How can I solve this ODE? $2(G')^2 - GG'' < 0$ I know it maybe not unique, but how can I tell the class of functions $G(x)$ for which
$2(G'(x))^2 - G(x)G''(x) < 0$
Suppose I also know
$G(0) = 1$
and
$G'(x) < 0$
 A: The function $G(x)$ is twise differentiantable in the interval $\Delta=(a,b)$ with continuous secod derivative and $0\in \Delta$. Also $G'(x)<0$, $G''(x)G(x)-2G'(x)^2>0$ and $G(0)=1$. Since $G(0)=1\neq 0$, then we may assume that $G(x)\neq 0$ in $\Delta$ (or better $G(x)>0$ in $\Delta$ from the continuity of $G$). However let us only assume that $G(x)\neq 0$. Set then
$$
f(x)=\frac{1}{G(x)}
$$
Then we have
$$
f'(x)=-\frac{G'(x)}{G(x)^2}\Rightarrow
$$
$$
f''(x)=-\frac{G''(x)G(x)^2-2G(x)G'(x)^2}{G(x)^3}
$$
Hence
$$
f''(x)=-\frac{G''(x)G(x)-2G'(x)^2}{G(x)^3}.
$$
and
$$
f''(x)G(x)<0\tag 1
$$
Also
$$
f'(x)>0\tag 2
$$
Hence $f$ is increasing in $\Delta$. Also $G(x)>0$ or $G(x)<0$ in $\Delta$ since if exists $x_0$ such that $G(x_0)=0$, then from (1) we have $f''(x_0)G(x_0)=0<0$, which is imposible. Hence $f''(x)>0,\forall x\in\Delta$ or $f''(x)<0,\forall x\in\Delta$.
If $f''(x)>0,\forall x\in\Delta$, (then it is from (1) $G(x)<0$). Hence $f(x)<0$ in $\Delta$. But $0>f(0)=1$ which is imposible. Hence we have $f''(x)<0$ and $f'(x)>0$ and $f(x)>0$.
Hence every $f$ that passes through the pont $\{0,1\}$ and is positive and increasing and have negative curvature ($f''(x)<0$) in any interval $(a,b)$ that contains $0$, define a $G(x)=1/f(x)$ with the proerties you want.
