The solution curve of the differential equati | vedclass

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(C) Given the differential equation: $y \frac{dx}{dy} = x(\ln(\frac{x}{y}) + 1)$.Divide by $y$: $\frac{dx}{dy} = \frac{x}{y}(\ln(\frac{x}{y}) + 1)$.Le

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$|\log_e \frac{x}{y}| = y$

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(C) Given the differential equation: $y \frac{dx}{dy} = x(\ln(\frac{x}{y}) + 1)$.Divide by $y$: $\frac{dx}{dy} = \frac{x}{y}(\ln(\frac{x}{y}) + 1)$.Let $v = \frac{x}{y}$,then $x = vy$. Differentiating with respect to $y$: $\frac{dx}{dy} = v + y \frac{dv}{dy}$.Substituting into the equation: $v + y \frac{dv}{dy} = v(\ln v + 1) = v \ln v + v$.$y \frac{dv}{dy} = v \ln v$.Separating variables: $\frac{dv}{v \ln v} = \frac{dy}{y}$.Integrating both sides: $\int \frac{dv}{v \ln v} = \int \frac{dy}{y}$.Let $u = \ln v$,then $du = \frac{1}{v} dv$. The integral becomes $\int \frac{du}{u} = \int \frac{dy}{y}$.$\ln|u| = \ln|y| + C \Rightarrow \ln|\ln v| = \ln y + C$.Substituting $v = \frac{x}{y}$: $\ln|\ln(\frac{x}{y})| = \ln y + C$.Passing through $(e, 1)$: $\ln|\ln(\frac{e}{1})| = \ln(1) + C \Rightarrow \ln(1) = 0 + C \Rightarrow C = 0$.Thus,$\ln|\ln(\frac{x}{y})| = \ln y$.Taking the exponential of both sides: $|\ln(\frac{x}{y})| = y$.

The solution curve of the differential equation $y \frac{dx}{dy} = x(\log_e x - \log_e y + 1)$,$x > 0, y > 0$ passing through the point $(e, 1)$ is

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