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

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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 $\frac{1+y}{y} dy = \frac{1+x}{x} dx$.This can be written as $(\frac{1}{y} + 1) dy = (\frac{1}{x} + 1) dx$.Integrating both sides, we get $\int (\frac{1}{y} + 1) dy = \int (\frac{1}{x} + 1) dx$.$\log|y| + y = \log|x| + x + C$.Rearranging the terms, $y - x = \log|x| - \log|y| + C$.$y - x = \log|\frac{x}{y}| + C$.Let $C = \log c$, then $y - x = \log|\frac{x}{y}| + \log c = \log|\frac{cx}{y}|$.Thus, the solution is $y - x = \log \left(\frac{cx}{y}\right)$.

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

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