The solution of ${y^5}x + y - x\frac{{dy}}{{dx}} = 0$ is

$A$ function $y = f(x)$ satisfies the differential equation $\frac{dy}{dx} - y = \cos x - \sin x$ with the initial condition that $y$ is bounded when $x \rightarrow \infty$. The area enclosed by $y = f(x)$,$y = \cos x$,and the $y$-axis is

The solution of the differential equation $\cos^2 x \frac{d^2y}{dx^2} = 1$ is

The solution of $(xy \cos xy + \sin xy)dx + x^2 \cos xy \, dy = 0$ is

If $x = \int_{-y}^{y} \frac{dt}{\sqrt{1 + 9t^2}}$ and $\frac{d^2y}{dx^2} = ky$,then $k$ equals

Match the statements/expressions in Column $I$ with the open intervals in Column $II$.

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