A condition for uniqueness of solution Assuming that $f$ is a continuous real function and $f(0)=0$ , $f(x)>0 $ when $x\neq 0$, prove that the differential equation $x'= f(x)$ with the initial value $x(0)=0$ has a unique solution if and only if $$\int_0^c \frac{1}{f(x)}dx$$ is not defined for all $c\in \mathbb{R}$. That is $$\int_0^c \frac{dx}{f(x)}= \infty.$$
 A: I am sharing a solution for one direction :
We note first that since $f$ is positive the integral $\int_0^c \frac{dx}{f(x)}$ can only be $+\infty$ when $c>0$ and $-\infty$ when $c<0$ as $\int_c^0 \frac{dx}{f(x)}= +\infty $ , moreover $\frac{1}{f}$ is continuous over $\mathbb{R}\setminus\{0\}$ and $\lim\limits_0 \frac{1}{f}= +\infty$
Let $y_1, y_2$ be two solutions of the differential equation, we seek to prove that $y_1= y_2$ under the hypothesis $\int_0^c \frac{dx}{f(x)}= \infty $  :
We have $y_1(0)= 0= y_2(0)$ and $$y_1(t)= \int_0^t f(y_1(s))ds \ \ ,\ \ y_2(t)= \int_0^t f(y_2(s))ds $$ for all $t$ in an interval $I= [-a,a]$ centered at $0$
The function $t\mapsto |y_1(t)- y_2(t)| $ is continuous and satisfies
$$\max\limits_{t\in I} |y_1(t)- y_2(t)| = \max\limits_{t\in I} |\int_0^t (\ f(y_1(s))- f(y_2(s))\ )ds|\leq \int_0^{|t_0|} |f(y_1(s))- f(y_2(s))|ds $$
for some $t_0 \in I$ as the maximum is reached.
By the hypothesis we have $\int_0^a \frac{dx}{f(x)}= +\infty$ , the divergence of this integral implies that $\forall M>0\ \exists\ x_0\in (0,a] $ such that $(\min\limits_{(0,x_0]} \frac{1}{f} )\times x_0> M$ (we can show it by considering first the limit of $\frac{1}{f}$ then the fact that $\int_\delta^a \frac{dx}{f(x)}$ can be as large as desired and taking the mean value)
So we can transform this to $(\frac{1}{\max\limits_{(0,x_0]} f})\times x_0> M \quad$ or  $\quad \max\limits_{(0,x_0]} f < x_0/M $ 
Using this in the previous inequality $$\int_0^{|t_0|} |f(y_1(s))- f(y_2(s))|ds \leq |t_0|\max\limits_{[0,t_0]} |f(y_1(s))- f(y_2(s))| \leq a\times 2\max\limits_J |f(s)|$$
where $J$ is the union $y_1([0,t_0])\cup y_2([0,t_0]) $ , $J$ is an interval (why?).
Now take $M$ to be $2a^2 N$ so that $\exists\ s_0\in J\setminus \{0\}$ for which
$ \max\limits_{(0,s_0]} f(s) < a/M= \frac{1}{2aN}$ and the inequality becomes
$$\max\limits_{I} |y_1(t)- y_2(t)|\leq 2a\max\limits_{(0,s_0]} f(s)< 1/N$$
As this hold for all $N>0$ then it can only be true when $\max\limits_{I} |y_1(t)- y_2(t)|= 0$ and we deduce that $y_1= y_2$ over the existence interval $I$.
