The correct option is C 2y−x=log(x−y+z)+c
The given equation can be written as
(2x−2y+5)dy=(x−y+3)dx
⇒dydx=x−y+32(x−y)+5
Put x−y=V⇒1−dydx=dVdx
Therefore, the given equation becomes
1−dVdx=V+32V+5 ⇒dVdx=V+22V+5
⇒dx=2V+5V+2dV ⇒dx=(2+1V+2)dV
On integrating, we get x=2V+log(V+2)+c
⇒x=2(x−y)+log(x−y+2)+c
Therefore, 2y−x=log(x−y+2)+c, is the required solution.