Power Series Solution y' = xy

1. Mar 5, 2014

vanceEE

1. The problem statement, all variables and given/known data
$$y' = xy$$

2. Relevant equations
$$y = a_{0} + a_{1}x + a_{2}x^{2}+... = \sum\limits_{n=0}^∞ a_{n}x^{n}$$
$$xy = a_{0}x + a_{1}x^{2} + a_{2}x^{3}+... = \sum\limits_{n=0}^∞ a_{n}x^{n+1}$$
$$y' = a_{1} + 2a_{2}x + 3a_{3}x^{2}... = \sum\limits_{n=1}^∞ n a_{n}x^{n-1}$$

3. The attempt at a solution

$$\sum\limits_{n=1}^∞ n a_{n}x^{n-1} = \sum\limits_{n=0}^∞ a_{n}x^{n+1}$$
$$\sum\limits_{n=1}^∞ n a_{n}x^{n-1} = \sum\limits_{n=2}^∞ a_{n-2}x^{n-1}$$
$$\sum\limits_{n=2}^∞ n a_{n}x^{n-1} = a_{1}x^{0} + \sum\limits_{n=2}^∞ a_{n-2}x^{n-1}$$
$$\sum\limits_{n=2}^∞ [n a_{n} - a_{n-2}]x^{n-1} = a_{1}x^{0}$$ (1)

$$n a_{n} - a_{n-2} = 0$$ for all n = 2,3,4,...
$$a_{1} = 0$$

$$a_{0} = 2a_{2}$$
let $$a_{0} = C$$
$$a_{2} = \frac{C}{2}$$
$$a_{1} = 3a_{3} = 0$$
$$a_{2} = 4a_{4}$$
$$a_{4} = \frac{C}{4*2} , a_{6} = \frac{C}{6*4*2}, a_{n} = \frac{C}{(2n)!!}$$
I end up with the solution $$y = C*\sum\limits_{n=0}^∞ \frac{x^{2n}}{(2n)!!}$$

What am I doing wrong? $$a_{1}$$ must be zero, correct?

Last edited: Mar 5, 2014
2. Mar 5, 2014

vanceEE

Please disregard, the solution is in fact $$y = C*\sum\limits_{n=0}^∞ \frac{x^{2n}}{(2n)!!}$$
$$C*\sum\limits_{n=0}^∞ \frac{x^{2n}}{(2n)!!} \equiv C*e^{\frac{x^{2}}{2}}$$

Last edited: Mar 5, 2014
3. Mar 5, 2014

D H

Staff Emeritus
That is correct.

Your original series, the one with the those an coefficients, was of the form $\sum_{n=0}^{\infty} a_n x^n$. Your new series is of the form $\sum_{n=0}^{\infty} b_n x^{2n}$. All of your $a_n$ for odd n are zero.