Solving $y'' y-2(y')^2\geq 0$ Here is an attempt

Let $y(x)=F^{a}(x)$ then 

$$y''y-2(y')^{2}=a(a-1)F^{a-2}(F')^{2}F^{a}+aF^{a-1}F''F^{a}-2(F^{a-1}F')^{2}$$

$$=(a(a-1)-2)F^{2(a-1)}(F')^{2}+aF^{2a-1}F''$$

so for $a(a-1)=2\Rightarrow a=-1,2$ we obtain the differential inequality $F^{2a-1}F''\geq 0$ and so the inequality is true for positive convex functions like $e^{2x}$.

Update

I apologize the previous attempt was wrong,as rightly pointed out by professor Israel.
That's wrong.  With $y(x) = e^{c g(x)}$ you should get

$$ y'' y - 2 (y')^2 = e^{2cg(x)} c (g''- c (g')^2) $$

BTW, with $g' = v$

$$ v' - c v^2 = q $$

is a Riccati D.E.  I don't think there is a known solution for general $q$.
$\frac {d}{dt} yy' = y'^2 + yy''\\
\frac {d}{dt} y^{-2}y' = -2y^{-3}y'^2 + y^{-2}y'' = y^{-3} (yy''-2y'^2)$

$yy''-2y'^2 \ge 0\\
y^{-3} (yy''-2y'^2) \ge 0\\
y^{-2} y' \ge C\\
-y^{-1} \ge Ct + D\\
y \ge \frac 1{Ct + D}$