# Thread: need help with ODEs (power series)

1. ## need help with ODEs (power series)

Find the general solution of the form $\displaystyle y = \sum_{n = 0}^{\infty}a_nx^n$ for the differential equation $\displaystyle y''+xy'+y=0$ .
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

2. Originally Posted by Awara
Find the general solution of the form $\displaystyle y = \sum_{n = 0}^{\infty}a_nx^n$ for the differential equation $\displaystyle y''+xy'+y=0$ .
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?

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

Then

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

if it looks like the last one you are done.

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

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

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

4. 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 $\displaystyle y = \sum_{n = 0}^{\infty}x^n[(n+2)(n+1)a_{n + 2}+na_n+(n+1)a_{n+1}]=0$ so we can that state that
$\displaystyle a_{n + 2}=(-na_n/(n+2)(n+1))-a_{n+1}/(n+2)$

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

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)

5. The solutions that I came up with are...

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

taking a closer look at the fist group we get

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

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