Using $y=\sum_{n = 0}^\infty a_nx^n$ in order to find recurrence relation of $a_n$'s for $y'=x-y^2$ I have to solve the differential equation $y'=x-y^2 => y''' = -2y'y' - 2yy''$. 
I plugged in the series $\sum_{n = 0}^\infty a_nx^n$ to get:
$$\sum_{n = 0}^\infty a_{n+3}(n+3)(n+2)(n+1)x^n + \sum_{n = 0}^\infty \sum_{n = 0}^\infty 2(a_{n+1}(n+1)x^n)^2 \\
+ \sum_{n = 0}^\infty \sum_{n = 0}^\infty 2(a_nx^n)(a_{n+2}(n+2)(n+1)x^n) = 0$$
How do I further solve this? The double summation is throwing me off. 
Thank you in advance.
 A: Note that $$ \left ( \sum_{n=0}^\infty a_n x^n \right )^2 = \sum_{n=0}^\infty\left ( \sum_{k=0}^na_ka_{n-k} \right ) x^n$$Thus $$y' = \sum_{n=1}^\infty n a_n x^{n-1} = \sum_{n=0}^\infty(n+1)a_{n+1}x^n$$ And you'll get$$y'+y^2 = \sum_{n=0}^\infty(n+1)a_{n+1}x^n + \sum_{n=0}^\infty\left ( \sum_{k=0}^na_ka_{n-k} \right ) x^n = \sum_{n=0}^\infty\left [ \left ( \sum_{k=0}^n a_ka_{n-k} \right )+(n+1)a_{n+1} \right ]x^n = x$$ Equating the coefficients on both sides, we have $$(n+1)a_{n+1} = - \left ( \sum_{k=0}^na_ka_{n-k} \right )$$ for $n \ne 1$ and $2a_2 + 2a_1a_0=1$. 
A: i dont know how you will find a recurrence relation for the nonlinear riccatti equation; recurrence is used for linear equations.
you can turn $y' = x - y^2$ into a linear equation of second order for $v$ defined by $$y = \frac {kv}{v'},\, y' =k\left( 1 - \frac{vv''}{v'^2}\right)+\frac {k'v}{v'}.$$  subbing in riccatti equation gives 
$$ k\left( 1 - \frac{vv''}{v'^2}\right)+\frac {k'v}{v'}=x- \frac{k^2v^2}{v'^2}\tag 1$$  choosing $k = x$ reduces $(1)$  to the linear equation
$$xv'' - v'- x^2v = 0 \tag 2$$
solve the linear equation by the series method and obtain $y$ by the transformation $y = \frac{xv}{v'}.$
