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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)`
Solution
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.
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