Prove that an IVP with discontinuous $f(t,x)$ has a solution for all $(t_0,x_0)$

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I'd like to find some theorem, lemma, proposition or intuitive explanation to this.

Prove that the function $f:\mathbb{R}\times \mathbb{R} \to \mathbb{R}$ defined as

$$f(t,x)=\begin{cases}\frac{tx}{t^2+x^2}\ \ \ if \ \ \ (t,x)\neq(0,0) \\ 0 \ \qquad if \ \ \ (t,x)=(0,0).\end{cases}$$

is not continuous on the origin, but the IVP $$\begin{cases}\dot{x}=f(t,x) \\ x(t_0)=x_0\end{cases}$$

Has a solution for all $(t_0,x_0)\in\mathbb{R^2}$.

Every theorem that I know uses continuity of $f(t,x)$ to say that there exists a solution to an IVP.

It's necessary to solve this problem? Or we can find some theorem to say that there exists a solution to this IVP?

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Caratheodory's existence theorem gives to answer to your question. It only needs continuity of $f$ in $x$ for fixed $t$.