If y = frac{x}{ln |c x|} (where c is an arbit | vedclass

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(D) Given $y = \frac{x}{\ln |c x|}$. Taking the reciprocal,we get $\frac{1}{y} = \frac{\ln |c x|}{x}$. This implies $\frac{x}{y} = \ln |c| + \ln |x|$.

Vedclass · Accepted Answer

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

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(D) Given $y = \frac{x}{\ln |c x|}$. Taking the reciprocal,we get $\frac{1}{y} = \frac{\ln |c x|}{x}$. This implies $\frac{x}{y} = \ln |c| + \ln |x|$. Differentiating both sides with respect to $x$: $\frac{d}{dx} \left( \frac{x}{y} \right) = \frac{d}{dx} (\ln |c| + \ln |x|)$. Using the quotient rule on the left side: $\frac{y(1) - x(y')}{y^2} = \frac{1}{x}$. $\frac{y - x \frac{dy}{dx}}{y^2} = \frac{1}{x}$. $y - x \frac{dy}{dx} = \frac{y^2}{x}$. $x \frac{dy}{dx} = y - \frac{y^2}{x}$. $\frac{dy}{dx} = \frac{y}{x} - \frac{y^2}{x^2}$. Comparing this with the given equation $\frac{dy}{dx} = \frac{y}{x} + \phi \left( \frac{x}{y} \right)$,we find that $\phi \left( \frac{x}{y} \right) = -\frac{y^2}{x^2}$.

Column $I$	Column $II$
$(A)$ The number of solutions of the equation $x e^{\sin x}-\cos x=0$ in the interval $(0, \frac{\pi}{2})$	$(p)$ $1$
$(B)$ Value$(s)$ of $k$ for which the planes $k x+4 y+z=0, 4 x+k y+2 z=0$ and $2 x+2 y+z=0$ intersect in a straight line	$(q)$ $2$
$(C)$ Value$(s)$ of $k$ for which $\|x-1\|+\|x-2\|+\|x+1\|+\|x+2\|=4 k$ has integer solution$(s)$	$(r)$ $3$
$(D)$ If $y^{\prime}=y+1$ and $y(0)=1$,then value$(s)$ of $y(\ln 2)$	$(s)$ $4$
	$(t)$ $5$

If $y = \frac{x}{\ln |c x|}$ (where $c$ is an arbitrary constant) is the general solution of the differential equation $\frac{dy}{dx} = \frac{y}{x} + \phi \left( \frac{x}{y} \right)$,then the function $\phi \left( \frac{x}{y} \right)$ is:

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