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.