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(B) Given the differential equation: $(1 + y^2)dx - xydy = 0$. Rearranging the terms to separate the variables,we get: $\frac{dx}{x} = \frac{y dy}{1 +

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$x^2 - y^2 = 1$

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(B) Given the differential equation: $(1 + y^2)dx - xydy = 0$. Rearranging the terms to separate the variables,we get: $\frac{dx}{x} = \frac{y dy}{1 + y^2}$. Integrating both sides: $\int \frac{dx}{x} = \int \frac{y dy}{1 + y^2}$. Let $u = 1 + y^2$,then $du = 2y dy$,so $y dy = \frac{du}{2}$. Thus,$\ln|x| = \frac{1}{2} \ln(1 + y^2) + C$. This can be written as $\ln|x| = \ln(\sqrt{1 + y^2}) + \ln c$,where $C = \ln c$. So,$|x| = c\sqrt{1 + y^2}$. Since the curve passes through $(1, 0)$,we substitute $x = 1$ and $y = 0$: $1 = c\sqrt{1 + 0^2} \implies c = 1$. Therefore,the equation is $|x| = \sqrt{1 + y^2}$,which simplifies to $x^2 = 1 + y^2$ or $x^2 - y^2 = 1$.

The equation of the curve passing through the point $(1, 0)$ which satisfies the differential equation $(1 + y^2)dx - xydy = 0$ is

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