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Question
For the given below, verify that the given function (implicit or explicit) is a solution to the corresponding differential equation.
`x^2 = 2y^2 log y : (x^2 + y^2) dy/dx - xy = 0`
Solution
We have, `x^2 = 2y^2 log y` ....(1)
Differentiating (1) w.r.t. x, we get
`2x = 2 [2y log y + y^2 xx 1/y] dy/dx`
`= 2 [2y log y + y] dy/dx`
⇒ `dy/dx = x/ (2y log y + y) = x/ (y (2 log y + 1))`
From (1), 2 log `y = x^2/y^2`
∴ `dy/dx = x/(y [x^2/y^2 + 1])`
`= (xy/(x^2 + y^2))`
⇒ `(x^2 + y^2) dy/dx - xy = 0`
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