Answer 1

`"x"^2 "dy"/"dx" = "x"^2 + "xy" + "y"^2`

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

Put y = vx

∴ `"dy"/"dx" = "v + x""dv"/"dx"`

∴ (1) becomes, `"v + x""dv"/"dx" = ("x"^2 + "x" * "vx" + "v"^2"x"^2)/"x"^2`

∴ `"v + x""dv"/"dx" = 1 + "v" + "v"^2`

∴ `"x" "dv"/"dx" = 1 + "v"^2`

∴ `1/(1 + "v"^2) "dv" = 1/"x" "dx"`

Integrating, we get

`int 1/(1 + "v"^2) "dv" = int 1/"x" "dx"`

∴ tan-1 v = log |x| + c

∴  tan^-1 `("y"/"x") = log |"x"| + "c"`

This is the general solution.