ODE Series Solution Near Regular Singular Point, x^2*y term?

[abalmos]

ODE Series Solution Near Regular Singular Point, x^2*y term? (fixed post body)

Homework Statement

Find the series solution (x > 0) corresponding to the larger root of the indicial equation.

$$5x^{2}y'' + 4xy' + 10x^{2}y = 0$$

Homework Equations

Solution form:

$$y = \sum_{i=0}^{\infty}a_{n}x^{r+n}$$

The Attempt at a Solution

$$\lim_{x\to\0}\frac{4x}{5x^{2}}x = \frac{4}{5} < \infty$$

$$\lim_{x\to\0}\frac{10x^{2}}{5x^{2}}x^{2} = 2x^{2} = 0 < \infty$$

The singular point is regular, and the solution should be in the form of (and its derivatives):

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

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

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

Plug the solution form into the ODE:

$$5x^{2}\sum_{n=0}^{\infty}a_{n}(r+n)(r+n-1)x^{r+n-2} + 4x\sum_{n=0}^{\infty}a_{n}(r+n)x^{r+n-1} + 10x^{2}\sum_{n=0}^{\infty}a_{n}x^{r+n}$$

Multiply the x terms into the series:

$$5\sum_{n=0}^{\infty}a_{n}(r+n)(r+n-1)x^{r+n} + 4\sum_{n=0}^{\infty}a_{n}(r+n)x^{r+n} + 10\sum_{n=0}^{\infty}a_{n}x^{r+n+2}$$

Here is where my trouble lies. My understanding is that I am trying to pull out the $$a_{0}$$ term of the series which will leave behind the indicial equation I am also to combine all of the series into one, factor out the x term, and then find the recurrence relation for the series term.

My problem specifically is the last term, which now has a $$x^{r+n+2}$$ that I can't fit into the rest of the series.

These are my attempts:

1) Blindly pull out $$a_{0}$$:

$$5a_{0}(r)(r-1)x^{r} + 4a_{0}r^{r} + 10a_{0}x^{r+2} + 5\sum_{n=1}^{\infty}a_{n}(r+n)(r+n-1)x^{r+n} + 4\sum_{n=1}^{\infty}a_{n}(r+n)x^{r+n} + 10\sum_{n=1}^{\infty}a_{n}x^{r+n+1}$$

Stuck because of the $$x^{r+2}$$ term in the indicial equation and the last series term $$x^{r+n+2}$$ is still an issue...

2) Try to combine series first

Can't find a way to shift the indexes so that they are all the same and have an x exponent that will allow me to factor them out after I combine the series.

3) Desperately ignore term 3 and pull out $$a_{0}$$ from just the first two terms, even though this seems very wrong.

$$5a_{0}(r)(r-1)x^{r} + 4a_{0}r^{r} + 5\sum_{n=1}^{\infty}a_{n}(r+n)(r+n-1)x^{r+n} + 4\sum_{n=1}^{\infty}a_{n}(r+n)x^{r+n} + 10\sum_{n=1}^{\infty}a_{n}x^{r+n+1}$$

leaving the indicial equation of:

$$a_{0}(5r^{2}-r) = 0$$
$$r = 0, \frac{1}{5}$$

This is somewhat rewarding, 1/5 is available choice for the r value (multiple choice question), and r = 0 seems to be correct for the version of the question which ask to find the solution for the lowest r.

But I am still not sure how to combine the series...

I have looked and many examples online but I am unable to find an example which ends up with $$r+n+2$$ exponent for x and I am thoroughly out of ideas.

Any help would be greatly appreciated.

Thanks!
- Andrew Balmos

 

[Brian-san]

Multiply the x terms into the series:

$$5\sum_{n=0}^{\infty}a_{n}(r+n)(r+n-1)x^{r+n} + 4\sum_{n=0}^{\infty}a_{n}(r+n)x^{r+n} + 10\sum_{n=0}^{\infty}a_{n}x^{r+n+2}$$

Here is where my trouble lies. My understanding is that I am trying to pull out the $$a_{0}$$ term of the series which will leave behind the indicial equation I am also to combine all of the series into one, factor out the x term, and then find the recurrence relation for the series term.

The main trick with this solution method is that you can rename the indices. In this case you change the index in the final sum so it matches the first two terms. Let n=n-2, then the lower index on the sum is n=2 and the third term is
$$10\sum_{n=2}^{\infty}a_{n-2}x^{r+n}$$

Now you can pull out the terms for n=0 and n=1 and combine all three sums into one summation (from n=2 to infinity), which will give the recurrence relation. You'll find that the r values from the indicial equation for $a_0$ is the same as what you calculated, since the third sum doesn't contribute an $a_0$ term.

 

[abalmos]

Brian-san:

Thank you very much for your help! Unfortunately I got stuck again.

Shifting the other index's to the right twice to match the last series term was a great idea, but I fear now you can not find a recurrence relation.

Currently this is what I got:

$$a_{0}r(5r-1)x^{r} + a_{1}(5r^{2} + 9r + 4)x^{r+1} + \sum_{n=2}^{\infty}(5a_{n}(r+n)(r+n-1) + 4a_{n}(r+n) + 10a_{n})x^{r+n}$$

So like before $$r = { 0, \frac{1}{5} }$$

In a attempt to find the recurrence relation I put $$r = \frac{1}{5}$$ in the final series:

$$\sum_{n=2}^{\infty}(5a_{n}(\frac{1}{5} + n)(n - \frac{4}{5}) + 4a_{n}(\frac{1}{5} + n) + 10a_{n})x^{n + \frac{1}{5}}$$

Which forces all $$a_{n} = 0$$

Which while it would be nice, but clearly I a still making a mistake somewhere.

Of course I realize that I need to be using the $$a_{1}(5r^{2} + 9r + 4)x^{r+1}$$ terms but I am not sure how to do that.

Maybe relate the $$a_{1}$$ to the $$a_{n}$$ terms? I am not sure how that could be done usefully...

Thank for any more help you can provide,
- Andrew Balmos

 

[abalmos]

Brain-san:

I missed a crucial detail in your post yesterday!

When I shifted the index I forgot to minus the index shift off the n's in the series, when you do that you have a $$a_{n-2}$$ term which can be used to find the recurrence relation.

Its always the obvious things I miss for me...

Thank you very much!

- Andrew Balmos