The curve passing through (3, 0) and satisfyi | vedclass

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Question

(B) Given differential equation: $(9 - x^2)(\frac{dy}{dx})^2 = 9 - y^2$$\Rightarrow (\frac{dy}{dx})^2 = \frac{9 - y^2}{9 - x^2}$$\Rightarrow \frac{dy}

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Circle

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(B) Given differential equation: $(9 - x^2)(\frac{dy}{dx})^2 = 9 - y^2$$\Rightarrow (\frac{dy}{dx})^2 = \frac{9 - y^2}{9 - x^2}$$\Rightarrow \frac{dy}{dx} = \pm \frac{\sqrt{9 - y^2}}{\sqrt{9 - x^2}}$Separating the variables:$\frac{dy}{\sqrt{9 - y^2}} = \pm \frac{dx}{\sqrt{9 - x^2}}$Integrating both sides:$\int \frac{dy}{\sqrt{3^2 - y^2}} = \pm \int \frac{dx}{\sqrt{3^2 - x^2}}$$\sin^{-1}(\frac{y}{3}) = \pm \sin^{-1}(\frac{x}{3}) + C$Since the curve passes through $(3, 0)$:$\sin^{-1}(0) = \pm \sin^{-1}(\frac{3}{3}) + C$$0 = \pm \frac{\pi}{2} + C \Rightarrow C = \mp \frac{\pi}{2}$Substituting $C$ back:$\sin^{-1}(\frac{y}{3}) = \pm \sin^{-1}(\frac{x}{3}) \mp \frac{\pi}{2}$Taking $\sin$ on both sides:$\frac{y}{3} = \sin(\pm \sin^{-1}(\frac{x}{3}) \mp \frac{\pi}{2}) = \mp \cos(\sin^{-1}(\frac{x}{3})) = \mp \sqrt{1 - \frac{x^2}{9}}$Squaring both sides:$\frac{y^2}{9} = 1 - \frac{x^2}{9} \Rightarrow x^2 + y^2 = 9$This is the equation of a circle with radius $3$ centered at the origin.

The curve passing through $(3, 0)$ and satisfying the differential equation $(9 - x^2)(\frac{dy}{dx})^2 = 9 - y^2$ represents:

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