The solution curve of the differential equation y(dx)/(dy) = x( _ e...

Given: $y\dfrac{dx}{dy} = x\left(\log_{e}x - \log_{e}y + 1\right), x > 0, y > 0$

$⇒ \dfrac{dx}{dy} = \left(\dfrac{x}{y}\right)\left[\log_{e}\left(\dfrac{x}{y}\right) + 1\right]$

Putting, $x = vy$

$⇒ \dfrac{dx}{dy} = v + y\dfrac{dv}{dx}$

$⇒ v + y\dfrac{dv}{dy} = \left(\dfrac{vy}{y}\right)\left[\log_{e}\left(\dfrac{vy}{y}\right) + 1\right]$

$⇒ v + y\dfrac{dv}{dy} = v\left[\log_{e}\left(v\right) + 1\right]$

$⇒ v + y\dfrac{dv}{dy} = v{\log}_{e}\left(v\right) + v$

$⇒ y\dfrac{dv}{dy} = v{\log}_{e}\left(v\right)$

$⇒ \dfrac{dv}{v{\log}_{e}v} = \dfrac{dy}{y}$

$⇒ ∫ \dfrac{dv}{v{\log}_{e}v} = ∫ \dfrac{dy}{y}$

Putting, $\log_{e}v = t$

$⇒ \dfrac{dv}{v} = dt$

$⇒ ∫ \dfrac{dt}{t} = ∫ \dfrac{dy}{y}$

$⇒ \log t = \log y + c$

$⇒ \log \left(\log_{e}\dfrac{x}{y}\right) = \log y + c$

Using point \left(e, 1\right)$</p><p>$⇒ \log \left(\log_{e}e\right) = \log \left(1\right) + c$</p><p>$⇒ c = 0$</p><p>$⇒ \log \left(\log_{e}\dfrac{x}{y}\right) = \log y$</p><p>$⇒ \left|\log_{e}\dfrac{x}{y}\right| = y$