Question

 Solve the following differential equation: (3xy + y²) dx + (x² + xy) dy = 0 

Answer 1

(3xy + y²) dx + (x² + xy) dy = 0 

`(dy)/(dx) = - ((3xy + y^2)/(x^2 + xy))`

Put y = vx ⇒ `(dy)/(dx) = v + x (dv)/(dx)` 

`v + x (dv)/(dx) = - ((3x . vx + v^2 x^2)/(x^2 + x . vx))`

` x (dv)/(dx) =  (-3v - v^2)/(1 + v) - v`

` x (dv)/(dx) =  (-3v - v^2 - v - v^2)/(1 + v) `

` x (dv)/(dx) = (-2v^2 - 4v)/(1 + v)`

`(1 + v)/(2v^2 + 4v)  "dv" = -(1)/(x) "dx"`

` int_  (1 + v)/(2v^2 + 4v)  "dv" = int_  -(1)/(x) "dx"`

` 1/4  int_  (2 + 2v)/(2v + v^2) "dv" = - int_  (1)/(x) "dx"`

`1/4 log  | v^2 + 2v| = - log  | x | + c`

`1/4 log ((y^2)/(x^2) + 2 (y)/(x)) . x = c `

`log ((y^2)/(x) + 2y) = 4c`