# 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.

-
@J.D. - yes, it's equivalent –  Belgi May 20 '12 at 21:02

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!

-

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.

-

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

-
$$t=2 e^{-6} \sqrt{6 \pi} \left(\text{erfi}(\sqrt{6}) - \text{erfi}\left(\sqrt {\ln \left( x/4 \right) +6} \right)\right)$$
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 $x'(t) \to 0-$.