y=c^{2}+frac{c}{x} is the solution of the dif | vedclass

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(B) Given the equation $y=c^{2}+\frac{c}{x}$.Differentiating both sides with respect to $x$,we get:$\frac{d y}{d x} = -\frac{c}{x^{2}}$From this,we ca

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

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(B) Given the equation $y=c^{2}+\frac{c}{x}$.Differentiating both sides with respect to $x$,we get:$\frac{d y}{d x} = -\frac{c}{x^{2}}$From this,we can express $c$ in terms of $x$ and $\frac{d y}{d x}$:$c = -x^{2} \frac{d y}{d x}$Now,substitute this value of $c$ back into the original equation:$y = (-x^{2} \frac{d y}{d x})^{2} + \frac{-x^{2} \frac{d y}{d x}}{x}$$y = x^{4} (\frac{d y}{d x})^{2} - x \frac{d y}{d x}$Rearranging the terms to form the differential equation:$x^{4} (\frac{d y}{d x})^{2} - x \frac{d y}{d x} - y = 0$Thus,the correct option is $B$.

$y=c^{2}+\frac{c}{x}$ is the solution of the differential equation

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