Second-order non-linear ODE $2tx'-x=lnx'$
I differentiated both sides with respect to x:
$x'+2tx''=\frac {x''}{x'}$
Substituting $p=x'$,
$p+2tp'=\frac{p'}{p}$
But I have no clue what can I do from here on.
EDIT: $t$ is the non-dependent variable.
 A: I would say that
$x=2tp-\ln p$
$x'=2tp'+ 2p-\frac{p'}{p}$
$2tp'+ p-\frac{p'}{p}=0 \Rightarrow t'+\frac{2t}{p}=\frac{1}{p^2},\,$ie linear equation for t, $\Rightarrow t=\frac{p+C}{p^2}$
Substituting the 't' in the first equation we have parametric equations for x, t (parameter = p):
$x=2\frac{p+C}{p}-\ln p,\,\,\,t = \frac{p+C}{p^2}$
A: If you are studying differential equations, you certainly should know how to solve a quadratic equation!
$tp^2- p- C= 0$ so, by the quadratic formula, $p= \frac{1\pm\sqrt{1+4Ct}}{2}$.
A: Apply the method in http://www.ae.illinois.edu/lndvl/Publications/2002_IJND.pdf#page=2:
Let $F(t,x,s)=2ts-\ln s-x$ ,
Then $\dfrac{dt}{ds}=-\dfrac{\dfrac{\partial F}{\partial s}}{\dfrac{\partial F}{\partial t}+s\dfrac{\partial F}{\partial x}}=-\dfrac{2t-\dfrac{1}{s}}{2s+s(-1)}=\dfrac{\dfrac{1}{s}-2t}{s}=\dfrac{1}{s^2}-\dfrac{2t}{s}$
$\dfrac{dt}{ds}+\dfrac{2t}{s}=\dfrac{1}{s^2}$
I.F.$=e^{\int\frac{2}{s}ds}=e^{2\ln s}=s^2$
$\therefore\dfrac{d(s^2t)}{ds}=1$
$s^2t=\int ds$
$s^2t=s+C_1$
$t=\dfrac{1}{s}+\dfrac{C_1}{s^2}$
$\therefore\dfrac{dx}{ds}=st=1+\dfrac{C_1}{s}$
$x=\int\left(1+\dfrac{C_1}{s}\right)ds$
$x=s+C_1\ln s+C_2$
Hence $\begin{cases}t=\dfrac{1}{s}+\dfrac{C_1}{s^2}\\x=s+C_1\ln s+C_2\end{cases}$
