$$y'= \left(\frac{y+2}{x+y-1}\right)^2$$
So, i've been struggling with this for about 4 days, and sure, call me names, math is not my stronger suit, but I'd really like to figure this out. I've tried all sorts of approaches, to separate it, to get it on homogeneous equation, but i always get stuck, and can't get around. I'd really appreciate your help and some tips on how should I approach it.
By using the sustitutions $t=x-3$ and $r=y+2$ the DE becomes $$\frac{dr}{dt}=\left(\frac{r}{t+r}\right)^2\tag{1}$$ Which is homogeneous. So lets put $r=ut$, then DE $(1)$ converts to $$u+t\frac{du}{dt}=\left(\frac{u}{1+u}\right)^2\tag{1'}$$ Now perform algebraic manipulations: \begin{align*} \frac{du}{dt}&=\frac{u^2-u(1+u)^2}{(1+u)^2t}\\[3pt] &=\frac{-u^3-u^2-u}{(1+u)^2t}\\[3pt] \frac{u^2+2u+1}{u(u^2+u+1)}\frac{du}{dt}&=-t^{-1}\tag{2} \end{align*} Integrating both sides of eqn $(2)$ we get \begin{align*} \int\frac{u^2+2u+1}{u(u^2+u+1)}\frac{du}{dt}dt&=\int-t^{-1}dt\\[3pt] \int\left(\frac1u+\frac1{u^2+u+1}\right)du&=\int-t^{-1}dt\qquad\\[3pt] \log|u|+\frac2{\sqrt{3}}\tan^{-1}\left(\frac{2u+1}{\sqrt{3}}\right)&=-\log|t|+c\\[3pt] \log\left|\frac{y+2}{x-3}\right|+\frac2{\sqrt{3}}\tan^{-1}\left(\frac{2(y+2)}{\sqrt{3}(x-3)}+\frac1{\sqrt3}\right)&=-\log|x-3|+c \end{align*}
So $$\log|y+2|+\frac2{\sqrt{3}}\tan^{-1}\left(\frac{2(y+2)}{\sqrt{3}(x-3)}+\frac1{\sqrt3}\right)=c$$