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