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### Frobenius Series Solutions

```Date: 05/01/2008 at 21:32:36
From: Judy
Subject: Frobenius series solution

Determine whether the differential equation:

(x^2)y'' + xy' + ((x^2)-8)y = 0

has two linearly independent Frobenius series solutions.

I am confused about the rest of the Frobenius series and how to show
that they are linearly independent.

I divided through by x^2 to get the general form of the equation:

y'' + (y'/x) + [((x^2)-8)y)/x^2]

Po=1 and Qo=1 hence the indicial eqn is: r(r-1) + r - 1
(r^2) - 1 = 0 therefore r=1,r=-1 and the Frobenius solutions are
y = xsigmaAnx^n.

```

```
Date: 05/02/2008 at 18:49:15
From: Doctor Jordan
Subject: Re: Frobenius series solution

Hi Judy,

There is a general theorem that tells us that the two functions y_1
and y_2 we get are linearly independent solutions.  This theorem tells
us what the solutions look like in the three different cases where r_1
- r_2 is not an integer, is 0 or is a positive integer, where r_1,r_2
are the roots of the indicial equation, with r_1 > r_2.

p(x) is the coefficient of y' and q(x) is the coefficient of the
constant term, thus here

p(x) = 1/x
q(x) = (x^2-1)/8

p_0 is equal to x*p(x) when we set x=0, thus p_0 = 1.
q_0 is equal to x^2*q(x) when we set x=0, thus q_0 = -8.

The indicial equation is

r(r-1) + p_0*r + q_0 = 0.

Thus the indicial equation is

r(r-1) + r - 8 =0

i.e.

r^2 - 8 = 0

which has the roots sqrt(8) and -sqrt(8).  Let r_1 = sqrt(8) and
r_2 = -sqrt(8).

r_1 - r_2 = sqrt(8) + sqrt(8) = 2*sqrt(8), which is not an integer.
Therefore two linearly independent solutions are given by

y_1 = sum_{n=0}^infinity a_n x^{n+r_1}

y_2 = sum_{n=0}^infinity b_n x^{n+r_n}

for x > 0, where a_n = c_n(r_1) and b_n = c_n(r_2).

y(x) = sum_{n=0}^infinity c_n(r) x^{n+r}.

We put y(x) into the given differential equation

(x^2)y'' + xy' + ((x^2)-8)y = 0

and find a formula for c_n(r).  Then we take r = sqrt(8) to find the
formula for a_n, and r = -sqrt(8) to find the formula for b_n, and
this tells us what the series for y_1 and y_2 are.

Please write back if you have difficulty finding a formula for c_n(r),
or if you have other questions about the Frobenius method or solving
differential equations with series.

- Doctor Jordan, The Math Forum
http://mathforum.org/dr.math/

```

```
Date: 05/04/2008 at 00:11:22
From: Judy
Subject: Thank you (Frobenius series solution)

Thanks for helping me answer this question.  I really appreciate it.
```
Associated Topics:
College Calculus

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