Differential equations Bernoulli? We have $$2y'=x+\ln(y')$$
I want to solve this by Bernoulli but I can't find the right variable to divide this :/ how can I solve this ? I thought about dividing by $y'$,but this is'nt taking me anywhere :/
 A: We want :
$$2\,y'=x+\ln(y')$$
Let's obtain an expression for $y'$ first :
$$-x=\ln\left(y'e^{-2y'}\right)=\ln\left(-2y'e^{-2y'}\right)-\ln(-2)$$
or
$$(-2\,y')e^{(-2y')}=e^{\ln(-2)-x}=-2\,e^{-x}$$
But the Lambert W function is defined by $z=W(z)e^{W(z)}$ so that (for $z=-2e^{-x}$) :
$$-2\,y'=W(-2e^{-x})$$
and we get :
$$y(x)=C-\frac 12\int W(-2e^{-x}) dx$$
$$y(x)=C-\frac 12\left(-\frac 12 W(-2e^{-x})^2-W(-2e^{-x})\right)$$
and the conclusion :
$$\boxed{\displaystyle y(x)=C+\frac 14 W(-2e^{-x})\left(W(-2e^{-x})+2\right)}$$

DETAIL of the Lambert W integration
The LambertW functions allowed some authors to rediscover the Inverse Function Integration i.e. following formula for $z:=f(y)$ :
$$\int f^{-1}(z)\,dz=y\,f(y)-\int f(y)\,dy$$
but we won't need this formula since a direct derivation is straightforward.
Let's set $\ z:=-2e^{-x}\;$ as well as $\ w:=W(z)\ $ and $\ z=w\,e^w\ $ and rewrite :
\begin{align}
\int W(-2e^{-x})\,dx&=-\int \frac{W(z)}z\,dz\\
&=-\int \frac 1{we^w}\,w\;(1+w)e^w\,dw\\
&=-\int (1+w)\,dw\\
&=-\frac{w^2}2-w\\
&=-\frac 12 W(-2e^{-x})^2-W(-2e^{-x})\\
\end{align}
