Existence and uniqueness of a solution of a differential equation with continuous function Let $y_0\in\mathbb{R}$ and $f\in C(\mathbb{R})$ such that $f(y_0)\neq0$. Show then that for any $x_0\in\mathbb{R}$, the point $(x_0,y_0)$ is a point of existence and local uniqueness $y'= f(y)$.
I had in mind to use Peano's existence theorem, but I don`t know how to rigorously pass from a function of one variable to a function of two variables. Also, another idea was to use the following function:
$$
F(y)=\int_{0}^{y}\frac{du}{f(u)},\quad y\in\mathbb{R}.
$$ We would write $y'=f(y)$ as $F(y)+c=x$, $c\in\mathbb{R}$ and its derivative (of $F$) would then be $\frac{1}{f(y)}$ which is either bigger than $0$, either smaller. Then $F$ is strictly monotone and from here I would say that we have a solution on the open interval
$$
I_{c}=\left(\int_{0}^{-\infty}\frac{du}{fu}+c,\int_{0}^{\infty}\frac{du}{f(u)}+c\right).
$$ I have proven that there exists a solution and it is saturated, but I am unsure how to prove the uniqueness part. Any help would be appreciated.
 A: As $f(y_0)\neq 0$ we can just solve the equation by separation of variables in the neighbourhood of $(x_0,y_0)$ and obtain
$$F(y)=\int_{y_0}^y \frac{\mathrm{d}u}{f(u)} = x-x_0. $$
Now as you said, the function $F$ is strictly monotone, therefore invertible. Hence we obtain explicit formula for the solution
$$y(x)=F^{-1}(x-x_0),$$
which is unique.
EDIT: Also some remarks about the interval of existence: since the question is about local existence and uniqueness, I suppose you don't have to compute the maximal interval on which $y$ exists and is unique, especially that the latter may get complicated, depending on $f$.
First of all, I think you should write $c$ in terms of $x_0,y_0$. Moreover, the integral limit should be $y_0$ instead of $0$ (simple example: take $f(y)=y^2$ and $y_0>0$ - then $\int_0^\infty \frac{\mathrm{d}u}{f(u)}=\infty$ but $y(x)$ blows up for $x=\int_{y_0}^\infty \frac{\mathrm{d}u}{f(u)}+x_0$, which is finite).
To obtain the interval of unique existence, you also need to consider the integral of $\frac{1}{f(u)}$ in the neighbourhoods of zeros of $f$ (another example: let $f(y)=-\sqrt{|y|}$ and $y_0>0$ - then $y(x)=0$ for $x=x_0+\int_0^{y_0}\frac{\mathrm{d}u}{\sqrt{u}}$ and loses its uniqueness).
