I am trying to find the power series solution of(adsbygoogle = window.adsbygoogle || []).push({});

[tex]y' = x^2y[/tex]

but dont know how to arrive at the answer of [tex]y = a_0exp(x^3/3)[/tex]. [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

[tex] y = \sum_{n=0}^\infty a_nx^n [/tex]

then

[tex] y' = \sum_{n=1}^\infty na_nx^{n-1} [/tex]

giving:

[tex] \sum_{n=1}^\infty na_nx^{n-1} = x^2 \sum_{n=0}^\infty a_nx^n [/tex]

[tex] \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty a_nx^{n+2} [/tex]

changing the index for the LHS to give [tex]x^n [/tex]

[tex] \sum_{n=0}^\infty (n + 1)a_{n + 1}x^n [/tex]

changing the index for the RHS to give [tex]x^n[/tex]:

[tex] \sum_{n=2}^\infty a_{n - 2}x^{n} [/tex]

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

[tex] a_1 + 2a_2x + \sum_{n=2}^\infty (n + 1)a_{n + 1}x^n = \sum_{n=2}^\infty a_{n - 2}x^{n} [/tex]

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

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Power series solution for 1st-order ODE

**Physics Forums | Science Articles, Homework Help, Discussion**