third-order nonlinear differential equation Solve the following differential question:
$$\frac{-1}{2}\frac{1}{(f'(x))^{2}} S(f)(x)=K,$$
where
$$ S(f)(x) = \frac{f'''}{f'} - \frac{3}{2}(\frac{f''}{f'})^{2}$$
is Schwartzian derivative of real valued function of one real variable $f=f(x)$ and $K\in \mathbb{R}$.
Thanks.
 A: First, as @Evgeny suggests, let $g(x)=f'(x)$. Then the differential equation becomes
$$
\frac{1}{g}\Bigl(\frac{g''}{g}-\frac{3}{2}\Bigl(\frac{g'}{g}\Bigr)^2\Bigr)=-2K.
$$
Next, the differential equation becomes a simpler if you let $h(x)=1/g(x)$. Note that then $g(x)=1/h(x)$, so (using the product rule and chain rule),
$$
g'=-\frac{1}{h^2}h'\quad \text{and}\quad g''=2\frac{(h')^2}{h^3}-\frac{h''}{h^2}.
$$
Inserting all this in the differential equation for $h$, we find that
$$
h^2\Bigl(2\frac{(h')^2}{h^2}-\frac{h''}{h}-\frac{3}{2}\Bigl(\frac{h'}{h}\Bigr)^2\Bigr)=-2K,
$$
or, simplified,
$$
\frac{1}{2}(h')^2-hh''=-2K.
$$
As mentioned in the comments, differentiating this leads to
$$
hh'''=0
$$
If we assume that $h\neq 0$, we find that $h'''=0$, and so $h=Ax^2+Bx+C$. If you insert this into the equation for $h$, you might determine some constant in term of $K$.
I leave it to you go to back to $f$ (via $g$, maybe).
A: Let $p = f'$ we obtain
$$
-\frac{1}{2}\frac{p''}{p^3} +\frac{3}{4}\left(\frac{p'}{p}\right)^2\frac{1}{p^2} = K
$$
we can rwerite as
$$
\frac{p''}{p^3}-\frac{3}{2}\left(\frac{p'}{p^2}\right)^2 = \\
\frac{p'}{p^3}\dfrac{d}{dp}p' -\frac{3}{2}\left(\frac{p'}{p^2}\right)^2 = 0
$$
now trying solutions of the form:
$$
p' = p^{\alpha}\phi(p) 
$$
we obtain
$$
\alpha p^{2(\alpha-2)}\phi^2 + p^{2\alpha-3}\phi\phi' -\frac{3}{2}p^{2(\alpha-2)}\phi^2 = -2K
$$
choosing $\alpha = 2$
$$
2\phi^2  + p\phi\phi' - \frac{3}{2}\phi^2 = p\phi\phi' + \frac{1}{2}\phi^2 = -2K 
$$
or 
$$
p\dfrac{d}{dp}\phi^2 = -4k - \phi^2
$$
separable or if we set $\phi = 2\sqrt{K}\sqrt{v}$
$$
4K p \frac{dv}{dp} = -4K(1+v)
$$
or
$$
\int \frac{1}{1+v}dv = -\int \frac{1}{p}dp
$$
$$
\ln(1+v) = -\ln p + C \implies 1+ v = \frac{A}{p}
$$
or
$$
\phi = \sqrt{4K} \sqrt{\frac{A}{p}-1}
$$
and finally
$$
p' = p^2\sqrt{4K} \sqrt{\frac{A}{p}-1} = \sqrt{4K}p^{3/2}\sqrt{A-p}
$$
now integrate that..as 
$$
\int p^{-3/2}(A-p)^{1/2} = \sqrt{4K}x + C
$$
where using the final sub of $p = Au$
we obtain
$$
\frac{1}{A}\int u^{-3/2}(1-u)^{-1/2} du = \frac{1}{A}\int \frac{2}{v^2\sqrt{1-v^2}}dv
$$
