The number of fish in a lake is modeled by the function $F$ that satisfies the logistic differential equation $\frac { d F } { d t } = 0.04 F \left( 1 - \frac { F } { 5000 } \right)$, where $t$ is the time in months and $F ( 0 ) = 2000$. What is $\lim _ { t \rightarrow \infty } F ( t )$?
(A) 10,000
(B) 5000
(C) 2500
(D) 2000

We consider the function $\theta : \mathbb{R} \rightarrow \mathbb{C}$ defined by $\theta(x) = \exp(-\pi x^{2})$, for $x \in \mathbb{R}$.
I.E.1) Justify that $\theta$ belongs to $\mathcal{S}$ and that $\mathcal{F}(\theta)$ is a solution of the differential equation
$$\forall \xi \in \mathbb{R}, \quad y'(\xi) = -2\pi\xi\, y(\xi)$$
I.E.2) Establish that $\mathcal{F}(\theta) = \theta$.
We will admit that $\int_{-\infty}^{+\infty} \theta(x) \mathrm{d}x = 1$.

Prove that $\forall x \in \mathbb{R}, \quad \int_{-\infty}^{+\infty} \frac{e^{2\pi\mathrm{i} x\xi}}{1+(2\pi\xi)^{2}} \mathrm{d}\xi = \frac{1}{2} e^{-|x|}$.

Let $f : \mathbb{R}_{+} \rightarrow \mathbb{C}$ be a continuous function and zero outside a segment. We define the function $\mathcal{L}(f)$ (Laplace transform of $f$) on $\mathbb{R}$ by
$$\forall x \in \mathbb{R}, \quad \mathcal{L}(f)(x) = \int_{0}^{+\infty} f(t) e^{-xt} \mathrm{d}t$$
We will admit that $\mathcal{L}(f)$ is of class $C^{\infty}$ on $\mathbb{R}$ and that
$$\forall x \in \mathbb{R}, \quad \forall n \in \mathbb{N}, \quad (\mathcal{L}(f))^{(n)}(x) = (-1)^{n} \int_{0}^{+\infty} f(t) t^{n} e^{-xt} \mathrm{d}t$$
In the remainder of this part, we shall assume that
$$\lim_{n \rightarrow +\infty} \sum_{0 \leqslant k \leqslant \lfloor nx \rfloor} \frac{(n\lambda)^{k}}{k!} e^{-n\lambda} = \frac{1}{2} \quad \text{if } x = \lambda$$
VI.C.1) Let $x \in \mathbb{R}_{+}$. Prove that
$$\lim_{n \rightarrow +\infty} \sum_{0 \leqslant k \leqslant \lfloor nx \rfloor} (-1)^{k} \frac{n^{k}}{k!} (\mathcal{L}(f))^{(k)}(n) = \int_{0}^{x} f(y) \mathrm{d}y$$
VI.C.2) Deduce that the map $\mathcal{L} : f \mapsto \mathcal{L}(f)$ is injective on the set of complex-valued functions, continuous on $\mathbb{R}_{+}$ and zero outside a bounded interval.

In subsection II.D, we assume that there exists a strictly positive real number $c$ such that the discrete real random variable $X$ satisfies $\mathbb{E}(X)=0$ and $\forall \omega \in \Omega,|X(\omega)| \leqslant c$.
We consider $Y$ the real random variable defined by $Y=\frac{1}{2}-\frac{X}{2c}$.
a) Verify that $X=-cY+(1-Y)c$.
b) Show that $\mathrm{e}^{X} \leqslant Y \mathrm{e}^{-c}+(1-Y) \mathrm{e}^{c}$.

The function $y = f(x)$ is the solution of the differential equation
$$\frac{dy}{dx} + \frac{xy}{x^2 - 1} = \frac{x^4 + 2x}{\sqrt{1 - x^2}}$$
in $(-1,1)$ satisfying $f(0) = 0$. Then
$$\int_{-\frac{\sqrt{3}}{2}}^{\frac{\sqrt{3}}{2}} f(x)\, dx$$
is
(A) $\frac{\pi}{3} - \frac{\sqrt{3}}{2}$
(B) $\frac{\pi}{3} - \frac{\sqrt{3}}{4}$
(C) $\frac{\pi}{6} - \frac{\sqrt{3}}{4}$
(D) $\frac{\pi}{6} - \frac{\sqrt{3}}{2}$

Let $y ( x )$ be a solution of the differential equation $\left( 1 + e ^ { x } \right) y ^ { \prime } + y e ^ { x } = 1$. If $y ( 0 ) = 2$, then which of the following statements is (are) true?
(A) $\quad y ( - 4 ) = 0$
(B) $\quad y ( - 2 ) = 0$
(C) $\quad y ( x )$ has a critical point in the interval $( - 1,0 )$
(D) $y ( x )$ has no critical point in the interval $( - 1,0 )$

Let $f:(0,\infty) \rightarrow \mathbb{R}$ be a differentiable function such that $f'(x) = 2 - \frac{f(x)}{x}$ for all $x \in (0,\infty)$ and $f(1) \neq 1$. Then
(A) $\lim_{x \rightarrow 0+} f'\left(\frac{1}{x}\right) = 1$
(B) $\lim_{x \rightarrow 0+} xf\left(\frac{1}{x}\right) = 2$
(C) $\lim_{x \rightarrow 0+} x^2 f'(x) = 0$
(D) $|f(x)| \leq 2$ for all $x \in (0,2)$

For any real numbers $\alpha$ and $\beta$, let $y _ { \alpha , \beta } ( x ) , x \in \mathbb { R }$, be the solution of the differential equation $$\frac { d y } { d x } + \alpha y = x e ^ { \beta x } , \quad y ( 1 ) = 1$$ Let $S = \left\{ y _ { \alpha , \beta } ( x ) : \alpha , \beta \in \mathbb { R } \right\}$. Then which of the following functions belong(s) to the set $S$ ?
(A) $f ( x ) = \frac { x ^ { 2 } } { 2 } e ^ { - x } + \left( e - \frac { 1 } { 2 } \right) e ^ { - x }$
(B) $f ( x ) = - \frac { x ^ { 2 } } { 2 } e ^ { - x } + \left( e + \frac { 1 } { 2 } \right) e ^ { - x }$
(C) $f ( x ) = \frac { e ^ { x } } { 2 } \left( x - \frac { 1 } { 2 } \right) + \left( e - \frac { e ^ { 2 } } { 4 } \right) e ^ { - x }$
(D) $f ( x ) = \frac { e ^ { x } } { 2 } \left( \frac { 1 } { 2 } - x \right) + \left( e + \frac { e ^ { 2 } } { 4 } \right) e ^ { - x }$

For $x \in \mathbb { R }$, let $y ( x )$ be a solution of the differential equation $$\left( x ^ { 2 } - 5 \right) \frac { d y } { d x } - 2 x y = - 2 x \left( x ^ { 2 } - 5 \right) ^ { 2 }$$ such that $y ( 2 ) = 7$. Then the maximum value of the function $y ( x )$ is

Let $y ( x )$ be the solution of the differential equation
$$x ^ { 2 } \frac { d y } { d x } + x y = x ^ { 2 } + y ^ { 2 } , \quad x > \frac { 1 } { e }$$
satisfying $y ( 1 ) = 0$. Then the value of $2 \frac { ( y ( e ) ) ^ { 2 } } { y \left( e ^ { 2 } \right) }$ is $\_\_\_\_$.

If $\frac { d y } { d x } + y \tan x = \sin 2 x$ and $y ( 0 ) = 1$, then $y ( \pi )$ is equal to
(1) - 1
(2) 5
(3) 1
(4) - 5

Let $y(x)$ be the solution of the differential equation $(x \log x) \frac{dy}{dx} + y = 2x \log x$, $(x \geq 1)$. Then $y(e)$ is equal to:
(1) $e$
(2) $0$
(3) $2$
(4) $2e$

Let $y ( x )$ be the solution of the differential equation $( x \log x ) \frac { d y } { d x } + y = 2 x \log x , ( x \geq 1 )$. Then $y ( e )$ is equal to
(1) $2 e$
(2) $e$
(3) 0
(4) 2

If a curve $y = f(x)$ passes through the point $(1,-1)$ and satisfies the differential equation, $y(1+xy)dx = x\,dy$, then $f\left(-\frac{1}{2}\right)$ is equal to: (1) $-\frac{2}{5}$ (2) $-\frac{4}{5}$ (3) $\frac{2}{5}$ (4) $\frac{4}{5}$

If a curve $y = f(x)$ passes through the point $(1, -1)$ and satisfies the differential equation, $y(1 + xy) dx = x\, dy$, then $f\left(-\frac{1}{2}\right)$ is equal to:
(1) $-\frac{2}{5}$
(2) $-\frac{4}{5}$
(3) $\frac{2}{5}$
(4) $\frac{4}{5}$

The solution of the differential equation $\frac{dy}{dx} + \frac{y}{2}\sec^2 x = \frac{\tan x \sec^2 x}{2y}$, where $y(0) = 1$, is given by:
(1) $y^2 = 1 + \frac{\tan x}{x}$
(2) $y^2 = 1 + \tan x$
(3) $y = 1 - \tan x$
(4) $y^2 = 1 - \tan x$

If $( 2 + \sin x ) \frac { d y } { d x } + ( y + 1 ) \cos x = 0$ and $y ( 0 ) = 1$, then $y \left( \frac { \pi } { 2 } \right)$ is equal to:
(1) $\frac { 1 } { 3 }$
(2) $- \frac { 2 } { 3 }$
(3) $- \frac { 1 } { 3 }$
(4) $\frac { 4 } { 3 }$

Let $y = y ( x )$ be the solution of the differential equation $\sin x \frac { d y } { d x } + y \cos x = 4 x , x \in ( 0 , \pi )$. If $y \left( \frac { \pi } { 2 } \right) = 0$, then $y \left( \frac { \pi } { 6 } \right)$ is equal to
(1) $- \frac { 4 } { 9 } \pi ^ { 2 }$
(2) $\frac { 4 } { 9 \sqrt { 3 } } \pi ^ { 2 }$
(3) $\frac { - 8 } { 9 \sqrt { 3 } } \pi ^ { 2 }$
(4) $- \frac { 8 } { 9 } \pi ^ { 2 }$

Let $y = y ( x )$ be the solution of the differential equation $\frac { d y } { d x } + 2 y = f ( x )$, where $f ( x ) = \left\{ \begin{array} { l l } 1 , & x \in [ 0,1 ] \\ 0 , & \text { otherwise } \end{array} \right.$. If $y ( 0 ) = 0$, then $y \left( \frac { 3 } { 2 } \right)$ is
(1) $\frac { e ^ { 2 } - 1 } { e ^ { 3 } }$
(2) $\frac { 1 } { 2 e }$
(3) $\frac { e ^ { 2 } + 1 } { 2 e ^ { 4 } }$
(4) $\frac { e ^ { 2 } - 1 } { 2 e ^ { 3 } }$

Let $y = y ( x )$ be the solution of the differential equation $\frac { d y } { d x } + 2 y = f ( x )$, where
$$f ( x ) = \left\{ \begin{array} { l c } 1 , & x \in [ 0,1 ] \\ 0 , & \text { otherwise } \end{array} \right.$$
If $y ( 0 ) = 0$, then $y \left( \frac { 3 } { 2 } \right)$ is
(1) $\frac { e ^ { 2 } - 1 } { 2 e ^ { 3 } }$
(2) $\frac { e ^ { 2 } - 1 } { e ^ { 3 } }$
(3) $\frac { 1 } { 2 e }$
(4) $\frac { e ^ { 2 } + 1 } { 2 e ^ { 4 } }$

If $\frac { d y } { d x } + \frac { 3 } { \cos ^ { 2 } x } y = \frac { 1 } { \cos ^ { 2 } x } , x \in \left( - \frac { \pi } { 3 } , \frac { \pi } { 3 } \right)$, and $y \left( \frac { \pi } { 4 } \right) = \frac { 4 } { 3 }$, then $y \left( - \frac { \pi } { 4 } \right)$ equals
(1) $\frac { 1 } { 3 }$
(2) $\frac { 1 } { 3 } + e ^ { 3 }$
(3) $\frac { 1 } { 3 } + e ^ { 6 }$
(4) $- \frac { 4 } { 3 }$

Let $y = y(x)$ be the solution of the differential equation, $\left(x^2 + 1\right)^2 \frac{dy}{dx} + 2x\left(x^2 + 1\right)y = 1$ such that $y(0) = 0$. If $\sqrt{a}\, y(1) = \frac{\pi}{32}$, then the value of $a$ is
(1) $\frac{1}{16}$
(2) $\frac{1}{2}$
(3) $\frac{1}{4}$
(4) $1$

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

If $y = y ( x )$ is the solution of the differential equation $\frac { d y } { d x } + ( \tan x ) y = \sin x , 0 \leq x \leq \frac { \pi } { 3 }$, with $y ( 0 ) = 0$, then $y \left( \frac { \pi } { 4 } \right)$ is equal to
(1) $\frac { 1 } { 4 } \log _ { e } 2$
(2) $\left( \frac { 1 } { 2 \sqrt { 2 } } \right) \log _ { e } 2$
(3) $\log _ { e } 2$
(4) $\frac { 1 } { 2 } \log _ { e } 2$