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(D) Given differential equation: $(1+y^2) dx - xy dy = 0$ Rearranging the terms to separate the variables: $\frac{dx}{x} = \frac{y}{1+y^2} dy$ Integra

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a part of hyperbola

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(D) Given differential equation: $(1+y^2) dx - xy dy = 0$ Rearranging the terms to separate the variables: $\frac{dx}{x} = \frac{y}{1+y^2} dy$ Integrating both sides: $\int \frac{dx}{x} = \int \frac{y}{1+y^2} dy$ $\ln |x| = \frac{1}{2} \ln(1+y^2) + C$ Multiplying by $2$: $2 \ln |x| = \ln(1+y^2) + 2C$ $\ln(x^2) - \ln(1+y^2) = C'$ (where $C' = 2C$) $\ln \left( \frac{x^2}{1+y^2} \right) = C'$ $\frac{x^2}{1+y^2} = e^{C'} = k$ Given $y(1) = 0$,substitute $x=1$ and $y=0$: $\frac{1^2}{1+0^2} = k \Rightarrow k = 1$ So,$\frac{x^2}{1+y^2} = 1 \Rightarrow x^2 = 1+y^2 \Rightarrow x^2 - y^2 = 1$ This equation represents a hyperbola.

The particular solution of the differential equation $(1+y^2) dx - xy dy = 0$ with the condition $y(1) = 0$ represents:

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