A simple polynomial differential equation Consider the differential equation
$$ \frac{d}{dx}y(x) = -(y(x))^3. $$
with initial condition $y(0) = 1$.
I know that it admits the unique solution
$$ y(x) = \sqrt{ \frac{ 1 }{ 1 + 2 x } }. $$
How can I prove that this solution is unique?
 A: It can be proved by the standard way. Let $y$ and $z$ be two solutions satisfying the given problem and set $w=y-z$. Then $w$ satisfies the equation $w'=-(y^3-z^3)=-w(y^2+yz+z^2)$ with zero initial condition. Multiply this equation by $w$ you will see that $ww'=-w^2(y^2+yz+z^2)$. Since the RHS of this equation is always non-positive ($y^2+yz+z^2\geq \frac{1}{2}(y^2+z^2)$) so we have $=(w^2/2)'= ww'\leq 0$. Integrating this inequality from $0$ to $x$ gives that $w^2(x)\leq 0$ and so you have $w=0$, or $y=z$.
A: You can try to apply the separation of variables. For your Cauchy problem it writes
$$-\int_{y(0)}^{y(t)}\frac{du}{u^3}=\int_0^t ds$$
or
$$ \frac{1}{2u^2}\Big|_{y(0)}^{y(t)}=t,$$
which gives
$$\frac{1}{(y(t))^2}=2t+1,$$
and finally $$y(t) = (2t+1)^{-1/2}.$$
All you need to do now is to prove that the method of separation of variables works. 
A: Another approach would be to prove that some bijective application of $y$ is unique.
In some neighbourhood of zero our function is positive, hence we can study the function $z(t) = y^{-2}(t)$. The application $y\to z$ is bijective,  hence we don't loose unicity.
Now we can write the differential equation on $z$:
$$z'= -2y^{-3}y' = 2,\quad z(0)=1.$$
Evidently, this equation has a unique solution $z=2t+1$. Note that, remarkably, we obtain that $y$ remains positive for all admissible $t$, hence our trick works everywhere.
