Solving the ODE $x''=\frac{1}{3x}$ I am studying ODE and I just started learning about second order ODE. The question I'm trying to solve is somewhat of a physics problem (but a very simple physics, the hard part is the ODE involved). It is $$x'' = \frac{1}{3x}.$$ Considering $x(t)$ as the distance between two bodies ($x'(t)$ is the velocity) I am giving the initial conditions : $x(0)=4,x'(0)=v(0)=-2$. I whish to find the value of $x(t)$ s.t $x'(t)=0$ (i.e _where_ $v(t)=0$) and when does this happen (i.e find t s.t $v(t)=0$). I would appriciate any help on this, I didn't really try anything because I don't know how to proceed.

Let me just expand on the answers that Peter and GEdgar have posted, which show that

$$ \frac{3}{2} v^2 = \log x + C. $$

We can get some mileage out of this expression, especially given the questions you ask about the ODE. Given that $x(0) = 4$ and $v(0) = -2$, we can find the value of $C$:

$$ \begin{align*} \frac{3}{2} (-2)^2 &= \log 4 + C \\\ C &= 6 - \log 4. \end{align*} $$

Moreover, we can use the equation to find $x$ such that $v = 0$:

$$ \begin{align*} \frac{3}{2} (0)^2 &= \log x + (6 - \log 4) \\\ \log x &= -6 + \log 4\\\ x &= \exp(-6 + \log 4)\\\ &= 4e^{-6}. \end{align*} $$

What I can't do is find the corresponding time ($t$) value, but perhaps $x$ is sufficient for your purposes.

Hope this helps!