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(D) Given the homogeneous differential equation $\frac{dy}{dx} = \frac{x - y}{x + y}$.Let $y = vx$, then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substit

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Pair of straight lines

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(D) Given the homogeneous differential equation $\frac{dy}{dx} = \frac{x - y}{x + y}$.Let $y = vx$, then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting into the equation: $v + x \frac{dv}{dx} = \frac{x - vx}{x + vx} = \frac{1 - v}{1 + v}$.$x \frac{dv}{dx} = \frac{1 - v}{1 + v} - v = \frac{1 - v - v - v^2}{1 + v} = \frac{1 - 2v - v^2}{1 + v}$.Separating variables: $\int \frac{1 + v}{1 - 2v - v^2} dv = \int \frac{1}{x} dx$.Let $u = 1 - 2v - v^2$, then $du = (-2 - 2v) dv = -2(1 + v) dv$, so $(1 + v) dv = -\frac{1}{2} du$.$-\frac{1}{2} \int \frac{1}{u} du = \ln|x| + C \implies -\frac{1}{2} \ln|1 - 2v - v^2| = \ln|x| + C$.$\ln|1 - 2v - v^2| = -2 \ln|x| + C' \implies 1 - 2(\frac{y}{x}) - (\frac{y}{x})^2 = \frac{K}{x^2}$.$x^2 - 2xy - y^2 = K$. Since it passes through $(0, 0)$, $K = 0$.$x^2 - 2xy - y^2 = 0$. This is a homogeneous second-degree equation of the form $Ax^2 + 2Hxy + By^2 = 0$ where $H^2 - AB = (-1)^2 - (1)(-1) = 2 > 0$, representing a pair of straight lines.

The solution of the differential equation $\frac{dy}{dx} = \frac{x - y}{x + y}$, with the initial condition $x = 0$ and $y = 0$, represents which of the following curves?

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