Prove that the general sln of the D.E (x+2y-4)dx-(2x+y-5)dy=0 can be expressed in the form (x-y-1)^3=c(x+y-3)?

plz quickly i need help

Prove that the general sln of the D.E (x+2y-4)dx-(2x+y-5)dy=0 can be expressed in the form (x-y-1)^3=c(x+y-3)?
(x+2y-4)dx-(2x+y-5)dy=0





dy/dx=(x+2y-4)/(2x+y-5)





x=x1+h


y=y1+k





dy/dx=dy1/dx1=(x1+2y1+h+2k-4)/





/(2x1+y1+2h+k-5)





h+2k-4=0


2h+k-5=0 h=2; k=1





x=x1+2; x1=x-2


y=y1+1 y1=y-1





dy1/dx1=(x1+2y1)/(2x1+y1)=





y1/x1=u; y1=u*x1





dy1/dx1=du/dx1*x1+u=





=(x1+2u*x1)/(2x1+u*x1)=





=(2u+1)/(u+2);





du/dx1*x1=(2u+1)/(u+2)-u=





=(1-u^2)/(u+2)





int (u+2)/(1-u^2)*du=int dx1/x1





(u+2)/(1-u^2)=a/(1-u)+b/(1+u); a=3/2; b=1/2





1/2*[3*int du/(1-u)+





+int du/(1+u)]=ln(x1)+lnc





-3*ln|u-1|+ln|1+u|=





=2*ln(cx1)





(1+u)/(1-u)^3=C*(x1)^2; x1=x-2; y1=y-1; u=y1/x1;





1+((y1/x1)=C(x1)^2*(1-y1/x1)^3





x1+y1=C(x1-y1)^3; x1=x-2; y1=y-1





x-2+y-1=C(x-2-y+1)^3





(x-y-1)^3=C(x+y-3)