Let $U \subseteq \mathbb R^{n+2}$ be an open set for some $n \geq 0$, and let $f: U \to \mathbb R$ be an analytic function. Then we say the equation $f(x,y,y′,…,y^{(n)})=0$ is an analytic differential equation.

Let $g(x)=y$ be a solution to an analytic differential equation.

Is $g$ necessarily analytic on some open set $V \subseteq \mathbb R$? What about if $U \subseteq \mathbb R^\infty$?

It is known that the answer to a more general question is no, namely, the solutions to a functional analytic differential equations need not be analytic anywhere. For example, the equation $f'(x)=2f(2x)$ uniquely defines the Fabius function, which is nowhere analytic.

Let $U \subseteq \mathbb R^{n+2}$ be an open set for some $n \geq 0$, and let $f: U \to \mathbb R$ be an analytic function. Then we say the equation $f(x,y,y',\ldots,y^{(n)})=0$ is an analytic differential equation.

Let $g(x)=y$ be a solution to an analytic differential equation.

Is $g$ necessarily analytic on some open set $V \subseteq \mathbb R$? What about if $U \subseteq \mathbb R^\infty$?

It is known that the answer to a more general question is no, namely, the solutions to a functional analytic differential equations need not be analytic anywhere. For example, the equation $f'(x)=2f(2x)$ uniquely defines the Fabius function, which is nowhere analytic.