How Do You Solve the Differential Equation y' = xy Using Power Series?

[vanceEE]

Homework Statement

$$y' = xy$$

Homework Equations

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

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?

 

[vanceEE]

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

 

[D H]

I end up with the solution $$y = C*\sum\limits_{n=0}^∞ \frac{x^{2n}}{(2n)!} $$

That is correct.

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

Your original series, the one with the those a_n 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.