Question

1. (x+3y-5)dx-(x-y-1)dy=0

Answer 1

Substitutions X=x-2, Y=y-1 is clear from solving stationary point equations:

$\begin{aligned} x+3y-5=0 \\ x-y-1=0 \end{aligned}$ => $\begin{aligned} x=2 \\ y=1 \end{aligned}$

Equation in terms of the variable z was integrated as follows:

$\frac{dz}{dX}X=\frac{(z+1)^2}{1-z}$ => $\frac{(1-z)dz}{(z+1)^2}=\frac{dX}{X}$ => $\frac{2-(1+z)}{(z+1)^2}dz=\frac{dX}{X}$

Integrating

$\int\frac{2-(1+z)}{(z+1)^2}dz=\int\frac{dX}{X}$ => $\int\frac{2dz}{(z+1)^2}-\int\frac{dz}{z+1}=\int\frac{dX}{X}$ =>$-\frac{2}{z+1}-ln|z+1|=ln|X|-C$ => $C=\frac{2}{z+1}+ln|z+1|+ln|X|$

Remember z=Y/X. So from here

$C=\frac{2}{\frac{Y}{X}+1}+ln|\frac{Y}{X}+1|+ln|X|$ => $C=\frac{2X}{X+Y}+ln|X+Y|$