I am not sure what you tried, but the usual method start like this
Assume the solution have the form
$\displaystyle \displaystyle y=\sum_{n=0}^{\infty}a_nx^n$
Then take some derivatives to get
$\displaystyle \displaystyle y'=\sum_{n=0}^{\infty}na_nx^{n-1}$
and
$\displaystyle \displaystyle y''=\sum_{n=0}^{\infty}n(n-1)a_nx^{n-2} $
Now plug these into the ode and work from here to solve for the $\displaystyle a_n$
Yeah I had those forms and subbed it back into the d.eq
\displaystyle y''=\sum_{n=0}^{\infty}n(n-1)a_nx^{n-2} + \displaystyle y'=\sum_{n=0}^{\infty}na_nx^{n-1} + \displaystyle y=\sum_{n=0}^{\infty}a_nx^n = 0
then
\displaystyle y''=\sum_{n=0}^{\infty}(n+2)(n+1)a_nx^{n} + \displaystyle y'=\sum_{n=1}^{\infty}na_nx^{n} + \displaystyle y=\sum_{n=0}^{\infty}a_nx^n = 0
Right?
My equation becomes 2a_2 + a_0 = 0 then to a_2 = -(1/2) a_0
EDIT: Ah I dunno how to get the formula source code up :/
you need to wrap them in math tags
[tex] [ /math] no space between the bracket and forward slash. [ /
If you change your editor to advanced mode you can just click
Tex button in the upper right
$\displaystyle \displaystyle \sum_{n=0}^{\infty}n(n-1)a_nx^{n-2} + \sum_{n=0}^{\infty}na_nx^{n-1} +\sum_{n=0}^{\infty}a_nx^n = 0$
$\displaystyle \displaystyle \sum_{n=0}^{\infty}(n+2)(n+1)a_nx^{n} + \sum_{n=1}^{\infty}na_nx^{n} + \sum_{n=0}^{\infty}a_nx^n = 0$
I think this is what you typed, but you are missing the $\displaystyle x$ and $\displaystyle x^2$ in front of the function and the first derivative
Even if 'non reccommended' the most confortable approach is to find y as...
$\displaystyle \displaystyle y(x)= \sum_{n=0}^{\infty} \frac{y^{(n)} (0)}{n!}\ x^{n}$ (1)
The values of $\displaystyle y(0)$ and $\displaystyle y^{(1)} (0)$ are given by the 'initial conditions'...
$\displaystyle y(0)= 5$
$\displaystyle y^{(1)}(0)= -2$
... and the others are given by the DE...
$\displaystyle y^{(2)} (0) = -x^{2} y^{(1)} (0) + x y(0)= 0$
$\displaystyle y^{(3)} (0) = y(0) -x y^{(1)} (0) -x^{2} y^{(2)}(0)= 5$
$\displaystyle y^{(4)} (0) = -3 x y^{(2)} (0) -x^{2} y^{(3)}(0)= 0$
$\displaystyle y^{(5)} (0) = -3 y^{(2)} (0) -5 x y^{(3)}(0) - x^{2} y^{(4)} (0) = 0$
$\displaystyle y^{(6)} (0) = -8 y^{(3)} (0) -7 x y^{(4)}(0) - x^{2} y^{(5)} (0) = -40$
... and so one, so that the power series is...
$\displaystyle \displaystyle y(x)= 5 -2 x + \frac{5}{6} x^{3} -\frac{1}{18} x^{6} + ...$ (2)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$