Find the value of $x(t)$ when $x'(t)=0$ given $x''(t) = a\cdot x'(t) / x(t)$ and some initial conditions I am considering the differential equation
$$ x''(t) = a\frac{x'(t)}{x(t)} $$
where $a>0$ is an arbitrary constant. I cannot solve this equation directly, but I am wondering if there's still any way I could answer the following below.
Given an initial condition $x(0) = x_{0}$ and $x'(0) = -v_{0}$ where $x_{0}, v_{0}>0$, how can I determine the position $x(T)$ for which $x'(T) = 0$? I don't care what the time $T$ is, just what is $x(T)$? Maybe there is a clever method to figure out the answer.
If a simple formula can't be found, then I am wonder if there are any numerical technique, even approximate ones at that.
 A: $\ddot x(t) = a\dfrac{\dot x(t)}{x(t)}; \; \; (1)$
$\ddot x(t) = a\dfrac{d}{dt}(\ln \; x(t)); \; \; (2)$
given that
$x(0) = x_0 > 0, \; \; (3)$
and
$\dot x(0) = -v_0 < 0, \; \; (4)$
(2) may be integrated with respect to $t$ 'twixt $0$ and $T$:
$\dot x(T) + v_0 = \dot x(T) - \dot x(0) = \displaystyle \int_0^T \ddot x(t) \; dt$
$= a \displaystyle \int_0^T \dfrac{d}{dt}(\ln \; x(t)) \; dt = a(\ln \; x(T) - \ln \; x(0))$
$ = a(\ln \; x(T) - \ln \; x_0) = a\ln \dfrac{x(T)}{x_0}. \; \; (5)$
Now if
$\dot x(T) = 0, \; \; (6)$
then (5) yields
$v_0 = a\ln \dfrac{x(T)}{x_0}, \; \; (7)$
whence
$x(T) = x_0 e^{v_0/a}, \; \; (8)$
a formula giving $x(T)$ in terms of $x_0$ and $v_0$.
One can also argue as follows:. from (2)
we write
$\dfrac{d}{dt}(\dot x(t) - a\ln \; x(t)) = 0, \; \; (9)$
or
$\dot x(t) - a\ln \; x(t) = C, \; \text{a constant}; \; \; (10)$
thus,
$\dot x(T) - a\ln \; x(T) = \dot x(0) - a\ln \; x(0), \; \; (11)$
or
$\dot x(T) - \dot x(0) = a(\ln \; x(T) - \ln \; x(0)), \; \; (12)$
and in light of (3), (4) this becomes
$\dot x(T) + v_0 = a(\ln \; x(T) - \ln \; x_0) = a\ln \dfrac{x(T)}{x_0}; \; \; (13)$
applying (6) to this equation begets (7), and we may continue as before.
A: Take $z = x'$. Since, $x'(0) < 0$, in a small interval around $0$, we can write $t$ as a function of $x$. Hence, the following equations makes sense
$$ x''=\frac{dx'}{dt} = \frac{dz}{dt} = \frac{dz}{dx}\frac{dx}{dt} = z \frac{dz}{dx} $$
Putting this in the given differential equation and noting that $x'(0) \neq 0 $, we get
$$ \frac{dz}{dx} = \frac{a}{x} $$
Since, $x_0 > 0$, so solving this equation we get
$$ z = a \ln(x) + C$$where $C$ is some suitable real number which satisfy the initial conditions. The new differential equation then take the form
$$ x' = a \ln(x) + C $$
From here you can determine your $x(T)$ such that $x'(T)=0$. If you try to solve it explicitly for $x$, I am not sure whether a solution in terms of elementary functions exist.
A: $$x''=a \frac {x'} x$$ Switching variables
$$-\frac {t''}{[t']^3}=\frac{a} {x \,t'} $$ Reduction of order $p=t'$ gives
$$t'=p=\frac{1}{a \log (x)+c_1}$$ Now, not elementary at all
$$t+c_2=\frac{e^{-\frac{c_1}{a}} }{a}\text{Ei}\left(\frac{c_1}{a}+\log (x)\right)$$ which cannot be inversed.
