Question

Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

xy = log y + C :  `y' = (y^2)/(1 - xy) (xy != 1)`

Answer 1

y = logy + C

xy’ + y = `1/y` · y’

y² + xyy’ = y

⇒ y² = y’ - xyy’

y² = y'(1 - xy)

y’ = `y^2/(1- xy)`

∴ xy = log y +  C is a solution of the given differential equation.