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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.

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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 $$ t - t_0 = \phi(y) - \phi(y_0). $$

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  • $\begingroup$ I understood up to here myself, this is where I get stuck. What I understand is because F is nonzero and continuous, f is continuous and bounded which implies its integrable over a fixed interval. $\endgroup$ – justaguy May 28 at 5:47
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    $\begingroup$ @justaguy see the update. Not sure what the problem is. $\endgroup$ – gt6989b May 28 at 13:15
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    $\begingroup$ @justaguy It is indeed true that every continuous bounded function on some interval has a continuous bounded antiderivative on the same interval. $\endgroup$ – K.defaoite May 28 at 13:20
  • $\begingroup$ why do we know this is unique though? is just because we have found the one solution? $\endgroup$ – justaguy May 28 at 15:48
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    $\begingroup$ @justaguy yes. Any other solution would either violate the integration (which gives you a form) or the initial condition (which forces the $C$ value) $\endgroup$ – gt6989b May 28 at 16:56

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