Uniqueness of the solution of $y'=y^{1/3}+1$ This Cauchy problem $$ \begin{cases}y'=y^{1/3}+1,\\ y (0)=0, \end{cases}$$ has a unique solution. How I can prove this?
 A: Hint: Let $$F(y) = \int_0^y \dfrac{dx}{x^{1/3}+1} $$
and show that $\dfrac{d}{dt} F(y(t)) = 1$ with $F(y(0)) = 0$.
A: I will expand on Robert Israel's hint. Letting $$F(y)=\int\limits_0^y \frac{1}{x^{1/3}+1}\,dx$$
we have that $F'(y)=\displaystyle\frac{1}{y^{1/3}+1}$ by the Fundamental Theorem of Calculus. Hence, if $y(t)$ is a solution to the Cauchy problem, by the chain rule we have that$$\frac{d}{dt}\, \Big[F(y(t))\Big] = F'(y(t))\cdot y'(t) = \frac{1}{{y(t)}^{1/3}+1}\cdot y'(t) = 1$$for all $t$ for which $y(t)$ is defined.
Moreover, $y(0)=0$ implies $F(y(0))=0$ (because the limits of integration are both zero). Thus, for any solution $y(t)$ to Cauchy problem defined near $0$, we have that $F(y(t))=t$.
Now, the integral in the definition of $F(y)$ can also be explicitly solved. It is a straightforward exercise to check that $$F(y)=3y^{1/3}\left(\frac12 y^{1/3}-1\right)+3\ln\left(\left\lvert y^{1/3}+1\right\rvert\right)$$
It follows that any solution to the Cauchy problem is implicitly defined by$$3{y(t)}^{1/3}\left(\frac12 {y(t)}^{1/3}-1\right)+3\ln\left(\left\lvert{y(t)}^{1/3}+1\right\rvert\right)-t=0$$
Of course, having an explicit equation that implicitly defines the solution is nice, but from the integral equation for $F$ we could already see the uniqueness of the solution.


Lemma: For $t>-1$, any solution $y(t)$ to the Cauchy problem satisfies $y(t)>-1$ and is strictly increasing.

Proof: Indeed, it is clear from the hypotheses that the derivative of $y(t)$ is positive whenever $y(t) > -1$, and in particular the initial value $y(0)=0$ guarantees that $y(t)>\geq 0$ whenever $t \geq 0$. To complete the proof, we show that for $-1<t<0$ we may have not $y(t) \leq -1$.
First, note that if $y(t_0)<-1$ for some $t_0<0$, then its derivative at $t_0$ is negative, so it is strictly decreasing in a neighborhood of $t_0$. It's easy to see that in this case, $y(t)$ will be strictly decreasing for all $t>t_0$, so that the initial condition $y(0)=0$ cannot be fulfilled.
Now, suppose $y(t)=-1$ for some $t<0$ and let $t_0$ be the greatest such $t$. We show that $t_0 \leq -1$, which completes the proof. Consider the integral:$$\int\limits_{t_0}^0y'(s)\, ds$$On one hand, by the Fundamental Theorem of Calculus, the integral is simply $y(0)-y(t_0)=1$. On the other hand, for all $t$ with $t_0<t<0$, $y(t) > -1$ and because $y$ is a solution the integral may also be expressed as $$1=\int\limits_{t_0}^0{y(s)}^{1/3}+1\, ds \leq |t_0| \cdot\left(1+ \max\limits_{s \in [t_0,0]}\left\{{y(s)}^{1/3}\right\}\right)=|t_0|$$
It follows that $t_0 \leq -1$ as claimed.

Finally, observe that for $y>-1$ the integrand of $F$ is always positive, so in that range $F$ is also strictly increasing. If there were two solutions $y_1,y_2$ to the Cauchy problem that disagreed at some $t$ in a neighborhood of $0$, then we'd have
\begin{aligned}
&y_1(t)>y_2(t)>-1\\
&F(y_1(t))=F(y_2(t))=t
\end{aligned}
which clearly violates $F$ being strictly increasing for $y > -1$. Notice the first line above is justified by the Lemma (and we take $y_1(t)>y_2(t)$ without loss of generality, since they're distinct).
EDIT: I've attached some pictures and observation. You can check that the solution looks something like this (the solution is graphed as $y(x)$):

You can see that as $x \to -\infty$, $y(x)$ appears to approach $-1$. This asymptotic behavior can be confirmed by looking at $F(y)$:

Because of the relation $F(y(t))=t$, this graph can be read as follows: any point $\big(y,F(y)\big)$ on the graph in the same connected component as that of $(0,0)$ -- remember that $y(0)=0$ -- corresponds to a point $\big(F(y)=t,y=y(t)\big)$ on the solution.
Now the asymptotic behavior is clear: on $F$, the logarithm has a vertical asymptote at $y=-1$, and the graph of the solution $y(t)$ is obtained simply by reflecting the corresponding connected component of the graph of $F$ about the identity line $y=x$. Also, because the graph of $F$ has no horizontal asymptotes (that is, $F(y) \to \infty$ as $y \to \infty$), this also shows that the solution $y(t)$ is more than locally unique: it is globally defined and unique (ie, it does not blow up in finite time).
