# Series Solution of Differential equations

1. Nov 9, 2014

### Husaaved

Working through Mathematical Methods in the Physical Sciences by Boas, on the chapter on Series Solutions of Differential Equations, Boas works the example:

$y' = 2xy$

Boas differentiates the series representation of y yielding y', substitutes both into the original equation, and expands, showing that

$na_n = 2a_{n-2}$

I see that this is true from expanding each series representation and equating coefficients and so on, but I am having trouble deriving this result from the series notation. Is expansion absolutely necessary to obtain this result, or is there something I'm missing?

2. Nov 9, 2014

### vela

Staff Emeritus
There's something you're missing. What trouble do you run into?

3. Nov 9, 2014

### Husaaved

Oops, miss post

4. Nov 9, 2014

### Husaaved

So, when I change the indices, I'm getting:

$\sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=1}^\infty 2a_{n+1}x^{n+2}$

or

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

-- which doesn't quite lead me to the $na_n = 2a_{n-2}$ that Boas ends up with.

I have solved these problems before by expansion and equating coefficients, so I understand these problems conceptually, I'm just not clear on how to manipulate the series notation to yield these more compact results.

Should I be manipulating the indices to match one another, and should I omit x from the expression altogether since it's the coefficients I'm worried about anyway?

Thank you.

5. Nov 9, 2014

### HallsofIvy

Staff Emeritus
I'm not sure what you mean by "expanding" each series representation.

If we have $y= \sum_{n=0}^\infty a_nx^n$ then
$y'= \sum_{n=1}^\infty n a_nx^{n-1}$
(Notice this sum starts at n= 1 because when n= 0, "$n a_nx^n= 0$".

So we have $\sum_{n=1}^\infty n a_nx^{n-1}= \sum_{n=0}^\infty 2a_nx^{n+1}$.

I think that by "expanding" the series you mean writing out the actual terms. No, you don't have to do that- if two such series are equal for all x, then the coefficients, on each side, of the same powers of x. But in order to use that, we need to change the "index" in each sum so that we have the same power of x.

That is, since "n" is "dummy index, in only has meaning inside the sum and is not part of the actual value, we can change the index on each side separately.
On the left, let j= n- 1. Then n= j+ 1, when n= 1, j= 0 so the sum becomes $\sum_{j=0}^\infty (j+ 1)a_{j+ 1}x^j$
On the right, let j= n+ 1. Then n= j- 1, when n= 0, j= 1 so the sum becomes $\sum_{j= 1}^\infty 2a_{j-1}x^j$.

First, we see that the left hand sum now starts at j= 0 while the left hand sum starts at j= 1. That is, for j= 0, we have $(0+1)a_{0+ 1}= a_1= 0$.

For j> 1, we have $(j+1)a_{j+1}= 2a_{j-1}$ or $a_{j+1}= \frac{2}{j+1}a_{j-1}$. For example, if j= 1, that says that $a_2= \frac{2}{2}a_0= a_0$. If j= 2, $a_{3}= \frac{2}{3}a_1= 0$. If j= 3, $a_4= \frac{2}{4}a_2= \frac{1}{2}a_2= \frac{1}{2}a_0$, etc. It should be clear that $a_n= 0$ for all odd n while $a_n$ , for every even n, will be a multiple of $a_0$ which is the "undetermined constant" we expect when solving a first order differential equation.

We already have $x_2= a_0$ and $x_4= \frac{1}{2}a_0$. Now, with n= 5, $a_6= \frac{2}{6}a_4= \frac{1}{3}a_4= \frac{1}{6}a_0$. With n= 7, $a_8= \frac{2}{8}a_6= \frac{1}{4}a_6= \frac{1}{24}a_0$. Notice that the denominators are factorials: in fact the denominator for the even index 2n is n!

Now, use "proof by induction", along with the equation $a_{j+1}= \frac{2}{j+ 1} a_j$ to show that $a_{2n}= \frac{1}{n!}a_0$.