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(B) Given differential equation is $(x^2 - y^2) \, dx + 2xy \, dy = 0$.This can be written as $\frac{dy}{dx} = \frac{y^2 - x^2}{2xy}$.Since this is a

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a circle of radius one

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(B) Given differential equation is $(x^2 - y^2) \, dx + 2xy \, dy = 0$.This can be written as $\frac{dy}{dx} = \frac{y^2 - x^2}{2xy}$.Since this is a homogeneous differential equation,we put $y = ux$,which implies $\frac{dy}{dx} = u + x \frac{du}{dx}$.Substituting these into the equation: $u + x \frac{du}{dx} = \frac{u^2 x^2 - x^2}{2x(ux)} = \frac{u^2 - 1}{2u}$.$x \frac{du}{dx} = \frac{u^2 - 1}{2u} - u = \frac{u^2 - 1 - 2u^2}{2u} = \frac{-(1 + u^2)}{2u}$.Separating variables: $\int \frac{2u}{1 + u^2} \, du = - \int \frac{1}{x} \, dx$.Integrating both sides: $\ln(1 + u^2) = -\ln|x| + \ln|C|$.$\ln(1 + u^2) = \ln\left(\frac{C}{x}\right) \Rightarrow 1 + u^2 = \frac{C}{x}$.Substituting $u = \frac{y}{x}$: $1 + \frac{y^2}{x^2} = \frac{C}{x} \Rightarrow \frac{x^2 + y^2}{x^2} = \frac{C}{x} \Rightarrow x^2 + y^2 = Cx$.Since the curve passes through $(1, 1)$,we have $1^2 + 1^2 = C(1) \Rightarrow C = 2$.Thus,the equation of the curve is $x^2 + y^2 = 2x$,which can be rewritten as $(x - 1)^2 + y^2 = 1$.This represents a circle with center $(1, 0)$ and radius $1$.

The curve satisfying the differential equation $(x^2 - y^2) \, dx + 2xy \, dy = 0$ and passing through the point $(1, 1)$ is

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