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.
 A: You have $$3x x''=1 $$
$$3 x''=\frac 1 x $$
Multiply by $x'$, to get
$$3x'x'' = \frac{x'}{x} $$
Now you have
$$\frac 3 2 d(x'^2) = d \log x$$
Upon integration you get
$$\frac 3 2 x'^2 = \log x +C$$ 
which quite a though ODE to solve. You might try an implicit solution:
$$\int \frac{dx}{\sqrt{2/3 \log x+c}}=t+K$$
then let $2/3 \log x +C=u$, and get
$$\int \frac{3}{2} \gamma \frac{\exp{\left[\frac{3}{2}u\right]}}{\sqrt u}du=t +K$$ 
$\gamma$ is $\exp -3/2C$. Then let $\sqrt {3u/2} =m$ and you get
an integral to express in terms of the error functions as Robert did.
A: Another method, but amounting to some similar calculations.  We have unknown function $x$ of variable $t$, but the DE does not actually mention $t$.  So we can do this:  Write $v=x'=dx/dt$ then convert to a DE for $v$ as a function of $x$.  Then it will be a first order DE.  
Details: $v = x'$ so
$$
\frac{dv}{dx} = \frac{dv/dt}{dx/dt} = \frac{x''}{v} = \frac{1}{3xv}
$$
The result is the same as Peter's:
$$
\frac{3}{2} v^2 = \log x + C
$$
A: 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!
A: I get the implicit solution
$$ t=2 e^{-6} \sqrt{6 \pi} \left(\text{erfi}(\sqrt{6}) - \text{erfi}\left(\sqrt {\ln  \left( x/4 \right) +6} \right)\right)$$
Note that $\ln(x/4) + 6 \to 0$ as $x \to 4/e^6+$, corresponding to $t \to 2 \sqrt{6 \pi} e^{-6} \text{erfi}(\sqrt{6}) \approx 2.244562992$.  Since the derivative of the implicit solution is
$$ 1=-{\frac {\sqrt {6}\ x' \left( t \right) }{2\sqrt {
\ln  \left( x \left( t \right)/4  \right) +6}}}
$$
this must correspond to $x'(t) \to 0-$.
