If $xdy = y(dx + ydy), y > 0$ and $y(1) = 1,$ then $y(-3)$ is equal to

If $y$ is a function of $x$,then $\frac{d^2y}{dx^2} + y \frac{dy}{dx} = 0$. If $x$ is a function of $y$,then the equation becomes:

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

Let $y$ be the solution of the differential equation $x \frac{dy}{dx} = \frac{y^2}{1 - y \log x}$ satisfying $y(1) = 1$. Then, $y$ satisfies:

If $f(x), f^{\prime}(x), f^{\prime \prime}(x)$ are positive functions and $f(0)=1, f^{\prime}(0)=2$,then the solution of the differential equation $\left|\begin{array}{ll}f(x) & f^{\prime}(x) \\ f^{\prime}(x) & f^{\prime \prime}(x)\end{array}\right|=0$ is

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