Question

Show that the general solution of differential equation `"dy"/"dx" + ("y"^2 + "y" + 1)/("x"^2 + "x" + 1) = 0` is given by (x + y + 1) = (1 - x - y - 2xy).

Answer 1

`"dy"/"dx" + ("y"^2 + "y" + 1)/("x"^2 + "x" + 1) = 0`

∴ `"dy"/"dx" = - (("y"^2 + "y" + 1)/("x"^2 + "x" + 1))`

∴ `1/("y"^2 + "y" + 1)"dy" = - 1/("x"^2 + "x" + 1)"dx"`

Integrating both sides, we get

`int 1/("y"^2 + "y" + 1)"dy" = - int 1/("x"^2 + "x" + 1)"dx"`

∴ `int 1/(("y"^2 + "y" + 1/4) + 3/4) "dy" = - int 1/(("x"^2 + "x" + 1/4) + 3/4)"dx"`

∴ `int 1/(("y" + 1/2)^2 + (sqrt3/2)^2)"dy" = - int 1/(("x" + 1/2)^2 + (sqrt3/2)^2)"dx"`

∴ `1/(sqrt3/2) tan^-1 [("y" + 1/2)/(sqrt3/2)] = - 1/((sqrt3/2)) tan^-1 [("x" + 1/2)/(sqrt3/2)] + "c"_1` 

∴ `2/sqrt3 tan^-1 (("2y" + 1)/sqrt3) + 2/sqrt3 tan^-1 (("2x" + 1)/sqrt3) = "c"_1`

∴ `2/sqrt3 tan^-1 [((("2y" + 1)/sqrt3) + (("2x" + 1)/sqrt3))/(1 - (("2y" + 3)/sqrt3)(("2x" + 1)/sqrt3))] = "c"_1`

∴ `tan^-1  ((("2y" + 1 + 2"x" + 1)/sqrt3))/(((3 - 4"xy" - 2"y" - 2"x" - 1)/3)) = sqrt3/2 "c"_1`

∴ `(("2y" + "2x" + 2)sqrt3)/(2 - "2x" - 2"y" - 4"xy") = tan (sqrt3/2  "c"_1)`

∴ `("x + y + z")/(1 - "x" - "y" - 2"xy") = 1/sqrt3 tan (sqrt3/2  "c"_1) = "c"`, where c = `1/sqrt3 tan (sqrt3/2  "c"_1)`

∴ (x  + y + 1) = c(1 - x - y - 2xy)

This is the general solution.