If the solution $y ( x )$ of the given differential equation $\left( \mathrm { e } ^ { y } + 1 \right) \cos x \mathrm {~d} x + \mathrm { e } ^ { y } \sin x \mathrm {~d} y = 0$ passes through the point $\left( \frac { \pi } { 2 } , 0 \right)$, then the value of $\mathrm { e } ^ { y \left( \frac { \pi } { 6 } \right) }$ is equal to $\_\_\_\_$

For a differentiable function $f : \mathbb { R } \rightarrow \mathbb { R }$, suppose $f ^ { \prime } ( x ) = 3 f ( x ) + \alpha$, where $\alpha \in \mathbb { R } , f ( 0 ) = 1$ and $\lim _ { x \rightarrow - \infty } f ( x ) = 7$. Then $9 f \left( - \log _ { \mathrm { e } } 3 \right)$ is equal to $\_\_\_\_$

If the solution curve $y = y ( x )$ of the differential equation $\left( 1 + y ^ { 2 } \right) \left( 1 + \log _ { e } x \right) d x + x d y = 0 , x > 0$ passes through the point $( 1,1 )$ and $y ( e ) = \frac { \alpha - \tan \left( \frac { 3 } { 2 } \right) } { \beta + \tan \left( \frac { 3 } { 2 } \right) }$, then $\alpha + 2 \beta$ is

Let $x = x ( y )$ be the solution of the differential equation $y ^ { 2 } \mathrm {~d} x + \left( x - \frac { 1 } { y } \right) \mathrm { d } y = 0$. If $x ( 1 ) = 1$, then $x \left( \frac { 1 } { 2 } \right)$ is :
(1) $\frac { 1 } { 2 } + \mathrm { e }$
(2) $3 + e$
(3) $3 - e$
(4) $\frac { 3 } { 2 } + e$

If for the solution curve $y = f ( x )$ of the differential equation $\frac { \mathrm { d } y } { \mathrm {~d} x } + ( \tan x ) y = \frac { 2 + \sec x } { ( 1 + 2 \sec x ) ^ { 2 } }$, $x \in \left( \frac { - \pi } { 2 } , \frac { \pi } { 2 } \right) , f \left( \frac { \pi } { 3 } \right) = \frac { \sqrt { 3 } } { 10 }$, then $f \left( \frac { \pi } { 4 } \right)$ is equal to :
(1) $\frac { \sqrt { 3 } + 1 } { 10 ( 4 + \sqrt { 3 } ) }$
(2) $\frac { 5 - \sqrt { 3 } } { 2 \sqrt { 2 } }$
(3) $\frac { 9 \sqrt { 3 } + 3 } { 10 ( 4 + \sqrt { 3 } ) }$
(4) $\frac { 4 - \sqrt { 2 } } { 14 }$

If $x = f ( y )$ is the solution of the differential equation $\left( 1 + y ^ { 2 } \right) + \left( x - 2 \mathrm { e } ^ { \tan ^ { - 1 } y } \right) \frac { \mathrm { d } y } { \mathrm {~d} x } = 0 , y \in \left( - \frac { \pi } { 2 } , \frac { \pi } { 2 } \right)$ with $f ( 0 ) = 1$, then $f \left( \frac { 1 } { \sqrt { 3 } } \right)$ is equal to :
(1) $e ^ { \pi / 12 }$
(2) $e ^ { \pi / 4 }$
(3) $e ^ { \pi / 3 }$
(4) $e ^ { \pi / 6 }$

Let $y = y(x)$ be the solution of the differential equation $\cos x\left(\log_e(\cos x)\right)^2 \mathrm{d}y + \left(\sin x - 3y\sin x\log_e(\cos x)\right)\mathrm{d}x = 0$, $x \in \left(0, \frac{\pi}{2}\right)$. If $y\left(\frac{\pi}{4}\right) = \frac{-1}{\log_e 2}$, then $y\left(\frac{\pi}{6}\right)$ is equal to:
(1) $\frac{1}{\log_e(3) - \log_e(4)}$
(2) $\frac{2}{\log_e(3) - \log_e(4)}$
(3) $\frac{1}{\log_e(4) - \log_e(3)}$
(4) $-\frac{1}{\log_e(4)}$

Let $y = y(x)$ be the solution of the differential equation $2\cos x \frac{\mathrm{d}y}{\mathrm{d}x} = \sin 2x - 4y \sin x$, $x \in \left(0, \frac{\pi}{2}\right)$. If $y\left(\frac{\pi}{3}\right) = 0$, then $y'\left(\frac{\pi}{4}\right) + y\left(\frac{\pi}{4}\right)$ is equal to $\_\_\_\_$.

If $y = y ( x )$ is the solution of the differential equation, $\sqrt { 4 - x ^ { 2 } } \frac { \mathrm {~d} y } { \mathrm {~d} x } = \left( \left( \sin ^ { - 1 } \left( \frac { x } { 2 } \right) \right) ^ { 2 } - y \right) \sin ^ { - 1 } \left( \frac { x } { 2 } \right) , - 2 \leq x \leq 2 , y ( 2 ) = \frac { \pi ^ { 2 } - 8 } { 4 }$, then $y ^ { 2 } ( 0 )$ is equal to

Problem 1

I. Find the general solution $y ( x )$ of the following differential equation:
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = y ( 1 - y ) ,$$
where $0 < y < 1$.
II. Find the value of the following definite integral, $I$ :
$$I = \int _ { - 1 } ^ { 1 } \frac { \arccos \left( \frac { x } { 2 } \right) } { \cos ^ { 2 } \left( \frac { \pi } { 3 } x \right) } \mathrm { d } x$$
where $0 \leq \arccos \left( \frac { x } { 2 } \right) \leq \pi$.
III. For any positive variable $x$, we define $f ( x )$ and $g ( x )$ respectively as
$$f ( x ) = \sum _ { m = 0 } ^ { \infty } \frac { 1 } { ( 2 m ) ! } x ^ { 2 m }$$
and
$$g ( x ) = \frac { \mathrm { d } } { \mathrm {~d} x } f ( x )$$
For any non-negative integer $n , I _ { n } ( x )$ is defined as
$$I _ { n } ( x ) = \int _ { 0 } ^ { x } \left\{ \frac { g ( X ) } { f ( X ) } \right\} ^ { n } \mathrm {~d} X$$
Here, you may use
$$\exp ( x ) = \sum _ { m = 0 } ^ { \infty } \frac { 1 } { m ! } x ^ { m }$$

1. Calculate $f ( x ) ^ { 2 } - g ( x ) ^ { 2 }$.
2. Express $I _ { n + 2 } ( x )$ using $I _ { n } ( x )$.