Consider the following initial value problem
$y'(t) = f(y(t)), $ $0 < t$
$y(0) = y_0$, where $y_0$ is a fixed constant.
Here, $y'(t)$ is given only for $t > 0$, not including $t = 0$. That is, $y'(0)$ is not prescribed.
Yet, if one tries to solve this numerically, say by Forward Euler, almost all textbooks that I saw starts by computing
$y(h) \approx y(0) + hy'(0) = y_0 + f(y(0)) = y_0 + f(y_0)$, as if $y'(t) = f(y(t)) $ holds at $t = 0$.
How can this be justified?
Is there a theorem that says that the above initial value problem is equivalent to the following initial value problem
$y'(t) = f(y(t)), $ $0 \leq t$
$y(0) = y_0$, where $y_0$
where we actually require a solution $y(t)$ to be differentiable and satisfy the ode at $t = 0$?