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(C) Given the differential equation: $\frac{dy}{dx} = \frac{(1+x)y}{(y-1)x}$.Separating the variables,we get:$\frac{y-1}{y} dy = \frac{1+x}{x} dx$Inte

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$\log(xy) + x - y = c$

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(C) Given the differential equation: $\frac{dy}{dx} = \frac{(1+x)y}{(y-1)x}$.Separating the variables,we get:$\frac{y-1}{y} dy = \frac{1+x}{x} dx$Integrating both sides:$\int (1 - \frac{1}{y}) dy = \int (1 + \frac{1}{x}) dx$$y - \log|y| = x + \log|x| + c$Rearranging the terms:$y - x - c = \log|x| + \log|y|$$y - x - c = \log|xy|$$\log|xy| + x - y = -c$Since $-c$ is also an arbitrary constant,we can write it as $C$:$\log|xy| + x - y = C$.

The solution of the differential equation $\frac{dy}{dx} = \frac{(1+x)y}{(y-1)x}$ is (where $c$ is the constant of integration):

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