# Unique Solution to 1st Order Autonomous ODE

Take the ODE $$y'=F(y)$$. Show it has a unique solution with initial condition $$y(t_0) = y_0$$ in a neighborhood of $$t_0$$ provided $$F$$ in continuous and $$F(y_0) \neq 0$$. I am trying to use the inverse function theorem by solving the ODE the inverse function satisfies but I am getting stuck.

The ODE $$\frac{dy}{dx} = F(y)$$ seems to separate into $$dx = \frac{dy}{F(y)} \iff x = \int \frac{dy}{F(y)} = \int f(y) dy,$$ where $$f(y) = 1/F(y)$$. Can you prove $$f$$ is integrable?
UPDATE After your response, we then understand that $$f$$ is integrable, therefore we conclude that there is some anti-derivative $$\phi$$ of $$f$$, so $$t = \phi(y) + C$$ and we enforce the initial condition, calculating $$C = t_0 - \phi(y_0),$$ so the final unique solution looks like $$t - t_0 = \phi(y) - \phi(y_0).$$
• @justaguy yes. Any other solution would either violate the integration (which gives you a form) or the initial condition (which forces the $C$ value) – gt6989b May 28 at 16:56