If x frac{dy}{dx} = y(log y - log x + 1),then | vedclass

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

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$\log \frac{y}{x} = cx$,where $c$ is the constant of integration

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(D) Given the differential equation: $x \frac{dy}{dx} = y(\log(\frac{y}{x}) + 1)$.Divide by $x$: $\frac{dy}{dx} = \frac{y}{x}(\log(\frac{y}{x}) + 1)$.Let $v = \frac{y}{x}$,so $y = vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting these into the equation: $v + x \frac{dv}{dx} = v(\log v + 1)$.$v + x \frac{dv}{dx} = v \log v + v$.$x \frac{dv}{dx} = v \log v$.Separating variables: $\frac{dv}{v \log v} = \frac{dx}{x}$.Integrating both sides: $\int \frac{1}{v \log v} dv = \int \frac{1}{x} dx$.Let $u = \log v$,then $du = \frac{1}{v} dv$. The integral becomes $\int \frac{1}{u} du = \int \frac{1}{x} dx$.$\log|u| = \log|x| + \log|c|$.$\log|\log v| = \log|cx|$.$\log v = cx$.Substituting $v = \frac{y}{x}$ back: $\log(\frac{y}{x}) = cx$.

If $x \frac{dy}{dx} = y(\log y - \log x + 1)$,then the solution of the equation is

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