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 $u$ .


One possible way is to assume $u'(s)=y(s)$ which implies $u''(s)=y'(s)$. Indeed, $$u''(s)+\rho(s)u'(s)=0\to y'(s)+\rho(s)y(s)=0$$ which is a separable one.

  • $\begingroup$ I am so sorry, I edited it $\endgroup$ – Mona Feb 7 '14 at 16:19

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

  • $\begingroup$ I did then what $\endgroup$ – Mona Feb 7 '14 at 16:09
  • $\begingroup$ Solve it like you would any other linear first order equation. Integrating factor in general. This particular equation is separable. $\endgroup$ – John Habert Feb 7 '14 at 16:14
  • $\begingroup$ I am so sorry, I edited it $\endgroup$ – Mona Feb 7 '14 at 16:16
  • $\begingroup$ Not a problem. You lost linear but it is still separable. If you don't have a specific function for $\rho$, then your final answer will have to contain an integral. $\endgroup$ – John Habert Feb 7 '14 at 16:19
  • $\begingroup$ I wand to describe or get all properties of $u$ in terms of $\rho$ $\endgroup$ – Mona Feb 7 '14 at 16:22

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