How to solve non-linear ODE $(v')^2+v'-\frac{v}{u}=0?$ Solving the certain problem,  I obtained non-differential equation
$(v')^2+v'-\frac{v}{u}=0,$ $v$ is function of $u.$ I'm not very familiar with differential equations, so I'm wondering is this to hard to solve or there is a certain method for solving it.
 A: $$\left(\frac{dv}{du}\right)^2+\frac{dv}{du}-\frac{v}{u}=0 \tag 1$$
The ODE is homogeneous. So, the convenient change of is :
$v(u)=uy(u) \quad\to\quad\frac{dv}{du}=y+u\frac{dy}{du} \quad\to\quad \left(y+u\frac{dy}{du}\right)^2+y+u\frac{dy}{du}-y=0$
$$\left(y+u\frac{dy}{du}\right)^2+u\frac{dy}{du}=0$$
Let $\quad u=e^x \quad\to\quad u\frac{dy}{du}=\frac{dy}{dx} $
$$\left(y+\frac{dy}{dx}\right)^2+\frac{dy}{dx}=0$$
This ODE is separable : $\quad \left(\frac{dy}{dx}\right)^2+(2y+1)\frac{dy}{dx}+y^2=0$
$$\frac{dy}{dx}=\frac12\left(2y+1 \pm\sqrt{4y+1}\right)$$
$$x=\int \frac{2dy}{2y+1 \pm\sqrt{4y+1}}$$
The next calculus has to be done two times distinctly, one with sign $+$ , one with sign $-$. 
$$x=\frac{1\pm\sqrt{4y+1} }{2y+1\pm\sqrt{4y+1}}\left(1+\left(1\pm\sqrt{4y+1}\right)\ln\left( 1\pm\sqrt{4y+1}\right) \right)+c$$
Let $\quad t=1\pm\sqrt{4y+1}\quad\to\quad y=\frac{t(t-2)}{4}\quad$ which leads in both cases to the same result :
$$x=\frac{2 }{t}+\ln(t^2)+c$$
$u=e^x=Ct^2e^{2/t}$
$v=yu=\frac{t(t-2)}{4}Ct^2e^{2/t} =\frac{C}{4}t^3(t-2)e^{2/t}$
The solution of Eq.$(1)$ is obtained on parametric form :
$$\begin{cases}
u=Ct^2e^{2/t}\\
v=\frac{C}{4}t^3(t-2)e^{2/t}
\end{cases}$$
The explicit form $v(u)$ cannot be expressed with a finite number of elementary functions. A special function is required : The Lambert W function.
$u=Ct^2e^{2/t} \quad\to\quad t= -\frac{1}{W\left(\pm\sqrt{\frac{C}{u}}\right)}$
$$v(u)=-\frac{C}{4}\left(\frac{1}{W\left(\pm\sqrt{\frac{C}{u}}\right)}\right)^3\left(-\frac{1}{W\left(\pm\sqrt{\frac{C}{u}}\right)}-2\right)\exp\left(-2W\left(\pm\sqrt{\frac{C}{u}}\right) \right)$$
