The differential equation whose solution is y | vedclass

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(A) Given the equation $y=c^2+\frac{c}{x}$.First,differentiate with respect to $x$:$\frac{dy}{dx} = 0 - \frac{c}{x^2} = -\frac{c}{x^2}$.From this,we c

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$x^4\left(\frac{dy}{dx}\right)^2-x\frac{dy}{dx}-y=0$

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(A) Given the equation $y=c^2+\frac{c}{x}$.First,differentiate with respect to $x$:$\frac{dy}{dx} = 0 - \frac{c}{x^2} = -\frac{c}{x^2}$.From this,we can express the constant $c$ in terms of $x$ and $\frac{dy}{dx}$:$c = -x^2\frac{dy}{dx}$.Now,substitute this value of $c$ back into the original equation:$y = (-x^2\frac{dy}{dx})^2 + \frac{-x^2\frac{dy}{dx}}{x}$.Simplifying the expression:$y = x^4\left(\frac{dy}{dx}\right)^2 - x\frac{dy}{dx}$.Rearranging the terms to form the differential equation:$x^4\left(\frac{dy}{dx}\right)^2 - x\frac{dy}{dx} - y = 0$.

The differential equation whose solution is $y=c^2+\frac{c}{x}$,where $c$ is a constant,is

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