# Math Help - nonlinear ODE

1. ## nonlinear ODE

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.

2. Originally Posted by zokomoko
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 $x'$

so

$x' x'' = -\ln x x' - x'$

and integrate giving

$\frac{1}{2} x'^2 = - x \ln x + c_1$ so $x' = \pm\, \sqrt{c_1 - 2 x \ln x}$

so (absorbing a 2 into c1)

$\int \frac{dx}{\sqrt{c_1 - 2 x \ln x}} = \pm \,t + c_2$