Exercise 1

PART A Consider the differential equation $$( E ) : \quad y ^ { \prime } + \frac { 1 } { 4 } y = 20 \mathrm { e } ^ { - \frac { 1 } { 4 } x } ,$$ with unknown $y$, a function defined and differentiable on the interval $[ 0 ; + \infty [$.

1. Determine the value of the real number $a$ such that the function $g$ defined on the interval $[ 0 ; + \infty [$ by $g ( x ) = a x \mathrm { e } ^ { - \frac { 1 } { 4 } x }$ is a particular solution of the differential equation $( E )$.
2. Consider the differential equation $$\left( E ^ { \prime } \right) : \quad y ^ { \prime } + \frac { 1 } { 4 } y = 0 ,$$ with unknown $y$, a function defined and differentiable on the interval $[ 0 ; + \infty [$. Determine the solutions of the differential equation ( $E ^ { \prime }$ ).
3. Deduce the solutions of the differential equation ( $E$ ).
4. Determine the solution $f$ of the differential equation ( $E$ ) such that $f ( 0 ) = 8$.

PART B Consider the function $f$ defined on the interval $[ 0 ; + \infty [$ by $$f ( x ) = ( 20 x + 8 ) \mathrm { e } ^ { - \frac { 1 } { 4 } x }$$ It is admitted that the function $f$ is differentiable on the interval $\left[ 0 ; + \infty \left[ \right. \right.$ and we denote $f ^ { \prime }$ its derivative function on the interval $\left[ 0 ; + \infty \left[ \right. \right.$. Moreover, it is admitted that $\lim _ { x \rightarrow + \infty } f ( x ) = 0$.

1. a. Justify that, for every positive real number $x$, $$f ^ { \prime } ( x ) = ( 18 - 5 x ) \mathrm { e } ^ { - \frac { 1 } { 4 } x }$$ b. Deduce the table of variations of the function $f$. The exact value of the maximum of the function $f$ on the interval $[ 0 ; + \infty [$ will be specified.
2. In this question we are interested in the equation $f ( x ) = 8$. a. Justify that the equation $f ( x ) = 8$ admits a unique solution, denoted $\alpha$, in the interval [14; 15]. b. Copy and complete the table below by running step by step the solution\_equation function opposite, written in Python language
   

   $a$	14
   $b$	15
   $b - a$	1
   $m$	14,5
   \begin{tabular}{ l } Condition
   $f ( m ) > 8$

   & FALSE & & & & \hline \end{tabular}
   \begin{verbatim} from math import exp def f(x) : return (20* x +8)*exp(-1/4* x) def solution_equation() : a,b = 14,15 while b-a>0.1: m = (a+b)/2 if f (m) > 8 : |a=m else : | b=m return a,b \end{verbatim}
   c. What is the objective of the solution\_equation function in the context of the question?

Exercise 1
Part A
Consider the differential equation $$\text{(E)} \quad y' + 0{,}4y = \mathrm{e}^{-0{,}4t}$$ where $y$ is a function of the real variable $t$. We seek the set of functions defined and differentiable on $\mathbb{R}$ that are solutions of this equation.

1. Let $u$ be the function defined on $\mathbb{R}$ by: $u(t) = t\mathrm{e}^{-0{,}4t}$.
   Verify that $u$ is a solution of (E).
   
2. Let $f$ be a function defined and differentiable on $\mathbb{R}$.
   We denote by $g$ the function defined on $\mathbb{R}$ by: $g(t) = f(t) - u(t)$. Let (H) be the differential equation $y' + 0{,}4y = 0$.
   1. [a.] Prove that if the function $g$ is a solution of the differential equation (H) then the function $f$ is a solution of the differential equation (E).
      We will admit that the converse is true.
   2. [b.] Solve the differential equation (H).
   3. [c.] Deduce the solutions of (E).
   4. [d.] Determine the solution $f$ of (E) such that $f(0) = 1$.

Part B
We are interested in blood glucose levels in a person who has just had a meal. The blood glucose level in $\mathrm{g.L}^{-1}$, as a function of time $t$, expressed in hours, elapsed since the end of the meal, is modelled by the function $f$ defined on $[0;6]$ by: $$f(t) = (t+1)\mathrm{e}^{-0{,}4t}$$

1. 1. [a.] Show that, for all $t \in [0;6]$, $f'(t) = (-0{,}4t + 0{,}6)\mathrm{e}^{-0{,}4t}$.
   2. [b.] Study the variations of $f$ on $[0;6]$ then draw up its variation table on this interval.
2. A person is hypoglycaemic when their blood glucose level is below $0{,}7\,\mathrm{g.L}^{-1}$.
   1. [a.] Prove that on the interval $[0;6]$ the equation $f(t) = 0{,}7$ admits a unique solution which we will denote $\alpha$.
   2. [b.] How long after having eaten does this person become hypoglycaemic? Express this time to the nearest minute.
3. We wish to determine the average blood glucose level in $\mathrm{g.L}^{-1}$ in this person during the six hours following the meal.
   1. [a.] Using integration by parts, show that: $$\int_0^6 f(t)\,\mathrm{d}t = -23{,}75\,\mathrm{e}^{-2{,}4} + 8{,}75$$
   2. [b.] Calculate the average blood glucose level in $\mathrm{g.L}^{-1}$ in this person during the six hours following the meal.
   3. [c.] By noting that the function $f$ is a solution of the differential equation (E), explain how we could have obtained this result differently.

Consider the differential equation
$$(E): \quad y' + 10y = \left(30x^2 + 22x - 8\right)\mathrm{e}^{-5x+1} \quad \text{with} \quad x \in \mathbb{R}$$
where $y$ is a function defined and differentiable on $\mathbb{R}$.

1. Solve on $\mathbb{R}$ the differential equation: $y' + 10y = 0$.
2. Let the function $f$ defined on $\mathbb{R}$ by $$f(x) = \left(6x^2 + 2x - 2\right)\mathrm{e}^{-5x+1}$$ We admit that $f$ is differentiable on $\mathbb{R}$ and we denote $f'$ the derivative function of the function $f$. Justify that $f$ is a particular solution of $(E)$.
3. Give the expression of all solutions of $(E)$.

Part A

We consider the differential equation $$\left(E_1\right): \quad y' + 0.48y = \frac{1}{250}$$ where $y$ is a function of the variable $t$ belonging to the interval $[0; +\infty[$.

1. We consider the constant function $h$ defined on the interval $[0; +\infty[$ by $h(t) = \frac{1}{120}$. Show that the function $h$ is a solution of the differential equation $(E_1)$.
2. Give the general form of the solutions of the differential equation $y' + 0.48y = 0$.
3. Deduce the set of solutions of the differential equation $(E_1)$.

Part B

We are now interested in the evolution of a population of bacteria in a culture medium. At an instant $t = 0$, an initial population of 30000 bacteria is introduced into the medium. We denote $p(t)$ the quantity of bacteria, expressed in thousands of individuals, present in the medium after a time $t$, expressed in hours. We therefore have $p(0) = 30$. We admit that the function $p$ defined on the interval $[0; +\infty[$ is differentiable, strictly positive on this interval and that it is a solution of the differential equation $(E_2)$: $$p' = \frac{1}{250} p \times (120 - p)$$ Let $y$ be the function strictly positive on the interval $[0; +\infty[$ such that, for all $t$ belonging to the interval $[0; +\infty[$, we have $p(t) = \frac{1}{y(t)}$.

1. Show that if $p$ is a solution of the differential equation $(E_2)$, then $y$ is a solution of the differential equation $\left(E_1\right): \quad y' + 0.48y = \frac{1}{250}$.
2. We admit conversely that, if $y$ is a strictly positive solution of the differential equation $(E_1)$, then $p = \frac{1}{y}$ is a solution of the differential equation $(E_2)$. Show that, for all $t$ belonging to the interval $[0; +\infty[$, we have: $$p(t) = \frac{120}{1 + K\mathrm{e}^{-0.48t}} \text{ with } K \text{ a real constant.}$$
3. Using the initial condition, determine the value of $K$.
4. Determine $\lim_{t \rightarrow +\infty} p(t)$. Give an interpretation of this in the context of the exercise.
5. Determine the time required for the bacterial population to exceed 60000 individuals. The result will be given in the form of a rounded value expressed in hours and minutes.

In this part, $\lambda(t) = t$ for all $t \in \mathbb{R}^+$. We assume that $f$ admits a finite limit $l$ at $+\infty$.
IV.D.1) Show that $E$ contains $\mathbb{R}^{+*}$.
IV.D.2) Show that $xLf(x)$ tends to $l$ at $0^+$.

In this part, $\lambda(t) = t$ for all $t \in \mathbb{R}^+$ and $f(t) = \dfrac{\sin t}{t}$ for all $t \in \mathbb{R}^{+*}$, $f$ being extended by continuity at 0.
Show that $E$ does not contain 0.

In this part, $\lambda(t) = t$ for all $t \in \mathbb{R}^+$ and $f(t) = \dfrac{\sin t}{t}$ for all $t \in \mathbb{R}^{+*}$, $f$ being extended by continuity at 0.
Show that $E = ]0, +\infty[$.

In this part, $\lambda(t) = t$ for all $t \in \mathbb{R}^+$ and $f(t) = \dfrac{\sin t}{t}$ for all $t \in \mathbb{R}^{+*}$, $f$ being extended by continuity at 0.
Show that $E^{\prime}$ contains 0.

In this part, $\lambda(t) = t$ for all $t \in \mathbb{R}^+$ and $f(t) = \dfrac{\sin t}{t}$ for all $t \in \mathbb{R}^{+*}$, $f$ being extended by continuity at 0.
Calculate $(Lf)^{\prime}(x)$ for $x \in E$.

In this part, $\lambda(t) = t$ for all $t \in \mathbb{R}^+$ and $f(t) = \dfrac{\sin t}{t}$ for all $t \in \mathbb{R}^{+*}$, $f$ being extended by continuity at 0.
Deduce from V.D the expression of $(Lf)(x)$ for $x \in E$.

In this part, $\lambda(t) = t$ for all $t \in \mathbb{R}^+$ and $f(t) = \dfrac{\sin t}{t}$ for all $t \in \mathbb{R}^{+*}$, $f$ being extended by continuity at 0.
We denote for $n \in \mathbb{N}$ and $x \geqslant 0$, $$f_n(x) = \int_{n\pi}^{(n+1)\pi} \frac{\sin t}{t} e^{-xt}\,dt.$$ Show that $\sum_{n \geqslant 0} f_n$ converges uniformly on $[0, +\infty[$.

In this part, $\lambda(t) = t$ for all $t \in \mathbb{R}^+$ and $f(t) = \dfrac{\sin t}{t}$ for all $t \in \mathbb{R}^{+*}$, $f$ being extended by continuity at 0.
What is the value of $Lf(0)$?

In this part, $\lambda(t) = t$ for all $t \in \mathbb{R}^+$.
Let $g$ be a continuous application from $[0,1]$ to $\mathbb{R}$. We assume that for all $n \in \mathbb{N}$, we have $$\int_0^1 t^n g(t)\,dt = 0.$$
VI.A.1) What can we say about $\displaystyle\int_0^1 P(t)g(t)\,dt$ for $P \in \mathbb{R}[X]$?
VI.A.2) Deduce from this that $g$ is the zero application.

In this part, $\lambda(t) = t$ for all $t \in \mathbb{R}^+$.
Let $f$ be fixed such that $E$ is non-empty, $x \in E$ and $a > 0$. We set $h(t) = \displaystyle\int_0^t e^{-xu} f(u)\,du$ for all $t \geqslant 0$.
VI.B.1) Show that $Lf(x+a) = a\displaystyle\int_0^{+\infty} e^{-at} h(t)\,dt$.
VI.B.2) We assume that for all $n \in \mathbb{N}$, we have $Lf(x + na) = 0$.
Show that, for all $n \in \mathbb{N}$, the integral $\displaystyle\int_0^1 u^n h\!\left(-\frac{\ln u}{a}\right)du$ converges and that it is zero.
VI.B.3) What do we deduce for the function $h$?

If $I$ is an interval of $\mathbb { R }$, we say that $u \in \mathcal { C } ^ { 1 } ( I , \mathbb { R } )$ satisfies (II.1) on $I$ if and only if $$\forall t \in I , \quad u ( t ) \left( u ( t ) + 2 t u ^ { \prime } ( t ) \right) = - 1$$
By stating precisely the theorem used, show that if $( t _ { 0 } , u _ { 0 } )$ is in $\left( \mathbb { R } ^ { * } \right) ^ { 2 }$, there exist an open interval $I$ of $\mathbb { R }$ containing $t _ { 0 }$ and a function $u \in \mathcal { C } ^ { 1 } ( I , \mathbb { R } )$ such that $u$ is a solution of (II.1) on $I$ and satisfies $u \left( t _ { 0 } \right) = u _ { 0 }$.

For $( x , y )$ in $\left( \mathbb { R } ^ { + * } \right) ^ { 2 }$, we define $\beta ( x , y ) = \int _ { 0 } ^ { 1 } t ^ { x - 1 } ( 1 - t ) ^ { y - 1 } \mathrm {~d} t$. We have the relation $(\mathcal{R})$: $\beta ( x , y ) = \frac { \Gamma ( x ) \Gamma ( y ) } { \Gamma ( x + y ) }$. We define $\psi ( x ) = \frac { \Gamma ^ { \prime } ( x ) } { \Gamma ( x ) }$.
From the relation $(\mathcal{R})$, justify that $\frac { \partial \beta } { \partial y }$ is defined on $\left( \mathbb { R } ^ { + * } \right) ^ { 2 }$.
Establish that for all real $x > 0$ and $y > 0 , \frac { \partial \beta } { \partial y } ( x , y ) = \beta ( x , y ) ( \psi ( y ) - \psi ( x + y ) )$.

For $( x , y )$ in $\left( \mathbb { R } ^ { + * } \right) ^ { 2 }$, we define $\beta ( x , y ) = \int _ { 0 } ^ { 1 } t ^ { x - 1 } ( 1 - t ) ^ { y - 1 } \mathrm {~d} t$.
Let $x > 0$ be fixed. What is the monotonicity on $\mathbb { R } ^ { + * }$ of the function $y \mapsto \beta ( x , y )$?

We define $\psi ( x ) = \frac { \Gamma ^ { \prime } ( x ) } { \Gamma ( x ) }$ on $\mathbb{R}^{+*}$.
Show that the function $\psi$ is increasing on $\mathbb { R } ^ { + * }$.

Throughout the problem, we denote for every integer $n \geqslant 1$, $H _ { n } = \sum _ { k = 1 } ^ { n } \frac { 1 } { k }$. We define $\psi ( x ) = \frac { \Gamma ^ { \prime } ( x ) } { \Gamma ( x ) }$ on $\mathbb{R}^{+*}$, satisfying $\psi ( x + 1 ) - \psi ( x ) = \frac{1}{x}$ for all $x > 0$.
Show that for every real $x > - 1$ and for every integer $n \geqslant 1$ $$\psi ( 1 + x ) - \psi ( 1 ) = \psi ( n + x + 1 ) - \psi ( n + 1 ) + \sum _ { k = 1 } ^ { n } \left( \frac { 1 } { k } - \frac { 1 } { k + x } \right)$$

Throughout the problem, we denote for every integer $n \geqslant 1$, $H _ { n } = \sum _ { k = 1 } ^ { n } \frac { 1 } { k }$. We define $\psi ( x ) = \frac { \Gamma ^ { \prime } ( x ) } { \Gamma ( x ) }$ on $\mathbb{R}^{+*}$, and $\psi$ is increasing on $\mathbb{R}^{+*}$.
Let $n$ be an integer $\geqslant 2$ and $x$ a real $> - 1$. We set $p = E ( x ) + 1$, where $E ( x )$ denotes the integer part of $x$. Prove that $$0 \leqslant \psi ( n + x + 1 ) - \psi ( n ) \leqslant H _ { n + p } - H _ { n - 1 } \leqslant \frac { p + 1 } { n }$$

We define $\psi ( x ) = \frac { \Gamma ^ { \prime } ( x ) } { \Gamma ( x ) }$ on $\mathbb{R}^{+*}$, satisfying $\psi ( x + 1 ) - \psi ( x ) = \frac{1}{x}$ for all $x > 0$.
Deduce that, for every real $x > - 1$, $$\psi ( 1 + x ) = \psi ( 1 ) + \sum _ { n = 1 } ^ { + \infty } \left( \frac { 1 } { n } - \frac { 1 } { n + x } \right)$$

We denote $\zeta ( x ) = \sum _ { n = 1 } ^ { + \infty } \frac { 1 } { n ^ { x } }$ for $x > 1$. We denote $g$ the function defined on $[ - 1 , + \infty [$ by $$g ( x ) = \sum _ { n = 2 } ^ { + \infty } \left( \frac { 1 } { n } - \frac { 1 } { n + x } \right)$$
Show that $g$ is of class $\mathcal { C } ^ { \infty }$ on $[ - 1 , + \infty [$.
Specify in particular the value of $g ^ { ( k ) } ( 0 )$ as a function of $\zeta ( k + 1 )$ for every integer $k \geqslant 1$.

We denote $\zeta ( x ) = \sum _ { n = 1 } ^ { + \infty } \frac { 1 } { n ^ { x } }$ for $x > 1$. We denote $g$ the function defined on $[ - 1 , + \infty [$ by $$g ( x ) = \sum _ { n = 2 } ^ { + \infty } \left( \frac { 1 } { n } - \frac { 1 } { n + x } \right)$$
Show that for every integer $n$ and for every $x$ in $] - 1,1 [$ $$\left| g ( x ) - \sum _ { k = 0 } ^ { n } \frac { g ^ { ( k ) } ( 0 ) } { k ! } x ^ { k } \right| \leqslant \zeta ( 2 ) | x | ^ { n + 1 }$$
Show that $g$ is expandable as a power series on $] - 1,1 [$.

We denote $\zeta ( x ) = \sum _ { n = 1 } ^ { + \infty } \frac { 1 } { n ^ { x } }$ for $x > 1$. We define $\psi ( x ) = \frac { \Gamma ^ { \prime } ( x ) } { \Gamma ( x ) }$ on $\mathbb{R}^{+*}$. We have shown that for every real $x > -1$, $\psi ( 1 + x ) = \psi ( 1 ) + \sum _ { n = 1 } ^ { + \infty } \left( \frac { 1 } { n } - \frac { 1 } { n + x } \right)$.
Prove that for every $x$ in $] - 1,1 [$, $$\psi ( 1 + x ) = \psi ( 1 ) + \sum _ { n = 1 } ^ { + \infty } ( - 1 ) ^ { n + 1 } \zeta ( n + 1 ) x ^ { n }$$

We denote $B$ the function defined on $\mathbb { R } ^ { + * }$ by $B ( x ) = \frac { \partial ^ { 2 } \beta } { \partial y ^ { 2 } } ( x , 1 )$, where $\beta ( x , y ) = \int _ { 0 } ^ { 1 } t ^ { x - 1 } ( 1 - t ) ^ { y - 1 } \mathrm {~d} t$ and $\psi ( x ) = \frac { \Gamma ^ { \prime } ( x ) } { \Gamma ( x ) }$. We have $\frac { \partial \beta } { \partial y } ( x , y ) = \beta ( x , y ) ( \psi ( y ) - \psi ( x + y ) )$.
Justify that $B$ is defined on $\mathbb { R } ^ { + * }$.
Using the relation found in III.B.1, establish that for every real $x > 0$ $$x B ( x ) = ( \psi ( 1 + x ) - \psi ( 1 ) ) ^ { 2 } + \left( \psi ^ { \prime } ( 1 ) - \psi ^ { \prime } ( 1 + x ) \right)$$
Deduce that $B$ is $\mathcal { C } ^ { \infty }$ on $\mathbb { R } ^ { + * }$.