`(2x -2y +3) dx - (x -y +1) dy = 0 , x -y =u`

Explanation

`(2x-2y+3) dx + (x-y+1)dy=0`
` (x-y+1) dy = (2x-2y+3) dx`
` (dy)/(dx) = (2x -2y+3)/(x-y+1) = (2(x-y)+3)/((x-y)+1)`
`Put x-y=u " " therefore 1-(dy)/(dx) =(du)/(dx) " " therefore (dy)/(dx) =1 , (du)/(dx)`
(1) becomes , `1-(du)/(dx) = (2u+3)/(u+1) `
`(du)/(dx) =1 - (2u+3)/(u+1) = (u +1 -2u-3)/(u+1)`
` (du)/(dx) = (-u-2)/(u+1) = - ((u+2)/(u+1))`
` (u+1)/(u+2) du= -dx`
Integrating we get ,
`int (u+1)/(u+2) du= intdx +c`
` int ((u+2)-1)/(u+2) du =-int dx +c , " " therefore int (1- 1/(u+2)) du = -f dx +c`
` int 1 du - int 1/(u+2) du = -int dx +c " " therefore u-log|u+2| = -x +c`
` therefore x-y-log|x-y+2| = -x +c " " therefore 2x-y-log|x-y+2|=c`
This is the general solution.