Unicity condition for an ODE Let $f$ be a continuous function on a neighborhood of $y_0$. If the differential equation: $$y'=f(y)$$ Has two solutions that pass through the point $(0,y_0)$, prove $f(y_0) = 0$ $$$$I have tried using the condition for unicity for equations of the form: $$F(x,y,y')=0$$ In which the non-unicity implies: $$\frac{\partial F}{\partial z}(0,y_0,f(y_0))=0$$ But $\frac{\partial F}{\partial z}=1$. Am I applying something wrong?

If $f(y_0) \
eq 0$, then there is only one local solution satisfying $y(0) = y_0$.

Namely, if $J$ is an open neighborhood of $y_0$ where $f\
eq 0$, then the function $$ F(y) := \int_{y_0}^y \frac{1}{f(s)}\, ds, \qquad y \in J $$ is locally invertible, being a bijection from $J$ to $F(J)$.

Hence the unique local solution of the Cauchy problem is given by $$ y(t) = F^{-1}(t), $$ for $t$ in a neighborhood of $0$.