If y = frac{x}{log_e|cx|} is the solution of | vedclass

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(B) Given $y = \frac{x}{\ln|cx|}$.Taking the derivative with respect to $x$ using the quotient rule:$\frac{dy}{dx} = \frac{\ln|cx| \cdot 1 - x \cdot \

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$-\frac{y^2}{x^2}$

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(B) Given $y = \frac{x}{\ln|cx|}$.Taking the derivative with respect to $x$ using the quotient rule:$\frac{dy}{dx} = \frac{\ln|cx| \cdot 1 - x \cdot \frac{1}{cx} \cdot c}{(\ln|cx|)^2} = \frac{\ln|cx| - 1}{(\ln|cx|)^2} = \frac{1}{\ln|cx|} - \frac{1}{(\ln|cx|)^2}$.Since $y = \frac{x}{\ln|cx|}$, we have $\frac{y}{x} = \frac{1}{\ln|cx|}$.Substituting this into the derivative expression:$\frac{dy}{dx} = \frac{y}{x} - \left(\frac{y}{x}\right)^2$.Comparing this with the given differential equation $\frac{dy}{dx} = \frac{y}{x} + \phi\left(\frac{x}{y}\right)$, we identify $\phi\left(\frac{x}{y}\right) = -\left(\frac{y}{x}\right)^2 = -\frac{y^2}{x^2}$.

If $y = \frac{x}{\log_e|cx|}$ is the solution of the differential equation $\frac{dy}{dx} = \frac{y}{x} + \phi\left(\frac{x}{y}\right)$, then $\phi\left(\frac{x}{y}\right)$ is given by

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