The differential equation corresponding to th | vedclass

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(D) Given the family of ellipses: $\frac{x^2}{a^2} + \frac{y^2}{4} = 1$. To find the differential equation, we differentiate both sides with respect t

Vedclass · Accepted Answer

$x y \frac{dy}{dx} = y^2 - 4$

Vedclass · Answer

(D) Given the family of ellipses: $\frac{x^2}{a^2} + \frac{y^2}{4} = 1$. To find the differential equation, we differentiate both sides with respect to $x$: $\frac{d}{dx} \left( \frac{x^2}{a^2} + \frac{y^2}{4} \right) = \frac{d}{dx} (1)$ $\frac{2x}{a^2} + \frac{2y}{4} \frac{dy}{dx} = 0$ $\frac{2x}{a^2} + \frac{y}{2} \frac{dy}{dx} = 0$ From the original equation, we have $\frac{x^2}{a^2} = 1 - \frac{y^2}{4} = \frac{4 - y^2}{4}$, so $\frac{1}{a^2} = \frac{4 - y^2}{4x^2}$. Substitute $\frac{1}{a^2}$ into the differentiated equation: $2x \left( \frac{4 - y^2}{4x^2} \right) + \frac{y}{2} \frac{dy}{dx} = 0$ $\frac{4 - y^2}{2x} + \frac{y}{2} \frac{dy}{dx} = 0$ Multiply by $2x$: $(4 - y^2) + xy \frac{dy}{dx} = 0$ $xy \frac{dy}{dx} = y^2 - 4$.

The differential equation corresponding to the family of ellipses $\frac{x^2}{a^2} + \frac{y^2}{4} = 1$, where '$a$' is an arbitrary constant, is:

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