From 467a1d6d5446284fdb4abfdde7116bc09822ed96 Mon Sep 17 00:00:00 2001
From: "flyingscorpio@arch-desktop" <tfranken@protonmail.com>
Date: Sun, 19 Sep 2021 14:50:19 +0200
Subject: [PATCH] Add exercices Analyse

---
 analyse/exercices/Makefile |  16 ++++
 analyse/exercices/main.tex | 178 +++++++++++++++++++++++++++++++++++++
 2 files changed, 194 insertions(+)
 create mode 100644 analyse/exercices/Makefile
 create mode 100644 analyse/exercices/main.tex

diff --git a/analyse/exercices/Makefile b/analyse/exercices/Makefile
new file mode 100644
index 0000000..4bce23d
--- /dev/null
+++ b/analyse/exercices/Makefile
@@ -0,0 +1,16 @@
+filename=$(shell basename $(shell pwd))
+timestamp=$(shell date +%Y-%m-%d_%H:%M)
+
+all: snapshot
+
+snapshot: main.tex
+	@latexmk -pdf main.tex >/dev/null 2>&1
+	@if ! cmp --silent build/main.pdf ${filename}_*.pdf; then \
+		touch ${filename}_tmp.pdf; \
+		rm ${filename}*.pdf; \
+		cp build/main.pdf ${filename}_${timestamp}.pdf; \
+		echo "Updated"; \
+	fi
+
+clean:
+	@rm -rf build 2>/dev/null
diff --git a/analyse/exercices/main.tex b/analyse/exercices/main.tex
new file mode 100644
index 0000000..17fb9bd
--- /dev/null
+++ b/analyse/exercices/main.tex
@@ -0,0 +1,178 @@
+\documentclass[a4paper,french,12pt]{article}
+
+\title{Analyse --- Exercices}
+\author{}
+\date{Dernière compilation~: \today{} à \currenttime}
+
+\usepackage{../../cours}
+\usepackage{enumitem}
+\usepackage{xfrac}
+
+\begin{document}
+
+\maketitle
+
+\section{Équations différentielles d'ordre 1}
+
+    \paragraph{$(E_1)$}
+
+        $y' - 2y = x^2$
+
+        \begin{enumerate}[label=\alph*)]
+
+            \item Solution homogène
+                \begin{align*}
+                    r = 2 \implies y_0 = \lambda e^{2x}
+                \end{align*}
+
+            \item Solution particulière
+                \begin{align*}
+                    \left\{
+                    \begin{array}{l}
+                        y_1 = ax^2 + bx + c \\
+                        y_1' = 2ax + b \\
+                    \end{array}
+                    \right.
+                    \quad y_1' - 2 y_1 = x^2 \Leftrightarrow
+                    2ax + b - 2(ax^2 + bx + c) = x^2 \\
+                    \left\{
+                    \begin{array}{l}
+                        -2a = 1 \\
+                        2a - 2b = 0 \\
+                        b - 2c = 0 \\
+                    \end{array}
+                    \right.
+                    \Leftrightarrow
+                    \left\{
+                    \begin{array}{l}
+                        a = \sfrac{-1}{2} \\
+                        b = \sfrac{-1}{2} \\
+                        c = \sfrac{-1}{4} \\
+                    \end{array}
+                    \right. \\
+                    \implies y_1 = -\frac{1}{2}x^2 -\frac{1}{2}x -\frac{1}{4}
+                \end{align*}
+
+            \item Solution générale
+
+                \begin{equation*}
+                    y = y_0 + y_1 = \boxed{-\frac{1}{2}x^2 -\frac{1}{2}x -\frac{1}{4} + \lambda e^{2x}}
+                \end{equation*}
+
+        \end{enumerate}
+
+    \paragraph{$(E_2)$}
+
+        $3y' - 9y = 7e^{3x}$
+
+        \begin{enumerate}[label=\alph*)]
+
+            \item Solution homogène
+                \begin{align*}
+                    r = 3 \implies y_0 = \lambda e^{3x}
+                \end{align*}
+
+            \item Solution particulière
+                \begin{align*}
+                    \left\{
+                    \begin{array}{l}
+                        y_1 = ax e^{3x} \\
+                        y_1' = ae^{3x} + 3axe^{3x} = (3ax + a)e^{3x} \\
+                    \end{array}
+                    \right.
+                    \quad 3y_1' - 9y_1 = 7e^{3x}
+                    &\Leftrightarrow 3(3ax + a)e^{3x} - 9axe^{3x} = 7e^{3x} \\
+                    &\Leftrightarrow (9ax + 3a)e^{3x} - 9axe^{3x} = 7e^{3x} \\
+                    &\Leftrightarrow 9ax + 3a - 9ax = 7 \\
+                    &\Leftrightarrow 3a = 7 \\
+                    &\Leftrightarrow a = \frac{7}{3} \\
+                    \implies y_1 = \frac{7}{3}xe^{3x}
+                \end{align*}
+
+            \item Solution générale
+
+                \begin{equation*}
+                    y = y_0 + y_1 = \lambda e^{3x} + \frac{7}{3}xe^{3x} = \boxed{(\frac{7}{3}x + \lambda)e^{3x}}
+                \end{equation*}
+
+        \end{enumerate}
+
+    \paragraph{$(E_3)$}
+
+        $2y' - 4y = 5e^{3x}$
+
+        \begin{enumerate}[label=\alph*)]
+
+            \item Solution homogène
+                \begin{align*}
+                    r = 2 \implies y_0 = \lambda e^{2x}
+                \end{align*}
+
+            \item Solution particulière
+                \begin{align*}
+                    \left\{
+                    \begin{array}{l}
+                        y_1 = ae^{3x} \\
+                        y_1' = 3ae^{3x} \\
+                    \end{array}
+                    \right.
+                    \quad 2y_1' - 4y_1 = 5e^{3x}
+                    &\Leftrightarrow 6ae^{3x} - 4ae^{3x} = 5e^{3x} \\
+                    &\Leftrightarrow 2a = 5 \\
+                    &\Leftrightarrow a = \frac{5}{2} \\
+                    \implies y_1 = \frac{5}{2}e^{3x}
+                \end{align*}
+
+            \item Solution générale
+
+                \begin{equation*}
+                    y = y_0 + y_1 = \boxed{\lambda e^{2x} + \frac{5}{2}e^{3x}}
+                \end{equation*}
+
+        \end{enumerate}
+
+    \paragraph{$(E_4)$}
+
+        $y' + 4y = 3\cos(2x)$
+
+        \begin{enumerate}[label=\alph*)]
+
+            \item Solution homogène
+                \begin{align*}
+                    r = -4 \implies y_0 = \lambda e^{-4x}
+                \end{align*}
+
+            \item Solution particulière
+                \begin{align*}
+                    \left\{
+                    \begin{array}{l}
+                        y_1 = a\cos(2x) + b\sin(2x) \\
+                        y_1' = -2a\sin(2x) + 2b\cos(2x) \\
+                    \end{array}
+                    \right.
+                \end{align*}
+                \begin{align*}
+                    y_1' + 4y_1 = 3\cos(2x)
+                    &\Leftrightarrow -2a\sin(2x) + 2b\cos(2x) + 4(a\cos(2x) + b\sin(2x)) = 3\cos(2x) \\
+                    &\Leftrightarrow
+                    \left\{
+                    \begin{array}{l}
+                        4a + 2b = 3 \\
+                        4b - 2a = 0 \\
+                    \end{array}
+                    \right.
+                    \Leftrightarrow
+                    \left\{
+                    \begin{array}{l}
+                        4a + 2b = 3 \\
+                        4a - 8b = 0 \\
+                    \end{array}
+                    \right.
+                    % TODO: finish
+                \end{align*}
+
+        \end{enumerate}
+
+\section{Équations différentielles d'ordre 2}
+
+\end{document}