Closed form solution for the (easy at first glance) IVP $wu' =(2-w) u$, $ww'=u-w$

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.

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.