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I'm currently studying autonomous equations : $y'=f(y)$

If I have a Cauchy problem such that : $$y'=f(y), \qquad y(t_0)=y_0$$ My first quesiont is the following : can I say that if $f(y_0)\neq 0$ and $f$ is localy Lipschitz continuous around $y_0$, then the solution is unique ? Then I would like to solve this type of equation. This is my reasoning :

$$\frac{dy(t)}{dt}=f(y(t))$$So, $$\frac{1}{f(y(t))}=\frac{dt}{dy(t)}$$ So, by integrating with respect to $y$ between $y$ and $y_0$, $$\int_{y_0}^{y}\frac{1}{f(s(t))}ds=\int_{y_0}^{y}\frac{dt}{ds(t)}dy=\int_{t_0}^{t}\frac{dt}{ds(t)}dt=t-t_0$$ The last equality was obtain by the change of variable $dt=dy$ and because $t$ is a function of $s$ $(t=t(s))$

$$t=t_0+\int_{y}^{y_0}\frac{1}{f(s(t))}ds$$ Which can give us the solution. Is that correct ? If it is, is that valid all the time ?

Dicordi
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  • There are lots of problems with this. For starters, $dy/dt$ is a symbol that means "first derivative of the function $y$ with respect to its only argument". However, you are manipulating it like a fraction. In particular, $dy/dt = f \circ y$ does not imply that $dt / dy = 1 / (f \circ y)$. In fact, it is not even clear what $dt / dy$ is in this context. – parsiad May 27 '19 at 18:48
  • My teacher wrote $dy/dt=f(y)$ implies $dy=f(y)dt$ so $$dy/f(y)=dt$$ so i don't understant why can't i inverse $dy/dt$ ? – Dicordi May 27 '19 at 18:56
  • $dy = f(y)dt$ is just short-form for $y(t) = y(t_0) + \int_{t_0}^t f(y(s)) ds$. You might want to read this question that explains why derivatives are not exactly fractions. I don't think it's a good idea to introduce this kind of syntactic sugar in introductory courses as it leads to confusion. – parsiad May 27 '19 at 18:59

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