Show that the general solution of the differentia equation `(dy)/(dx)+(y^2+ y+1)/(x^2+x+1)=0` is given by `x+y+1=A(1-x-y-2x y)` where A is a parameter.

Explanation

Given differntial equation is :
`(dy)/(dx)+(y^(2)+y+1)/(x^(2)+x+1)=0`
`implies (dy)/(y^(2)+y+1)+(dx)/(x^(2)+x+1)=0`
On integration
`int(dy)/(y^(2)+y+1)+int(dx)/(x^(2)+x+1)=C`
`impliesint(dy)/(y^(2)+y+1+((1)/(2))^(2)-((1)/(2))^(2))+int(dx)/(x^(2)+x+1+((1)/(2))^(2)-((1)/(2))^(2))=C`
`impliesint(dy)/((y+(1)/(2))^(2)+(1-(1)/(4)))+int(dx)/((x+(1)/(2))^(2)+(1-(1)/(4)))=C`
`impliesint(dy)/((y+(1)/(2))^(2)+((sqrt(3))/(2))^(2))+int(dx)/((x+(1)/(2))^(2)+((sqrt(3))/(2))^(2))=C`
`implies(2)/(sqrt(3))tan^(-1)((y+(1)/(2))/((sqrt(3))/(2)))+(2)/(sqrt(3))tan^(-1)((x+(1)/(2))/((sqrt(3))/(2)))=C`
`implies tan^(-1)(2y+1)/(sqrt(3))+tan^(-1)(2x+1)/(Sqrt(3))=(sqrt(3)C)/(2)=k` (say)
`impliestan^(-1)[((2y+1)/(sqrt(3))+(2x+1)/(sqrt(3)))/(1-((2y+1)/(sqrt(3)))((2x+1)/(sqrt(3))))]=k`
`impliestan^(-1)[((2y+1+2x+1)/(sqrt(3)))/(1-((4xy+2x+2y+1)/(3)))]=k`
`implies(2sqrt(3)(x+y+1))/(3-(4xy+2x+2y+1))=tank`
`implies(2sqrt(3)(x+y+1))/(2(1-x-y-2xy))=tank`
`implies x+y+1=(1)/(sqrt(3))tank(1-x-y-2xy)`
Let `A=(1)/(sqrt(3))tank` which is an arbitrary constant.
`implies x+y+1=A(1-x-y-2xy)`