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(D) Given differential equation is $\frac{dy}{dx} = \frac{xy+y}{xy+x}$.Separating the variables,we get $\frac{dy}{dx} = \frac{y(x+1)}{x(y+1)}$.Rearran

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$y-x = \log \left(\frac{cx}{y}\right)$

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(D) Given differential equation is $\frac{dy}{dx} = \frac{xy+y}{xy+x}$.Separating the variables,we get $\frac{dy}{dx} = \frac{y(x+1)}{x(y+1)}$.Rearranging the terms,we have $\frac{1+y}{y} dy = \frac{1+x}{x} dx$.Integrating both sides,we get $\int (\frac{1}{y} + 1) dy = \int (\frac{1}{x} + 1) dx$.This results in $\log |y| + y = \log |x| + x + C$.Rearranging the terms,$y - x = \log |x| - \log |y| + C$.$y - x = \log \left|\frac{x}{y}\right| + C$.Let $C = \log c$,then $y - x = \log \left|\frac{cx}{y}\right|$.

The solution of the differential equation $\frac{dy}{dx} = \frac{xy+y}{xy+x}$ is

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