The differential equation frac{dx}{dy} = frac | vedclass

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(D) Given the differential equation $\frac{dx}{dy} = \frac{3y}{2x}$.Separating variables,we get $2x \, dx = 3y \, dy$.Integrating both sides,we get $x

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both $(A)$ and $(B)$

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(D) Given the differential equation $\frac{dx}{dy} = \frac{3y}{2x}$.Separating variables,we get $2x \, dx = 3y \, dy$.Integrating both sides,we get $x^2 = \frac{3}{2}y^2 + C$,which can be rewritten as $2x^2 - 3y^2 = 2C$.Case $1$: If $C > 0$,let $2C = k^2$. Then $2x^2 - 3y^2 = k^2$,or $\frac{x^2}{k^2/2} - \frac{y^2}{k^2/3} = 1$. This is a hyperbola with $a^2 = \frac{k^2}{2}$ and $b^2 = \frac{k^2}{3}$. The eccentricity $e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{k^2/3}{k^2/2}} = \sqrt{1 + \frac{2}{3}} = \sqrt{\frac{5}{3}}$.Case $2$: If $C Thus,the eccentricity can be either $\sqrt{\frac{5}{3}}$ or $\sqrt{\frac{5}{2}}$.

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	$(s)$ $(0, \frac{\pi}{8})$
	$(t)$ $(-\pi, \pi)$

The differential equation $\frac{dx}{dy} = \frac{3y}{2x}$ represents a family of hyperbolas (except when it represents a pair of lines) with eccentricity:

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