The differential equation of the family of cu | vedclass

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Question

(A) Given the equation of the family of curves: ${x^2}y = a$. Differentiating both sides with respect to $x$ using the product rule: $\frac{d}{dx}({x^

Vedclass · Accepted Answer

$\frac{dy}{dx} + \frac{2y}{x} = 0$

Vedclass · Answer

(A) Given the equation of the family of curves: ${x^2}y = a$. Differentiating both sides with respect to $x$ using the product rule: $\frac{d}{dx}({x^2}y) = \frac{d}{dx}(a)$ ${x^2} \frac{dy}{dx} + y \frac{d}{dx}({x^2}) = 0$ ${x^2} \frac{dy}{dx} + y(2x) = 0$ ${x^2} \frac{dy}{dx} + 2xy = 0$ Dividing the entire equation by ${x^2}$ (assuming $x 
eq 0$): $\frac{dy}{dx} + \frac{2xy}{x^2} = 0$ $\frac{dy}{dx} + \frac{2y}{x} = 0$. Thus,the correct option is $A$.

The differential equation of the family of curves represented by the equation ${x^2}y = a$ is:

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