Solution of y''-(1+x)y=0 using power series ..

**Liverpool** · January 14th 2011, 07:09 AM

Hello

Obtain the solution of the following differential equation as a power series about the origin :

$y''-(1+x)y=0$

Solution:

Let $\displaystyle y=\sum_{n=0}^{\infty} a_n x^n$
then :

$\displaystyle y'=\sum_{n=1}^{\infty} n a_n x^{n-1}$

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

Rewrite the equation as : $y''-y -xy=0$

and substitute the values of y & y'' gives:

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

or:

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

My problem is in the last summation
I want to make the summations start with the same n and the powers of the x's are n in order to get the recursive relation, but I can not do that .

the last summation makes my life hard

any help?

**Ackbeet** · January 14th 2011, 07:36 AM

Try breaking off the first term of the last summand thus:

$\displaystyle \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^{\infty} a_n x^n - a_{0}x-\sum_{n=1}^{\infty} a_n x^{n+1}=0,$ then re-indexing gives

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

You might need to break off other terms as well, in order to get all the powers of x to be the same. Those are your two tricks: re-indexing and breaking off terms. You should be able to get all the starting indices and powers of x to be the same in all three summations, which would enable you to combine them into one summation.

**HallsofIvy** · January 14th 2011, 10:11 AM

I'm not sure why Ackbeet put the last sum to power $x^{n+2}$ . I would think you would want them all as $x^n$ so you could combine the sums. One way of doing that is, recognizing that you want to make " $x^{n+1}$ " look like " $x^j$ ", set j= n+1. Then n= j- 1 so that $a_n= a_{j-1}$ . Also when n= 0, j= 1 so the sum $\sum_{n=0}^\infty a_nx^{n+1}$ becomes $\sum_{j= 1}^\infty a_{j-1}x^j$ . Relabel "j" to be "n" and you have
$\displaystyle \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^{\infty} a_n x^n - a_{0}x-\sum_{n=1}^{\infty} a_{n-1} x^{n}=0$
Since the last sum does not start until n= 1, as Ackbeet suggests, do n= 0 separately:
With n= 0,
$2a_2- a_0= 0$
and then
$\displaystyle \sum_{n=1}^{\infty} (n+2)(n+1) a_{n+2} x^n - \sum_{n=1}^{\infty} a_n x^n - a_{0}x-\sum_{n=1}^{\infty} a_{n-1} x^{n}=0$
so that for each n, greater than or equal to 1,
$(n+2)(n+1)a_{n+2}- a_n- a_{n-1}= 0$
or
$a_{n+2}= \frac{a_n+ a_{n-1}}{(n+2)(n+1)}$ .

Presumably, you have some "initial conditions" that will give you $a_0$ ( $y(0)= a_0$ ) and $a_1$ ( $y'(0)= a_1$ ). Then the first equation tells you that $a_2= a_0/2$ so that you have $a_0$ , $a_1$ , and $a_2$ . Then you can find $a_3$ by taking n= 1 in the last equation, find $a_4$ by taking n= 2, etc. With luck, you might be able to find a general formula but that is rare.

**zzzoak** · January 14th 2011, 10:54 AM

I may be wrong, but for n=1 (as there is term $-a_0x$ ) it must be
written separately

$ n=0: \; \; 2a_2-a_0=0 $

$ n=1: \; \; 6a_3-a_1-2a_0=0 $

$ n>=2: \; \; (n+2)(n+1)a_{n+2}-a_n-a_{n-1}=0 $

**HallsofIvy** · January 14th 2011, 01:48 PM

The " $a_0x$ " term is my mistake. Being basically lazy, I copied the LaTex version of the equation from Ackbeet's post but, being basically careless, I copied the wrong equation! I intended to copy the first equation before he had separated that part out. What I should have had was:
$\displaystyle \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^{\infty} a_n x^n - \sum_{n=1}^{\infty} a_{n-1} x^{n}=0$

I just copied it again, but this time I took out the " $-a_0x$ " part!

**Liverpool** · February 9th 2011, 02:22 AM

Thanks all.

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