • #2
MathematicalPhysicist
Gold Member
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You just need to equate powers of $x$ by changing dumby variables of the different sums.

I can do this, though I will get another infraction from the moderators, so I better not.
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,833
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(Working out of Boas chapter 12, section 11)

[tex]
3xy'' + (3x + 1)y' + y = 0
[/tex]

I'm asked to solve the differential equation using the method of Frobenius but I'm finding the way Boas introduces/explains/exemplifies the method to be incredibly confusing. So, I used some google-fu and was even more confused. Seems like everyone has a different plan of attack for these problems.

What I've done so far is to assume
[tex]
y = \sum_{n=0}^{\infty} c_n x^{n+s}
[/tex]

*I know in a normal expansion it's simpy [itex]x^n[/itex] but from my understanding we're to multiply the summation by another factor of [itex]x^s[/itex], or whatever variable we choose.

...doing like wise for the respective derivatives:
[tex]
y' = \sum_{n=0}^{\infty} c_n (n+s) x^{n+s-1}
[/tex]
So that [itex](3x+1)y'= \sum_{n=0}^\infty 3c_n(n+s)x^{n+s}+ \sum_{n=0}^\infty c_n(n+s)x^{n+s- 1}[/itex]

[tex]
y'' = \sum_{n=0}^{\infty} c_n (n+s-1) (n+s) x^{n+s-2}
[/tex]
So that [itex]3xy''= \sum_{n=0}^\infty 3c_n(n+s-1)(n+s)x^{n+s-1}[/itex].

then we substitute these expansions into the respective places within the original equation.

Now this is where I'm getting REALLY REALLY confused. Boas says to make a table and then from there find the indicial equation. From various pdf's and youtube videos I'm getting different information. Can anyone please point me onto the right path?

Thanks.
Putting those into your equations and combinging like powers will give sums in [itex]x^{n+s}[/itex] and [itex]x^{n+s-1}[/itex]. The crucial point is that we could choose "s" to be anything- we choose it so that the lowest power of x is [itex]x^0= 1[/itex]. That is, we choose s to be such that this infinite series contains a constant term. The lowest power of x in each sum is with n=0 so that the lowest power term is [tex]x^{s-1}[/itex]. You want to determine s so that coefficient is not 0. That is the "indicial equation".
 

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