Consider the differential equation $\sec x \frac{dy}{dx} - 2y = 2 + 3\sin x$
If $y(0) = -\frac{7}{4}$, then find $y\left(\frac{\pi}{6}\right)$.
(A) $-5/2$
(B) $0$
(C) $1$
(D) $3/2$

The solution of the differential equation $\mathbf { x d y } - \mathbf { y d x } = \sqrt { \mathbf { x } ^ { 2 } + \mathbf { y } ^ { 2 } } \mathbf { d x }$ is (where c is integration constant)
(A) $\sqrt { \mathrm { x } ^ { 2 } + \mathrm { y } ^ { 2 } } = \mathrm { cx } ^ { 2 } - \mathrm { y }$
(B) $\sqrt { \mathrm { x } ^ { 2 } + \mathrm { y } ^ { 2 } } = \mathrm { cx } ^ { 2 } + \mathrm { y }$
(C) $\sqrt { \mathrm { x } ^ { 2 } + \mathrm { y } ^ { 2 } } = \mathrm { cx } - \mathrm { y }$
(D) $\sqrt { \mathrm { x } ^ { 2 } + \mathrm { y } ^ { 2 } } = \mathrm { cx } + \mathrm { y }$

For each of Q , S , V in the following sentences, choose the appropriate expression from among (0) $\sim$ (7) at the bottom of this page. For the other $\square$, enter the correct number.
Suppose we have a differentiable function $f ( x )$ which satisfies the equation
$$\int _ { 0 } ^ { x } f ( t ) d t = \left( 1 + e ^ { - x } \right) f ( x ) + 2 x - 4 \log 2 \tag{1}$$
We are to find $f ( x )$ and the value of $\lim _ { x \rightarrow \infty } f ( x )$.
When we differentiate each side of (1) with respect to $x$ and transform the equation, we have
$$\left( 1 + e ^ { - x } \right) ( \mathbf { Q } ) = \mathbf { R } . \tag{2}$$
Next we set $f ( x ) = e ^ { x } g ( x )$, and using (2), we obtain
$$g ^ { \prime } ( x ) = \frac { \mathbf { S } } { 1 + e ^ { - x } }$$
and hence
$$g ( x ) = \mathbf { T } \log \left( 1 + e ^ { - x } \right) + C ,$$
where $C$ is an integral constant. Furthermore, since $g ( 0 ) = f ( 0 )$, we see that $C = \mathbf { U }$. Thus we obtain $g ( x )$ and from that,
$$f ( x ) = \mathbf { V } \log \left( 1 + e ^ { - x } \right) .$$
Finally, we set $e ^ { - x } = t$ and obtain
$$f ( x ) = \mathbf { W } \log ( 1 + t ) ^ { \frac { 1 } { t } }$$
and hence
$$\lim _ { x \rightarrow \infty } f ( x ) = \lim _ { t \rightarrow \mathbf { X } } \mathbf { W } \log ( 1 + t ) ^ { \frac { 1 } { t } } = \mathbf { Y }$$
Choices: (0) $f ^ { \prime } ( x ) - f ( x )$
(1) $f ( x ) - f ^ { \prime } ( x )$
(2) $f ^ { \prime } ( x ) - 2 f ( x )$
(3) $f ( x ) - 2 f ^ { \prime } ( x )$
(4) $2 e ^ { x }$
(5) $- 2 e ^ { x }$ (6) $2 e ^ { - x }$ (7) $- 2 e ^ { - x }$

The solution to the differential equation
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = | - 6 x | \quad \text { for all } x$$
is $y = f ( x ) + c$, where $c$ is a constant.
Which one of the following is a correct expression for $f ( x )$ ?

Problem 1

I. Find the value of the following definite integral:
$$I = \int _ { 2 } ^ { 4 } \frac { d x } { \sqrt { ( x - 2 ) ( 4 - x ) } }$$
II. Find the general solution and the singular solution of the following differential equation:
$$y = x \frac { d y } { d x } + \frac { d y } { d x } + \left( \frac { d y } { d x } \right) ^ { 2 }$$
III. Find the general solution of the following differential equation:
$$x ^ { 2 } \frac { d ^ { 2 } y } { d x ^ { 2 } } - x \frac { d y } { d x } - 8 y = x ^ { 2 }$$

Problem 1
I. Obtain the general solution of the following differential equation: $$x ^ { 2 } \frac { d ^ { 2 } y } { d x ^ { 2 } } - x \frac { d y } { d x } + y = x ^ { 3 }$$
II. Obtain the general solution of the following differential equation: $$x ^ { 2 } \frac { d y } { d x } - x ^ { 2 } y ^ { 2 } + x y + 1 = 0$$ Note that $y = \frac { 1 } { x }$ is a particular solution.
III. Let $I _ { n }$ be defined by: $$I _ { n } = \int _ { 0 } ^ { \frac { \pi } { 4 } } \tan ^ { n } x \, d x$$ where $n$ is a non-negative integer.

1. Calculate $I _ { 0 } , I _ { 1 }$, and $I _ { 2 }$.
2. Calculate $I _ { n }$ for $n \geq 2$.

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 $I _ { n } ( x )$.

Let a be a real number, and $$f ( x ) = \ln ( 2 x + 8 )$$ The vertical asymptote of the function $$g ( x ) = \frac { \sin x } { x ^ { 2 } + a x }$$ is also a vertical asymptote of the function.\ Accordingly, what is a?\ A) 0\ B) 1\ C) 2\ D) 3\ E) 4