Differential equation: (x+2y1)dx(x+2y5)dy=0

(x+2y-1)dx-(x+2y-5)dy=0

(x+2y1)dx(x+2y5)dy=0

Let u=x+2y1du=dx+2dy which gives dx=du2dy
Substituting to the equation
u(du2dy)(u4)dy=0udu(3u4)dy=0udu3u4dy=013(143u4)dudy=013(143u4)dudy=C13u43ln|3u4|y=Cu4ln|3u4|3y=C4ln|3u4|=u3y+C4ln|3(x+2y1)4|=x+2y3y+C4ln|3x+6y7|=xy+C3x+6y7=Cexy4(ANSWER)

P.S. There can also be other forms of the general solution