# Thread: Power series - general solution

1. ## Power series - general solution

Assuming a power series solution of the form

Show that

2(dy/dx) + xy = x

has general solution y(x) = 1 + Ae^[(x^2)/4] where A is a constant.

I know I have to substitute y = a(0) + a(1) x + a(2) x^2 + ....
into the equation and then equate the powers of x on both sides. After that solve for the sequence a(n). but i don't seem to be getting what I should? If anyone could help me that would be brilliant! xxx

2. Hello,

$\displaystyle y=\sum_{n=0}^\infty a_n x^n=a_0+ \sum_{n=1}^\infty a_n x^n$

hence $\displaystyle \frac{dy}{dx}=\sum_{n=1}^\infty na_n x^{n-1}$

$\displaystyle 2 \frac{dy}{dx}+xy=x$

\displaystyle \begin{aligned} 2 \frac{dy}{dx}+xy &=2 \sum_{n=1}^\infty na_nx^{n-1}+x \sum_{n=0}^\infty a_nx^n \\ &=\sum_{n=1}^\infty 2na_nx^{n-1}+\sum_{n=0}^\infty a_nx^{n+1} \end{aligned}

Now change the indices so that you get $\displaystyle x^n$ in the series :

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

Now make the starting indice be the same for the two series :

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

$\displaystyle =2a_1+\sum_{n=1}^\infty \left(2(n+1)a_{n+1}+a_{n-1}\right) x^n$

You want this to be equal to x.
Hence :

$\displaystyle x=2a_1+\sum_{n=1}^\infty \left(2(n+1)a_{n+1}+a_{n-1}\right) x^n$

Make appear the factors of x.

$\displaystyle x=2a_1+(4a_2+a_0)x+\sum_{n=2}^\infty \left(2(n+1)a_{n+1}+a_{n-1}\right) x^n$

By identification, we get :

$\displaystyle \left\{\begin{array}{lll} a_1=0 \\ 4a_2+a_0=1 \\ 2(n+1)a_{n+1}+a_{n-1}=0 \end{array} \right.$

Solve for the recursive relation :P
Note that if n is odd, then the coefficient of the power series is 0, since a1=0.

I've already made a big part of the working, so now it's you to go ! Good luck