Analytic solutions to algebraic differential equation Dear Colleagues and Friends,
Here I need to find some good reference on a subject that seems very much studied: sorry, if the rest of this question is too naive.
I believe that it's known that if a function $f(z)$ satisfies an equation $P(z, f(z)) = 0$ with $P(z, \xi) \in \mathbb{C}[[z, \xi]]$ a non-zero analytic function with $P(0, 0)=0$, then $f(z)$ is analytic in some neighbourhood of zero (for my purposes $P(z, \xi)$ can be taken to be just a polynomial).
What happens if instead of $P$ above we have, say, $P(z, \xi_0, \xi_1, \dots, \xi_n)$ (also a polynomial in $\mathbb{C}[z,\xi_0,\dots,\xi_n]$) and the equation now becomes $P(z, f(z), f'(z), ..., f^{(n)}(z)) = 0$?  
Is anything known about $f(z)$ being analytic in this case, and under which conditions on $P$?
Any discussion and especially a good reference will be much appreciated!
Sasha
 A: Your belief is not correct, as Robert Israel pointed in his comment. Same for
differential equations: $yy'-(3/2)z^2=0$ has a solution $y(z)=z^{3/2}$ which is not
analytic at $0$. The correct theorem is:
If $P(z_0,y_0,y_1,\ldots,y_{n})=0$, where $P$ is a polynomial,
(or an analytic function in a neighborhood of $(z_0,\ldots,y_n)$),
and $\partial P/\partial y_{n}\neq 0$ at the same point, then
there exists a unique analytic function $y(z)$ in a neighborhood of $z_0$, which satisfies
$P(z,y,y',\ldots,y^{(n)})=0$ in this neighborhood, and $y(z_0)=y_0,\; y'(z_0)=y_0,\ldots,y^{(n)}(z_0)=y_{n}$. One does not need to assume a priori that $y(z)$ is analytic. If it is $n$ times differentiable in a neighborhood
of $z_0$, it is automatically analytic. For the proof, see, for example H. Cartan, Elementary theory of analytic functions of one and several complex variables,
or Coddington and  Levinson, Ordinary differential equations. 
EDIT. The second question (asked in the comment) seems to be: if $f(z)$ is a formal power series satisfying a polynomial equation, does it follow that $f$ is convergent (so $f$ is analytic). The answer is yes in the case of implicit function theorem ($P(z,f(z))=0$), and no in the case of differential equations: 
the divergent series 
$$f(z)=\sum_{n=1}^\infty (n-1)!z^n$$
satisfies the Euler equation $z^2f'(z)-f(z)-z=0$, and $f(0)=0$.
