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# A simple ordinary differential equationsequation

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Consider an entire function $f : \mathbb{C} \rightarrow \mathbb{C}$! We search the function $$g: \mathbb{R} (a,b) \rightarrow \mathbb{C},$$ which solves the following equation locally: $g'(t)=f(g(t))$ and $g(0)=f(x_0)$.

I can compute the inverse $G$ of $g$, if $f(x_0) \neq 0$, i.e. $$G(y) = \int\limits_{f(x_0)}^y \frac{d s}{f(s)}.$$

I also known how to compute the Taylor expansion recursively, whose radius of convergence is positive (see below). Also we can give suitable approximations of the solution in terms of Picard iterations. I am not interested in such a solution!

Is there an alternative to this integral expression?

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Consider an entire function $f : \mathbb{C} \rightarrow \mathbb{C}$! We search the function $$g: \mathbb{R} \rightarrow \mathbb{C},$$ which solves the following equation: $g'(t)=f(g(t))$ and $g(0)=f(x_0)$.

I can compute the inverse $G$ of $g$, if $f(x_0) \neq 0$, i.e. $$G(y) = \int\limits_{f(x_0)}^y \frac{d s}{f(s)}.$$

The motivation is that the function $g$ will parametrize a curve, where the argument of the primitive of $f$ is constant. This primitive of f is well defined in some neighborhood, if $f(x_0) \neq 0$. Hence,

I suspect that the number of solutions can be given in terms of the vanishing order of $f$ at $x_0$.

I also known how to compute the Taylor expansion recursively, whose radius of convergence is positive (see below). Also we can give suitable approximations of the solution in terms of Picard iterations. I am not interested in such a solution!

Is there anything nicer than just the inverse of an alternative to this integral expressionfor the solution?

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