Closed form solution for the (easy at first glance) IVP $wu' =(2-w) u$, $ww'=u-w$ Consider the nonlinear autonomous 1st order IVP:
\begin{align*}
u'(t) & = \frac{(2-w) u}{w}  \\ 
w'(t) & = \frac{u-w}{w}
\end{align*} $t > 0$, with $u(0)=w(0)=0$. We know that $t = 0$ is a singular point of the system and the asymptotic behaviors $u = 3w = 6t$ are valid for $t \ll 1$.
Looking at the phase portrait, we see that the critical point $u = w = 2$ is a stable focus of the system. That allows me to say that for $t \to \infty$, $u$ and $w$ converges to the value $2$. I can support this numerically but trying to derive an analytical solution for all $t$ seems impossible.
Both Matlab and Mathematica tell me that there does not exist such a solution to the system. I have tried a variety of change of variables, which led me to nowhere, and divide formally the equations to get: 
$$ \frac{\mathrm{d} u}{\mathrm{d} w}  = \frac{(2-w) u}{u-w}, \quad u(0) = 0$$
I am (and so is Mr. Mathematica) completely unable to solve this equation although it looks easier (at least to the untrained eye). 
My question is, 

  
*
  
*Should I give up on solving this equation algebraically and embrace numerics? 
  
*Is there any change of variables that would make my life easier?
  

Any thoughts are deeply appreciated.
 A: It is possible to separate $u$ and $w$ quite easily for your system:
$\begin{cases} wu'=(2-w)u \\ ww'=u-w \end{cases}\implies
\begin{cases} u'=(2-w)(w'+1) \\ u=w(w'+1) \end{cases}$
We can then isolate $w$ like this : $\quad w'(w'+1)+ww''=(2-w)(w'+1)$
But I'm afraid this ODE is not much simpler to solve even though it is in single variable.
A: There are two different attempts to attack the equation. But both of them are unsuccessful.
$${\bf Attempt\ 1.\ To\ full\ differencial}$$
Let $\ v=w-2,\ $ then
$$
\begin{cases}
u'=-\,\dfrac{v}{v+2}\,u\\[4pt]
u = (v'+1)(v+2), 
\end{cases}\rightarrow
\begin{cases}
v''(v+2) + v'(v'+v+1) + v = 0\\
u = (v'+1)(v+2), 
\end{cases}
$$
$$\left(\left(v'+\frac{v}2\right)(v+2)\right)' + v = 0,$$
and this seems as the maximum of possible.
$${\bf Attempt\ 2.\ Substitution.}$$
Let $\dot a = \frac{da}{dt},$ then as shown
$$u = w(\dot w +1),\quad \dot u = (2-w)(\dot w+1),$$
$$w\ddot w+ (\dot w + w - 2)(\dot w+1) = 0.$$
That means that
$$w\dot y + (y+w-2)(y+1) = 0,$$
where
$$y=\dot w.$$
If to consider $y$ as function of $w,$ then
$$\dot y = \dfrac{dy}{dt} = \dfrac{dy}{dw}\cdot\dfrac{dw}{dt} = y'y,$$
$$wyy'+(y-w-2)(y+1) = 0,$$
$$y(wy'+ y-w-1) = w+2,$$
$$y((wy)'-w-1) = w+2.\qquad(1)$$
Homogenius equation is:
$$y((wy)'-w-1) = 0.$$
Non-trivial solution:
$$wy = \frac12w^2 + w + C_1,$$
$$y = \frac12w+1+\dfrac {C_1}w.$$
Using the constant variation method:
$$y=\frac12w+1+\dfrac zw.$$
Substitution to $(1)$ gives:
$$\left(\frac12w+1+\dfrac zw\right)z' = w+2,$$
$$(2z+w^2+2w)z' = 2(w^2+2w),$$
but further simplification seems impossible.
