If frac{dy}{dx} = frac{xy + y}{xy + x},then t | vedclass

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

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$y = Ax{e^{x - y}}$

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(C) Given the differential equation: $\frac{dy}{dx} = \frac{y(x + 1)}{x(y + 1)}$Separating the variables,we get:$\frac{y + 1}{y} dy = \frac{x + 1}{x} dx$This can be written 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$$y + \log|y| = x + \log|x| + C$Rearranging the terms:$y - x = \log|x| - \log|y| + C$$y - x = \log|\frac{x}{y}| + C$Let $C = -\log|A|$,then:$y - x = \log|\frac{x}{y}| - \log|A| = \log|\frac{x}{Ay}|$Taking the exponential of both sides:$e^{y - x} = \frac{x}{Ay}$$Ay e^{y - x} = x$$y = \frac{x}{A} e^{x - y}$Replacing $\frac{1}{A}$ with a new constant $A$,we get:$y = Ax e^{x - y}$

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

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