Singular points

[JJ007]

$\displaystyle x^2(x-1)y" - xy' + (x+1)y = 0$

So my singular points are x=0, x=1
x=0 is regular
x=1 is regular

Given:
$\displaystyle y= x^r \sum_0^{\infty}C_nx^n$ C_0 is not equal to 0.

How do you find r, C1, C2 and verify two solutions given by series are linearly independent ?

 

[galactus]

If $\displaystyle r_{1}$ and $\displaystyle r_{2}$ are distinct and do not differ by an integer, then there are two linearly independent solutions of the form:

$\displaystyle y_{1}=\sum_{n=0}^{\infty}c_{n}x^{n+r_{1}}, \;\ c_{0}\neq 0$

$\displaystyle y_{2}=\sum_{n=0}^{\infty}b_{n}x^{n+r_{2}}, \;\ b_{0}\neq 0$

Solving DE's using series is rather tedious to type out here.

But, we start with:

$\displaystyle y=\sum_{n=0}^{\infty}c_{n}x^{n+r}$

$\displaystyle y'=\sum_{n=0}^{\infty}(n+r)c_{n}x^{n+r-1}$

$\displaystyle y''=\sum_{n=0}^{\infty}(n+r)(n+r-1)c_{n}x^{n+r-2}$

make the subs into the given DE:

$\displaystyle x^{2}(x-1)\sum_{n=0}^{\infty}(n+r)(n+r-1)c_{n}x^{n+r-2}-x\sum_{n=0}^{\infty}(n+r)c_{n}x^{n+r-1}+(x+1)\sum_{n=0}^{\infty}c_{n}x^{n+r}=0$

$\displaystyle \sum_{n=0}^{\infty}(n+r)(n+r-1)c_{n}x^{n+r+1}-\sum_{n=0}^{\infty}(n+r)(n+r-1)c_{n}x^{n+r}-\sum_{n=0}^{\infty}(n+r)c_{n}x^{n+r}+\sum_{n=0}^{\infty}c_{n}x^{n+r+1}+\sum_{n=0}^{\infty}c_{n}x^{n+r}=0$

$\displaystyle x^{r}\left[\sum_{n=0}^{\infty}\left(r^{2}+(2n-1)r+n^{2}-n+1\right)c_{n}x^{n+1}-\sum_{n=0}^{\infty}\left(r^{2}+2nr+n^{2}-1\right)c_{n}x^{n}\right]=0$

Man, this is tedious business. It has been awhile since I delved into this stuff. But, continue by adjusting so the indices are the same.

The C's can be found by using initial conditions. r can be found when the indices are adjusted. That is, find the n=0 case.

Letting k=n+1 will aid in setting the indices equal.

In this problem, I am pretty sure we get

$\displaystyle r^{2}-1=0$. Which means $\displaystyle r=\pm 1$

Then, the iterations come in.