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 $\phi\left(\frac{x}{y}\right) = -\left(\frac{y}{x}\right)^2 = -\frac{y^2}{x^2}$.

Column $I$	Column $II$
$(A)$ Interval contained in the domain of definition of non-zero solutions of the differential equation $(x-3)^2 y^{\prime}+y=0$	$(p)$ $(-\frac{\pi}{2}, \frac{\pi}{2})$
$(B)$ Interval containing the value of the integral $\int_1^5(x-1)(x-2)(x-3)(x-4)(x-5) dx$	$(q)$ $(0, \frac{\pi}{2})$
$(C)$ Interval in which at least one of the points of local maximum of $\cos^2 x+\sin x$ lies	$(r)$ $(\frac{\pi}{8}, \frac{5\pi}{4})$
$(D)$ Interval in which $\tan^{-1}(\sin x+\cos x)$ is increasing	$(s)$ $(0, \frac{\pi}{8})$
	$(t)$ $(-\pi, \pi)$

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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