solve the following ODE:
d^2(x)
------ = -ln(x)-1
dt^2
I've tried a couple of things and got this:
d^2(x)
------ = -ln(x*e)
dt^2
but couldn't solve it, your help is appriciated.
The best I can do is to quadrature. Multiply your ODE by $\displaystyle x'$
so
$\displaystyle x' x'' = -\ln x x' - x'$
and integrate giving
$\displaystyle \frac{1}{2} x'^2 = - x \ln x + c_1$ so $\displaystyle x' = \pm\, \sqrt{c_1 - 2 x \ln x}$
so (absorbing a 2 into c1)
$\displaystyle \int \frac{dx}{\sqrt{c_1 - 2 x \ln x}} = \pm \,t + c_2$