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(C) Given differential equation is $(1+y^2) dx - xy dy = 0$. Rearranging the terms,we get $(1+y^2) dx = xy dy$. Separating the variables,we have $\fra

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hyperbola

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(C) Given differential equation is $(1+y^2) dx - xy dy = 0$. Rearranging the terms,we get $(1+y^2) dx = xy dy$. Separating the variables,we have $\frac{1}{x} dx = \frac{y}{1+y^2} dy$. Integrating both sides: $\int \frac{1}{x} dx = \int \frac{y}{1+y^2} dy$. This gives $\log |x| = \frac{1}{2} \log (1+y^2) + C$. Given $x=1$ and $y=0$,we substitute these values: $\log(1) = \frac{1}{2} \log(1+0^2) + C$,which implies $0 = 0 + C$,so $C=0$. The equation becomes $\log x = \frac{1}{2} \log (1+y^2)$. Multiplying by $2$,we get $2 \log x = \log (1+y^2)$,which is $\log(x^2) = \log(1+y^2)$. Taking the exponential of both sides,$x^2 = 1+y^2$,or $x^2 - y^2 = 1$. This equation represents a rectangular hyperbola.

The particular solution of the differential equation $(1+y^2) dx - xy dy = 0$ at $x=1, y=0$,represents

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