1. ## Power Series

$\displaystyle y''+3y'+(1-x)y=0$

I used these:
$\displaystyle y=\sum_{n=0}^{\infty} a_{n}x^{n}$

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

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

I Substitued the values in the origional equation:
$\displaystyle \sum_{n=2}^{\infty} (n)(n-1)(a_{n})x^{n-2}+3\sum_{n=1}^{\infty} (n)(a_{n})x^{n-1}+(1-x)\sum_{n=0}^{\infty} a_{n}x^{n}=0$

Next, I multiplied the $\displaystyle 3$ in, and the $\displaystyle (1-x)$:
$\displaystyle \sum_{n=2}^{\infty} (n)(n-1)(a_{n})x^{n-2}+ \sum_{n=1}^{\infty} (3n)(a_{n})x^{n-1}+ \sum_{n=0}^{\infty} a_{n}x^{n}- \sum_{n=0}^{\infty} a_{n}x^{n+1}=0$

Now, I need all of these with the same starting value for $\displaystyle n$ and have $\displaystyle x$ to the same power.
*Here is where i get messed up*
$\displaystyle \sum_{n=0}^{\infty} (n+2)(n+1)(a_{n+2})x^{n}+ \sum_{n=0}^{\infty} (3)(n+1)(a_{n+1})x^{n}+ \sum_{n=0}^{\infty} a_{n}x^{n}- \sum_{n=0}^{\infty} a_{n}x^{n+1}=0$

The last term in is sum has the $\displaystyle x^{n+1}$ not $\displaystyle x^{n}$, so did I do this wrong or how do I find the recursive relation?

Questions:
How do i find the recursive relation?
What is the recursive relation?

This is my FIRST time trying to use power series with differential equations so in the explanation if you could explain everything with a lot of detail(not skipping steps) that would be great.

EDIT: I used http://www.stewartcalculus.com/data/...lveDEs_Stu.pdf for reference.

2. When you are changing the indexing, set the exponents of x equal to k.
$\displaystyle k=n-2$; $\displaystyle k=n-1$; $\displaystyle k=n$; and $\displaystyle k=n+1$

Now all you exponents are just k.

When you solve for n, you simple sub what n equals in for n of the giving sum.

For instance, in the first sum, $\displaystyle n=k+2$

3. I dont understand what you mean, could you show me? The exponents are all fine except for the last of the sums. If I change the exponent of x, then the starting value of the sum will change, which i dont want.

4. Mr. Fantastic, who is an admin, deletes my post if I post the solution.

I made a mistake in my first comment but here is what you need to do.

In your final sum where you have all $\displaystyle x^n$ and one $\displaystyle x^{n+1}$, say $\displaystyle k=0$ and $\displaystyle k=n+1$.

Sub into summations.

You will now have 3 $\displaystyle \sum_{k=0}^{\infty}$ and $\displaystyle \sum_{k=1}^{\infty}$.

However, by making the sub, you will now have the final $\displaystyle x^k$ and all the other xs will be of the k power.

For the 3 sums that don't start at 1, just take the first term of each one. this bump them from $\displaystyle k=0$ to $\displaystyle k=1$.

Here is an example so you see:

$\displaystyle 2a_{2}+\sum_{k=1}^{\infty}(k+2)(k+1)a_{k+2}x^k$

That was your first sum. Now do that for all others that start at 0.

5. ## Thanks

THANKS!!! Now I have one last question...I hope.

How do I: Use the recursion relation to find the series expansion through $\displaystyle x^{5}$ for $\displaystyle y(0)=a_{0}$ and $\displaystyle y'(0)=a_{1}$

Here are my recursion relations that I found by plugging the n values from 0 to 3 in my recursion relation. Then I got them all in terms of just $\displaystyle a_{0}$ and $\displaystyle a_{1}$. From here how do i find the series expanion mentioned above?:

For $\displaystyle n=0$

$\displaystyle a_{2}=-\dfrac{3}{2}a_{1}-\dfrac{1}{2}a_{0}$
--------------------------------------
For $\displaystyle n=1$

$\displaystyle a_{3}=\dfrac{4}{3}a_{1}+\dfrac{2}{3}a_{0}$
--------------------------------------
For $\displaystyle n=2$

$\displaystyle a_{4}=-\dfrac{19}{24}a_{1}-\dfrac{11}{24}a_{0}$
--------------------------------------
For $\displaystyle n=3$

$\displaystyle a_{5}=-\dfrac{14}{15}a_{1}-\dfrac{31}{60}a_{0}$

6. You need to show $\displaystyle a_0$ and $\displaystyle a_1$ since they are the key.

Also, I would list a few more maybe down to n=8

7. How does and come into play? Is this saying that:

$\displaystyle a_{0}=1$ because $\displaystyle y=\sum_{n=0}^{\infty} a_{n}x^{n}$ (sub in 0 for x)

$\displaystyle a_{1}=0$ because $\displaystyle y=\sum_{n=1}^{\infty} (n)a_{n}x^{n-1}$ (sub in 0 for x)

8. Originally Posted by dwsmith

$\displaystyle 2a_{2}+\sum_{k=1}^{\infty}(k+2)(k+1)a_{k+2}x^k$

That was your first sum. Now do that for all others that start at 0.
You should have, I think, two other constants out side the sum. You need to take them and set them equal to zero.

9. I got it! Thanks for all of the help!