Power series solution for 1st-order ODE

[Luminous Blob]

I am trying to find the power series solution of

$$y' = x^2y$$

but don't know how to arrive at the answer of $$y = a_0exp(x^3/3)$$. [I know that it's an easily solved separable equation, I'm just trying to figure out how to find the power series solution]

My solution so far:

Assume
$$y = \sum_{n=0}^\infty a_nx^n$$

then
$$y' = \sum_{n=1}^\infty na_nx^{n-1}$$

giving:

$$\sum_{n=1}^\infty na_nx^{n-1} = x^2 \sum_{n=0}^\infty a_nx^n$$
$$\sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty a_nx^{n+2}$$

changing the index for the LHS to give $$x^n$$

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

changing the index for the RHS to give $$x^n$$:

$$\sum_{n=2}^\infty a_{n - 2}x^{n}$$

Then taking the first two terms out of the LHS sum, so that both sums start from the same point:

$$a_1 + 2a_2x + \sum_{n=2}^\infty (n + 1)a_{n + 1}x^n = \sum_{n=2}^\infty a_{n - 2}x^{n}$$

I don't know what to do after this (I'm not entirely sure if what I've done so far is right, either).

If the y term in the initial equation didn't have the $$x^2$$ in front of it, it would be easy to equate the coefficients of $$x^n$$ to get the recursion formula. But having the terms $$a_1$$ and $$2a_2x$$ in front of the sum on the LHS throws me - can anyone explain clearly to me the correct steps required to solve the problem?

 

[MathNerd]

We know that $$y'(0) = a_1$$ and $$y''(0) = 2 a_2$$

We also know that $$y'(x) = x^2 \ y(x)$$ so that $$y'(0) = 0 = a_1$$

Differentiating both sides we get $$y''(x) = 2x \ y(x) + x^2 \ y'(x)$$ and it is seen that $$y''(0) = 0 = 2 a_2$$ so now we know that $$a_1=a_2=0$$.

Hope this helps!

 

[Luminous Blob]

We know that $$y'(0) = a_1$$ and $$y''(0) = 2 a_2$$

We also know that $$y'(x) = x^2 \ y(x)$$ so that $$y'(0) = 0 = a_1$$

Differentiating both sides we get $$y''(x) = 2x \ y(x) + x^2 \ y'(x)$$ and it is seen that $$y''(0) = 0 = 2 a_2$$ so now we know that $$a_1=a_2=0$$.

Hope this helps!

Thanks, that was a great help.