Solving the ODE $\;\;y'^2-yy'+e^x=0$ 
Solve,
$$y'^2-yy'+e^x=0$$

By seeing it as a quadratic equation in $y'$, we have $y'=\dfrac{y\pm\sqrt{y^2-4e^x}}{2}$. But from here I don't know how to continue since both $y^2$ and $e^x$ are under square root.
Another approach I've tried is using $y'=\frac1{x'}$,
$$\frac1{x'^2}-\frac{y}{x'}+e^x=0\;\Rightarrow\; e^x x'^2-y\;x'+1=0$$
I'm not sure if it helps.
 A: Use the substitution $y = 2e^{x/2}v$
$$\begin{cases}y = 2e^{x/2}v \\ y' = 2e^{x/2}v' + e^{x/2}v\end{cases} \implies e^x(2v'+v)(2v'-v) + e^x = 0$$
or
$$4{v'}^2-v^2+1=0$$
which can be solved by integrating directly
$$\pm x/2 + C = \int\frac{dv}{\sqrt{v^2-1}}  = \cosh^{-1}(v)$$
which means
$$v = \sqrt{1+C^2}\cosh\left(\frac{x}{2}\right)+C\sinh\left(\frac{x}{2}\right)$$
and finally
$$y = 2\sqrt{1+C^2}e^{x/2}\cosh\left(\frac{x}{2}\right)+2Ce^{x/2}\sinh\left(\frac{x}{2}\right)$$
or
$$y = (\sqrt{1+C^2}+C)e^x + (\sqrt{1+C^2}-C)$$

$\textbf{Edit}$: Even more simply, the original found answer can be simplified to
$$\boxed{y = Ce^x + \frac{1}{C}}$$
The missing solutions can be found by going back to the equation
$$4{v'}^2-v^2+1 = 0$$
The original solution assumed $v^2-1\neq 0$ so that we could divide by it later. In that case though, notice that $v=\pm 1$ is the constant solution
$$4(0)^2-(\pm1)^2+1 = 0$$
so we also have
$$\boxed{y = \pm 2e^{x/2}}$$
Moral of the story, always look for constant solutions in nonlinear DEs.
