Non-linear first order differential equations What technique should I use to solve the following first-order nonlinear ordinary differential equation:
$$y'(x)+\log (y(x))=-x-1$$
Thanks in advance

Hint:
$y'+\log y=-x-1$
$\dfrac{dy}{dx}=-x-\log y-1$
$(x+\log y+1)\dfrac{dx}{dy}=-1$
Let $u=x+\log y+1$ ,
Then $x=u-\log y-1$
$\dfrac{dx}{dy}=\dfrac{du}{dy}-\dfrac{1}{y}$
$\therefore u\left(\dfrac{du}{dy}-\dfrac{1}{y}\right)=-1$
$u\dfrac{du}{dy}-\dfrac{u}{y}=-1$
$u\dfrac{du}{dy}=\dfrac{u}{y}-1$
$u\dfrac{du}{dy}=\dfrac{u-y}{y}$
$(u-y)\dfrac{dy}{du}=uy$
This belongs to an Abel equation of the second kind.
Let $v=u-y$ ,
Then $y=u-v$
$\dfrac{dy}{du}=1-\dfrac{dv}{du}$
$\therefore v\left(1-\dfrac{dv}{du}\right)=u(u-v)$
$v-v\dfrac{dv}{du}=u^2-uv$
$v\dfrac{dv}{du}=(u+1)v-u^2$
Let $s=u+1$ ,
Then $\dfrac{dv}{du}=\dfrac{dv}{ds}\dfrac{ds}{du}=\dfrac{dv}{ds}$
$\therefore v\dfrac{dv}{ds}=sv-(s-1)^2$
$v\dfrac{dv}{ds}=sv-s^2+2s-1$
Let $t=\dfrac{s^2}{2}$ ,
Then $\dfrac{dv}{ds}=\dfrac{dv}{dt}\dfrac{dt}{ds}=s\dfrac{dv}{dt}$
$\therefore sv\dfrac{dv}{dt}=sv-s^2+2s-1$
$v\dfrac{dv}{dt}=v-s+2-\dfrac{1}{s}$
$v\dfrac{dv}{dt}=v\pm\sqrt{2t}+2\pm\dfrac{1}{\sqrt{2t}}$
This belongs to an Abel equation of the second kind in the canonical form.
Please follow the method in https://arxiv.org/ftp/arxiv/papers/1503/1503.05929.pdf
