y dx +x(ln x −ln y−1)dy=0 where y(1)=0


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Question Number 125958 by bramlexs22 last updated on 15/Dec/20

$y d x + x (\ln x - \ln y - 1) d y = 0$ $w h e r e y (1) = 0$
	Answered by Olaf last updated on 16/Dec/20

	$y d x + x (\ln x - \ln y - 1) d y = 0$ $Let y = x u$ $x u d x + x (\ln x - \ln x - \ln u - 1) (u d x + x d u) = 0$ $x u d x - x (\ln u + 1) (u d x + x d u) = 0$ $- x (u \ln u d x + x \ln u d u + x d u) = 0$ $u \ln u d x + x (\ln u + 1) d u = 0$ $(\frac{\ln u + 1}{u \ln u}) d u = - \frac{d x}{x}$ $\frac{d u}{u} + \frac{d u}{u \ln u} = - \frac{d x}{x}$ $\ln ∣ u ∣ + \ln ∣ \ln u ∣ = - \ln ∣ x ∣ + C_{1}$ $\ln x and \ln y exist \Rightarrow x > 0 and u > 0$ $\ln u + \ln ∣ \ln u ∣ = - \ln x + C_{1}$ $\ln (u ∣ \ln u ∣) = \ln \frac{C_{2}}{x}$ $u ∣ \ln u ∣ = \frac{C_{2}}{x}$ $\frac{y}{x} ∣ \ln \frac{y}{x} ∣ = \frac{C_{2}}{x}$ $y ∣ \ln \frac{y}{x} ∣ = C_{2}$ $∣ \ln \frac{y}{x} ∣ = \frac{C_{2}}{y}$ $C_{2} > 0 and y > 0 \Rightarrow \ln \frac{y}{x} = \frac{C_{2}}{y}$ $\frac{y}{x} = e^{\frac{C_{2}}{y}}$ $x = {y e}^{- \frac{C_{2}}{y}}$ $x = {y e}^{\frac{C_{3}}{y}}$
	Answered by liberty last updated on 17/Dec/20

	$y d x = (1 + \ln y - \ln x) x d y$ $\frac{d y}{d x} = \frac{y}{x (1 + \ln y - \ln x)}; l e t y = z x$ $\Leftrightarrow z + x \frac{d z}{d x} = \frac{z x}{x (1 + \ln z x - \ln x)}$ $z + x \frac{d z}{d x} = \frac{z}{1 + \ln z}; x \frac{d z}{d x} = \frac{- z \ln z}{1 + \ln z}$ $\Leftrightarrow \frac{(1 + \ln z) d z}{z \ln z} = - \frac{d x}{x}$ $\int \frac{d z}{z \ln z} + \int \frac{d z}{z} = - \int \frac{d x}{x}$ $\Leftrightarrow \ln z + \ln (\ln z) + \ln x = C$ $\Leftrightarrow \ln (z x \ln z) = C \Rightarrow z x \ln z = C_{1}$ $(\frac{y}{x}) x \ln (\frac{y}{x}) = C_{1} \Rightarrow y \ln (\frac{y}{x}) = C_{1}$ $\Leftrightarrow \frac{y}{x} = e^{\frac{C_{1}}{y}}; y (1) = 0 \Rightarrow n o t c o r r e c t$
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