Solution of a differential equation? Please, could someone help me to solve the differential equation$$u''(s)+\rho(s)\cdot [u'(s)]^2=0$$
for any arbitrary function $\rho$. We assume that the solution is a function of $\rho,u(0),u'^(0)$
My trial: I put $z=u'(s)$, then
$$\frac{dz}{ds}=-\rho z^2
\\ \frac{-1}{z}+\frac{1}{u'(0)}=-\int_0^s \rho ds
\\ \frac{1}{z}=\sigma(s)+c
\\ u(s)=\int_0^s\frac{1}{\sigma(s)+c}ds+u(0)$$
Where $\sigma(s)=\int_0^s \rho ds$
Now, the main point is to find the conditions on $\rho$ or else that grantee the existence of a non-trivial solution $u$ .

Try letting $y(s) = u'(s)$ and $y'(s) = u''(s)$.
