The foci of the curve which satisfies the dif | vedclass

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(A) Given differential equation: $(1 + y^2) dx = xy dy$Rearranging the terms to separate variables: $\frac{dx}{x} = \frac{y dy}{1 + y^2}$Integrating b

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$(\pm \sqrt{2}, 0)$

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(A) Given differential equation: $(1 + y^2) dx = xy dy$Rearranging the terms to separate variables: $\frac{dx}{x} = \frac{y dy}{1 + y^2}$Integrating both sides: $\int \frac{1}{x} dx = \int \frac{y}{1 + y^2} dy$Multiply by $2$ to simplify the integral on the right: $2 \int \frac{1}{x} dx = \int \frac{2y}{1 + y^2} dy$$2 \ln|x| = \ln(1 + y^2) + C$$ln(x^2) = \ln(1 + y^2) + ln(c) \implies x^2 = c(1 + y^2)$Since the curve passes through $(1, 0)$,substitute $x = 1$ and $y = 0$: $1^2 = c(1 + 0^2) \implies c = 1$The equation of the curve is $x^2 = 1 + y^2$,which simplifies to $x^2 - y^2 = 1$.This is a rectangular hyperbola of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with $a^2 = 1$ and $b^2 = 1$.For this hyperbola,$a = 1$ and $b = 1$. The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + 1} = \sqrt{2}$.The foci are given by $(\pm ae, 0) = (\pm 1 \cdot \sqrt{2}, 0) = (\pm \sqrt{2}, 0)$.

The foci of the curve which satisfies the differential equation $(1 + y^2) dx - xy\, dy = 0$ and passes through the point $(1, 0)$ are:

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