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(D) Given differential equation: $x y \, dy - (1 + y^2) \, dx = 0$. Separating the variables, we get $\frac{y}{1 + y^2} \, dy = \frac{1}{x} \, dx$. In

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(D) Given differential equation: $x y \, dy - (1 + y^2) \, dx = 0$. Separating the variables, we get $\frac{y}{1 + y^2} \, dy = \frac{1}{x} \, dx$. Integrating both sides: $\int \frac{y}{1 + y^2} \, dy = \int \frac{1}{x} \, dx$. $\frac{1}{2} \ln(1 + y^2) = \ln|x| + C$. Since the curve passes through $(1, 0)$, we have $\frac{1}{2} \ln(1 + 0) = \ln(1) + C \Rightarrow C = 0$. Thus, $\ln(1 + y^2) = 2 \ln|x| \Rightarrow 1 + y^2 = x^2$. For the curve $x^2 - y^2 = 1$, differentiating w.r.t. $x$: $2x - 2y \frac{dy}{dx} = 0 \Rightarrow m_1 = \frac{dy}{dx} = \frac{x}{y}$. For the curve $x^2 + 3y^2 = 3$, differentiating w.r.t. $x$: $2x + 6y \frac{dy}{dx} = 0 \Rightarrow m_2 = \frac{dy}{dx} = -\frac{x}{3y}$. At the intersection point, $x^2 = 1 + y^2$. Substituting into $x^2 + 3y^2 = 3$: $(1 + y^2) + 3y^2 = 3 \Rightarrow 4y^2 = 2 \Rightarrow y^2 = \frac{1}{2}$. Then $x^2 = 1 + \frac{1}{2} = \frac{3}{2}$. At the intersection point, $m_1 m_2 = (\frac{x}{y})(-\frac{x}{3y}) = -\frac{x^2}{3y^2} = -\frac{3/2}{3(1/2)} = -1$. Since the product of slopes is $-1$, the curves intersect at an angle $	heta = \frac{\pi}{2}$. Therefore, $\frac{2	heta}{\pi} = \frac{2(\pi/2)}{\pi} = 1$.

The curve that satisfies the differential equation $x y \, dy - (1 + y^2) \, dx = 0$ passes through $(1, 0)$ and intersects the curve $x^2 + 3y^2 = 3$ at an angle $\theta$. Then $\frac{2\theta}{\pi} =$

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