Let $y(x)$ be a solution for the initial value problem for the Bernoulli equation $$y'=y^2-x,\ \ \ y(0)=1.$$ I want to find $a_n$ for $n\leq5$ in the sequence $\sum_{n=0}^{\infty} a_n x^n$.
So I know that $$y'=\sum_{n=1}^{\infty} na_n x^{n-1}=\sum_{n=0}^{\infty} (n+1)a_{n+1} x^n$$ but I have no clue how to get any further from here... Any tips?
This is not a Bernoulli equation but a Riccati equation. There is a transformation to a second order linear ODE that makes computing power series solutions easy, which give rational expressions for this equation that can also be expanded as power series.
Set $y=-\frac{u'}{u}$ then $$ 0=y'-y^2+x=-\frac{u''}u+x $$ so you have to solve the Airy equation $$ u''(x)=xu(x) $$ with $u(0)=-1$, $u'(0)=1$. Apart from expressing the solution via Airy functions, the power series $u=\sum a_kx^k$ satisfies $$ a_0=-1\\ a_1=1\\ a_2=0\\ (k+2)(k+1)a_{k+2}=a_{k-1}~\text{ for }~ k\ge 1\\ a_3=-\frac16\\ a_4=\frac1{12}\\ a_5=0,\dots $$