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Question

(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}

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given the differential equation: $\frac{dy}{dx} = \frac{(1 + x)y}{(y - 1)x}$.Separating the variables,we get:$\frac{y - 1}{y} dy = \frac{1 + x}{x} dx$.This can be rewritten as:$(1 - \frac{1}{y}) dy = (1 + \frac{1}{x}) dx$.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:$x - y + \log|x| + \log|y| = -c$.Using the property $\log a + \log b = \log(ab)$:$x - y + \log|xy| = C$ (where $C = -c$ is a constant).Thus,the solution is $\log(xy) + x - y = c$.

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

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