The general solution of $\frac{dy}{dx} = \frac{x+y+1}{y-x+1}$ is

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 $y\,dx - x\,dy + 3x^2y^2e^{x^3}dx = 0$ is

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

If $2xy^3dx + x^2y^2dy = ydx - xdy$ and $y(2) = 1$,then the value of $y(-1)$ will be (where $y(x)$ denotes the value of $y$ for a given $x$):

The curve amongst the family of curves represented by the differential equation,$(x^2 - y^2)dx + 2xy\, dy = 0$ which passes through $(1, 1)$,is