Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
@J.D. - yes, it's equivalent – Belgi May 20 '12 at 21:02
up vote 3 down vote accepted

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!

share|cite|improve this answer

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.

share|cite|improve this answer

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

share|cite|improve this answer

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 $x'(t) \to 0-$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.