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(B) Given $\frac{dy}{dx} = \frac{x + y}{x - y}$.Let $y = vx$, then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.Substituting into the equation: $v + x\frac{dv

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$c(x^2 + y^2)^{\frac{1}{2}} = e^{	an^{-1}(\frac{y}{x})}$, where $c$ is an arbitrary constant

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(B) Given $\frac{dy}{dx} = \frac{x + y}{x - y}$.Let $y = vx$, then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.Substituting into the equation: $v + x\frac{dv}{dx} = \frac{x + vx}{x - vx} = \frac{1 + v}{1 - v}$.$x\frac{dv}{dx} = \frac{1 + v}{1 - v} - v = \frac{1 + v - v + v^2}{1 - v} = \frac{1 + v^2}{1 - v}$.Separating variables: $\int \frac{1 - v}{1 + v^2} dv = \int \frac{1}{x} dx$.$\int \frac{1}{1 + v^2} dv - \int \frac{v}{1 + v^2} dv = \ln|x| + C$.$	an^{-1}(v) - \frac{1}{2} \ln(1 + v^2) = \ln|x| + C$.$	an^{-1}(\frac{y}{x}) = \ln|x| + \frac{1}{2} \ln(1 + \frac{y^2}{x^2}) + C = \ln|x| + \frac{1}{2} \ln(\frac{x^2 + y^2}{x^2}) + C$.$	an^{-1}(\frac{y}{x}) = \ln|x| + \frac{1}{2} \ln(x^2 + y^2) - \ln|x| + C = \frac{1}{2} \ln(x^2 + y^2) + C$.$2	an^{-1}(\frac{y}{x}) = \ln(x^2 + y^2) + 2C$.Taking exponential on both sides: $e^{2	an^{-1}(\frac{y}{x})} = e^{\ln(x^2 + y^2)} \cdot e^{2C} = k(x^2 + y^2)$.This matches the form $c(x^2 + y^2)^{\frac{1}{2}} = e^{	an^{-1}(\frac{y}{x})}$.

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

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