The differential equation whose solution repr | vedclass

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Question

(B) Given the family of curves: $x^2 y = 4e^x + c$. To find the differential equation,we differentiate both sides with respect to $x$: $\frac{d}{dx}(x

Vedclass · Accepted Answer

$x^2 \frac{dy}{dx} + (2xy - 4e^x) = 0$

Vedclass · Answer

(B) Given the family of curves: $x^2 y = 4e^x + c$. To find the differential equation,we differentiate both sides with respect to $x$: $\frac{d}{dx}(x^2 y) = \frac{d}{dx}(4e^x + c)$. Using the product rule on the left side: $x^2 \frac{dy}{dx} + y \frac{d}{dx}(x^2) = 4e^x$. $x^2 \frac{dy}{dx} + 2xy = 4e^x$. Rearranging the terms,we get: $x^2 \frac{dy}{dx} + (2xy - 4e^x) = 0$. Thus,the correct option is $B$.

The differential equation whose solution represents the family $x^2 y = 4e^x + c$,where $c$ is an arbitrary constant,is

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