need help with ODEs (power series)

**Awara** · March 6th 2008, 04:52 PM

Find the general solution of the form $ y = \sum_{n = 0}^{\infty}a_nx^n$ for the differential equation $y''+xy'+y=0$ .
Derive from the general solution that e^(-(x^2)/2) is a solution of the differential equation.

I know how to solve the DE using the series, but do not know specifically how to show that one of the solutions is e^(-(x^2)/2) as asked by my teacher. please help in here

**TheEmptySet** · March 6th 2008, 05:01 PM

Originally Posted by Awara

Find the general solution of the form $ y = \sum_{n = 0}^{\infty}a_nx^n$ for the differential equation $y''+xy'+y=0$ .
Derive from the general solution that e^(-(x^2)/2) is a solution of the differential equation.

I know how to solve the DE using the series, but do not know specifically how to show that one of the solutions is e^(-(x^2)/2) as asked by my teacher. please help in here

What do your series solutions look like?

$f(x)=e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$

Then

$f(\frac{-x^2}{2})=e^{\frac{-x^2}{2}}=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{((2^{n})n!)}$

if it looks like the last one you are done.

**Awara** · March 6th 2008, 05:19 PM

the problem is that i cant solve the ode to match that condition that you've stated.. i end up with $\sum_{n = 0}^{\infty}x^n[(n+2)(n+1)a_{n + 2}+na_n+(n+1)a_{n+1}]=0 $ so we can that state that
$a_{n + 2}=(-na_n/(n+2)(n+1))-a_{n+1}/(n+2)$

i tried using n values 0,1,2,3 but do not know how to put them together to match $\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{((2^{n})n!)}$

id appreciate it if you can help me on this because its a killer quetsion

**Jhevon** · March 6th 2008, 05:22 PM

Originally Posted by Awara

the problem is that i cant solve the ode to match that condition that you've stated.. i end up with $ y = \sum_{n = 0}^{\infty}x^n[(n+2)(n+1)a_{n + 2}+na_n+(n+1)a_{n+1}]=0 $ so we can that state that
$a_{n + 2}=(-na_n/(n+2)(n+1))-a_{n+1}/(n+2)$

i tried using n values 0,1,2,3 but do not know how to put them together to match $\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{((2^{n})n!)}$

id appreciate it if you can help me on this because its a killer quetsion

what you have given us is the recursive formula. tell us your full solution and then we can help. (you should be able to simplify one of your solutions to get it in the form of the series posted before)

**TheEmptySet** · March 6th 2008, 05:55 PM

The solutions that I came up with are...

$y=c_1[1-\frac{1}{2}x^2+\frac{1}{8}x^4-\frac{1}{48}x^6+...]+c_2[x-\frac{1}{3}x^3+\frac{1}{15}x^5-\frac{1}{105}x^7+...]$

taking a closer look at the fist group we get

$c_1[1-\frac{1}{2}x^2+\frac{1}{8}x^4-\frac{1}{48}x^6+...]=c_1[\frac{1}{2^0 0!}x^{2 \cdot 0}-\frac{1}{2^1 1!}x^{2 \cdot 1}+\frac{1}{2^22!}x^{2 \cdot2}-\frac{1}{2^33!}x^{2 \cdot3}+\frac{1}{2^44!}x^{2 \cdot4}- ...]$

Does this look familiar. It is the power series for $e^{\frac{-x^2}{2}}$

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