If a and b are arbitrary constants,then the d | vedclass

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(C) Given the equation: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Differentiating with respect to $x$: $\frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} = 0 \

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$x y \frac{d^2 y}{d x^2}+x\left(\frac{d y}{d x}\right)^2-y \frac{d y}{d x}=0$

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(C) Given the equation: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Differentiating with respect to $x$: $\frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} = 0 \implies \frac{y}{b^2} \frac{dy}{dx} = -\frac{x}{a^2} \implies \frac{y}{x} \frac{dy}{dx} = -\frac{b^2}{a^2}$. Differentiating again with respect to $x$: $\frac{d}{dx} \left( \frac{y}{x} \frac{dy}{dx} \right) = \frac{d}{dx} \left( -\frac{b^2}{a^2} \right) = 0$. Using the product rule: $\frac{y}{x} \frac{d^2y}{dx^2} + \frac{dy}{dx} \left( \frac{x \frac{dy}{dx} - y}{x^2} \right) = 0$. Multiplying by $x^2$: $xy \frac{d^2y}{dx^2} + x \left( \frac{dy}{dx} \right)^2 - y \frac{dy}{dx} = 0$.

If $a$ and $b$ are arbitrary constants,then the differential equation having $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ as its general solution is

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