Existence of solution of ODE $y^{\prime}=f(y,t)$ where $f(y,t)$ is not defined in initial value. Consider a differential separable equation $$y^{\prime}=f(y,t)$$ with initial solution $y(t_0)=y_0$. Suppose that $f(y_0,t_0)$ is not defined. Is there a theorem which can be used to prove the existence and the uniqueness of the solution of this Differential Equation?
The trouble is because $f(y_0,t_0)$ is not defined (much worse than discontinuous where we can still use Carathéodory's existence theorem)
For example a separable differential equation $y^{\prime}=\frac{1}{y-1}+2$ with initial solution $y(0)=1$.
 A: I think what you are after is the concept of weak solution, that is precisely used to give a sense to differential equations without having to worry about single points in the domain. You can check intuition behind weak solution , https://en.wikipedia.org/wiki/Weak_solution
The idea is to use the integral formulation of the equation, that in your case is:
\begin{equation}
y(t)-y(t_0)=\int^t_{t_0}f(s,y)d s
\end{equation}
if you are willing to adopt this as a definition of your solution then you can dispense with the C1 assumption and even the "definition" of the function on a set of measure zero.
A: The answer is no. At least there is no such definable solution, because, according to the definition, a solution of the IVP
$$
y'=f(t,y), \quad y(t_0)=y_0,
$$
is a  $C^1$ function $\psi$ which is defined in an open interval $I$, with $t_0\in I$, and satisfying the both the initial condition and the ODE, i.e.,
$$
\psi(t_0)=y_0\,\,\,\text{and}\,\,\,\psi'(t)=f\big(t,\psi(t)\big),
$$
for all $t\in I$. Satisfaction of the ODE implies that
$$
\big(t,\psi(t)\big) \in D, \quad \text{for all $t\in I$},
$$
where $D$ is the domain of definition (and continuity) of $f$. Hence $(t_0,y_0)\in D$.
