Question

Find the general solution of y²dx + (x² – xy + y²) dy = 0.

Find the general solution of the differential equation:

y²dx + (x² – xy + y²) dy = 0

Answer 1

The given equation is y²dx + (x² – xy + y²) dy = 0

⇒ y²dx = – (x² – xy + y²)dy

⇒ `"dx"/"dy" = - (x^2 - xy + y^2)/y^2`

Since it is a homogeneous differential equation

∴ Put x = vy

⇒ `"dx"/"dy" = "v" + y * "dv"/"dy"`

So, `"v" + y * "dv"/"dy" = - (("v"^2y^2 - "v"y^2 + y^2)/y^2)`

⇒ `"v" + y * "dv"/"dy" = -(y^2("v"^2 - "v" + 1))/y^2`

⇒ `"v" + y * "dv"/"dy" = (-"v"^2 + "v" - 1)`

⇒ `y * "dv"/"dy" = - "v"^2 + "v" - 1 = "v"`

⇒ `y * "dv"/"dy" = - "v"^2 - 1`

⇒ `"dv"/(("v"^2 + 1)) = - "dy"/y`

Integrating both sides, we get

⇒ `int "dv"/(("v"^2 + 1)) = -int "dy"/y`

⇒ `tan^-1"v" = - log y + "c"`

⇒  `tan^-1(x/y) + log y + "c"`

Hence the required solution is `tan^-1(x/y) + log y + "c"`.