If the curve $y = y(x)$ represented by the solution of the differential equation $(2xy^2 - y)dx + xdy = 0$ passes through the intersection of the lines $2x - 3y = 1$ and $3x + 2y = 8$,then $|y(1)|$ is equal to ...... .

Match the statements/expressions given in Column $I$ with the values given in Column $II$.

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 $y(\ln 2)$	$(s)$ $4$
	$(t)$ $5$

$A$ 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

If the differential equation $\begin{vmatrix} f(x) & f'(x) \\ f'(x) & f''(x) \end{vmatrix} = 0$ holds for all $x$, with initial conditions $f(0) = 1$ and $f'(0) = 2$, then which of the following is true?

Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx}=1+xe^{y-x}$,where $-\sqrt{2} < x < \sqrt{2}$ and $y(0)=0$. Then,the minimum value of $y(x)$ for $x \in(-\sqrt{2}, \sqrt{2})$ is equal to:

$A$ continuously differentiable function $\phi (x)$ in $(0, \pi)$ satisfying $y' = 1 + y^2$ and $y(0) = 0 = y(\pi)$ is