728 lines
28 KiB
TeX
728 lines
28 KiB
TeX
\documentclass[a4paper,french,11pt]{article}
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\title{Analyse --- Exercices}
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\author{}
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\date{Dernière compilation~: \today{} à \currenttime}
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\usepackage{../../cours}
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\usepackage{enumitem}
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\usepackage{xfrac}
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\begin{document}
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\maketitle
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\section{Équations différentielles d'ordre 1}
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\paragraph{$(E_1)$}
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$y' - 2y = x^2$
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\begin{enumerate}[label=\alph*)]
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\item Solution homogène
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\begin{align*}
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r = 2 \implies y_0 = \lambda e^{2x}
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\end{align*}
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\item Solution particulière
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\begin{align*}
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\left\{
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\begin{array}{l}
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y_1 = ax^2 + bx + c \\
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y_1' = 2ax + b \\
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\end{array}
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\right.
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\end{align*}
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\begin{align*}
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\text{Dans } (E_1)
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&\implies
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2ax + b - 2(ax^2 + bx + c) = x^2 \\
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&\iff
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\left\{
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\begin{array}{l}
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-2a = 1 \\
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2a - 2b = 0 \\
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b - 2c = 0 \\
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\end{array}
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\right.
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\iff
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\left\{
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\begin{array}{l}
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a = \sfrac{-1}{2} \\
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b = \sfrac{-1}{2} \\
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c = \sfrac{-1}{4} \\
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\end{array}
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\right.
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\end{align*}
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\begin{align*}
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\implies y_1 = -\frac{1}{2}x^2 -\frac{1}{2}x -\frac{1}{4}
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\end{align*}
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\item Solution générale
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\begin{equation*}
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y = y_0 + y_1 = \boxed{-\frac{1}{2}x^2 -\frac{1}{2}x -\frac{1}{4} + \lambda e^{2x}}
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\end{equation*}
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\end{enumerate}
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\paragraph{$(E_2)$}
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$3y' - 9y = 7e^{3x}$
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\begin{enumerate}[label=\alph*)]
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\item Solution homogène
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\begin{align*}
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r = 3 \implies y_0 = \lambda e^{3x}
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\end{align*}
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\item Solution particulière
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\begin{align*}
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\left\{
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\begin{array}{l}
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y_1 = ax e^{3x} \\
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y_1' = ae^{3x} + 3axe^{3x} = (3ax + a)e^{3x} \\
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\end{array}
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\right.
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\end{align*}
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\begin{align*}
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\text{Dans } (E_2)
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&\implies 3(3ax + a)e^{3x} - 9axe^{3x} = 7e^{3x} \\
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&\iff (9ax + 3a)e^{3x} - 9axe^{3x} = 7e^{3x} \\
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&\iff 9ax + 3a - 9ax = 7 \\
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&\iff 3a = 7 \\
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&\iff a = \frac{7}{3}
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\end{align*}
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\begin{align*}
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\implies y_1 = \frac{7}{3}xe^{3x}
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\end{align*}
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\item Solution générale
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\begin{equation*}
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y = y_0 + y_1 = \lambda e^{3x} + \frac{7}{3}xe^{3x} = \boxed{(\frac{7}{3}x + \lambda)e^{3x}}
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\end{equation*}
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\end{enumerate}
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\paragraph{$(E_3)$}
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$2y' - 4y = 5e^{3x}$
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\begin{enumerate}[label=\alph*)]
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\item Solution homogène
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\begin{align*}
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r = 2 \implies y_0 = \lambda e^{2x}
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\end{align*}
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\item Solution particulière
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\begin{align*}
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\left\{
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\begin{array}{l}
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y_1 = ae^{3x} \\
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y_1' = 3ae^{3x} \\
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\end{array}
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\right.
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\end{align*}
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\begin{align*}
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\text{Dans } (E_3)
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&\iff 6ae^{3x} - 4ae^{3x} = 5e^{3x} \\
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&\iff 2a = 5
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\iff a = \frac{5}{2}
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\end{align*}
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\begin{align*}
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\implies y_1 = \frac{5}{2}e^{3x}
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\end{align*}
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\item Solution générale
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\begin{equation*}
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y = y_0 + y_1 = \boxed{\lambda e^{2x} + \frac{5}{2}e^{3x}}
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\end{equation*}
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\end{enumerate}
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\paragraph{$(E_4)$}
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$y' + 4y = 3\cos(2x)$
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\begin{enumerate}[label=\alph*)]
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\item Solution homogène
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\begin{align*}
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r = -4 \implies y_0 = \lambda e^{-4x}
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\end{align*}
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\item Solution particulière
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\begin{align*}
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\left\{
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\begin{array}{l}
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y_1 = a\cos(2x) + b\sin(2x) \\
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y_1' = -2a\sin(2x) + 2b\cos(2x) \\
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\end{array}
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\right.
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\end{align*}
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\begin{align*}
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\text{Dans } (E_4)
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&\implies -2a\sin(2x) + 2b\cos(2x) + 4(a\cos(2x) + b\sin(2x)) = 3\cos(2x) \\
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&\iff
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\left\{
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\begin{array}{l}
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4a + 2b = 3 \\
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4b - 2a = 0 \\
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\end{array}
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\right.
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\iff
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\left\{
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\begin{array}{l}
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4a + 2b = 3 \\
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4a - 8b = 0 \\
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\end{array}
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\right.
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\iff
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\left\{
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\begin{array}{l}
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10b = 3 \\
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4b - 2a = 0 \\
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\end{array}
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\right. \\
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&\iff
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\left\{
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\begin{array}{l}
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b = \sfrac{3}{10} \\
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a = \sfrac{12}{20} = \sfrac{3}{5} \\
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\end{array}
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\right.
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\end{align*}
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\begin{align*}
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\implies y_1 = \frac{3}{5}\cos(2x) + \frac{3}{10}\sin(2x)
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\end{align*}
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\item Solution générale
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\begin{equation*}
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y = y_0 + y_1 = \boxed{\lambda e^{-4x} + \frac{3}{5}\cos(2x) + \frac{3}{10}\sin(2x)}
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\end{equation*}
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\end{enumerate}
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\section{Équations différentielles d'ordre 2}
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\paragraph{$(E_5)$}
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$y'' - 7y' + 10 y = (x + 3) e^{2x}$
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\begin{enumerate}[label=\alph*)]
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\item Solution homogène
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\begin{align*}
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r^2 - 7r + 10 = 0
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\implies \Delta = 9
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\implies
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\left\{
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\begin{array}{l}
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r_1 = \frac{7 - 3}{2} = 2 \\\\
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r_2 = \frac{7 + 3}{2} = 5 \\
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\end{array}
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\right.
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\end{align*}
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\begin{align*}
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\implies
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y_0 = \lambda e^{2x} + \mu e^{5x}
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\end{align*}
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\item Solution particulière
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second membre~: $(x + 3)e^{\alpha x}$ avec $\alpha = 2$ $\implies \alpha$ racine de l'équation caractéristique.
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\begin{align*}
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&\left\{
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\begin{array}{l}
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y_1 = xe^{2x} (ax + b) \\
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y_1' = (2x + 1)e^{2x}(ax + b) + axe^{2x} \\
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y_1'' = (4ax + 2a + 2b)e^{2x} + (4ax^2 + 4ax + 4bx + 2b)e^{2x} \\
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\end{array}
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\right. \\
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\implies
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&\left\{
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\begin{array}{l}
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y_1 = xe^{2x} (ax + b) \\
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y_1' = (2ax^2 + 2ax + 2bx + b)e^{2x} \\
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y_1'' = (4ax^2 + 8ax + 4bx + 2a + 4b)e^{2x} \\
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\end{array}
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\right.
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\end{align*}
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Dans $(E_5)$~:
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$(4ax^2 + 8ax + 4bx + 2a + 4b)e^{2x} - 7(2ax^2 + 2ax + 2bx + b)e^{2x} + 10(ax + b)xe^{2x} = (x + 3) e^{2x}$
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\begin{align*}
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\implies
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-6ax + 2a - 3b = x + 3
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\implies
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\left\{
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\begin{array}{l}
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-6a = 1 \\
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2a - 3b = 3 \\
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\end{array}
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\right.
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\implies
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\left\{
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\begin{array}{l}
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a = \frac{-1}{6} \\\\
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b = \frac{-10}{9} \\
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\end{array}
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\right.
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\end{align*}
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\begin{align*}
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\implies
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y_1 = xe^{2x}(-\frac{1}{6}x - \frac{10}{9})
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\end{align*}
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\item Solution générale
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\begin{equation*}
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y = y_0 + y_1 = \boxed{\lambda e^{2x} + \mu e^{5x} + xe^{2x}(-\frac{1}{6}x - \frac{10}{9})}
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\end{equation*}
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\end{enumerate}
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\paragraph{$(E_6)$}
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$y'' - y = x^3$
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\begin{enumerate}[label=\alph*)]
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\item Solution homogène
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\begin{align*}
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r^2 - 1 = 0
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\implies \Delta = 4
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\implies
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\left\{
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\begin{array}{l}
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r_1 = \frac{0 - 2}{2} = -1 \\\\
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r_2 = \frac{0 + 2}{2} = 1 \\
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\end{array}
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\right.
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\end{align*}
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\begin{align*}
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\implies
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y_0 = \lambda e^{-x} + \mu e^{x}
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\end{align*}
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\item Solution particulière
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second membre~: $x^3e^{\alpha x}$ avec $\alpha = 0$ $\implies \alpha$ non racine de l'équation caractéristique.
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\begin{align*}
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&\left\{
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\begin{array}{l}
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y_1 = ax^3 + bx^2 + cx + d \\
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y_1' = 3ax^2 + 2bx + c \\
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y_1'' = 6ax + 2b \\
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\end{array}
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\right.
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\end{align*}
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Dans $(E_6)$~:
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$6ax + 2b - ax^3 - bx^2 - cx - d = x^3$
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\begin{align*}
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\implies
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-ax^3 - bx^2 + 6ax - cx + 2b - d = x^3
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\implies
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\left\{
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\begin{array}{l}
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-a = 1 \\
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-b = 0 \\
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6a - c = 0 \\
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2b - d = 0 \\
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\end{array}
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\right.
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\implies
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\left\{
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\begin{array}{l}
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a = -1 \\
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b = 0 \\
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c = -6 \\
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d = 0 \\
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\end{array}
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\right.
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\end{align*}
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\begin{align*}
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\implies
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y_1 = -x^3 - 6x
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\end{align*}
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\item Solution générale
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\begin{equation*}
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y = y_0 + y_1 = \boxed{\lambda e^{-x} + \mu e^{x} - x^3 - 6x}
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\end{equation*}
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\end{enumerate}
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\paragraph{$(E_7)$}
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$y'' + y = \cos{x}$
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\begin{enumerate}[label=\alph*)]
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\item Solution homogène
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\begin{align*}
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r^2 + 1 = 0
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\implies \Delta = -4
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\implies
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\left\{
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\begin{array}{l}
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r_1 = \frac{0 - 2i}{2} \\\\
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r_2 = \frac{0 + 2i}{2} \\
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\end{array}
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\right.
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\implies
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\left\{
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\begin{array}{l}
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\alpha = 0 \\
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\beta = 1
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\end{array}
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\right.
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\end{align*}
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\begin{align*}
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\implies
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y_0 = e^{0}(\lambda \cos{x} + \mu \sin{x})
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= \lambda \cos{x} + \mu \sin{x}
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\end{align*}
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\item Solution particulière
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second membre~: $e^{\alpha x}(P_0(x)\cos{\beta x} + 0 \times \sin{\beta x})$ avec $\alpha = 0, \beta = 1, P_0 = 1$ $\implies \alpha + i\beta$ racine de l'équation caractéristique.
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\begin{align*}
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y_1 &= xe^{\alpha x}(a\cos{\beta x} + b\sin{\beta x}) \\
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&= x(a\cos{x} + b\sin{x}) \\
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\implies
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&\left\{
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\begin{array}{l}
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y_1 = x(a\cos{x} + b\sin{x}) \\
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y_1' = a\cos{x} + b\sin{x} - x(a\sin{x} - b\cos{x}) \\
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y_1'' = -a\sin{x} + b\cos{x} - a\sin{x} + b\cos{x} - x(a\cos{x} + b\sin{x}) \\
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\end{array}
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\right. \\
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\implies
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&\left\{
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\begin{array}{l}
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y_1 = x(a\cos{x} + b\sin{x}) \\
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y_1' = a\cos{x} + b\sin{x} - x(a\sin{x} - b\cos{x}) \\
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y_1'' = -2a\sin{x} + 2b\cos{x} - x(a\cos{x} + b\sin{x}) \\
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\end{array}
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\right.
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\end{align*}
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Dans $(E_7)$~:
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\begin{align*}
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-2a\sin{x} + 2b\cos{x} - x(a\cos{x} + b\sin{x}) + x(a\cos{x} + b\sin{x}) = \cos{x} \\
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\iff
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-2a\sin{x} + 2b\cos{x} = \cos{x} \\
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\iff
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2b\cos{x} - 2a\sin{x} = \cos{x} \\
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\implies
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\left\{
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\begin{array}{l}
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2b = 1 \\
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-a = 0 \\
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\end{array}
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\right.
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\implies
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\left\{
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\begin{array}{l}
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a = 0 \\
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b = \frac{1}{2} \\
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\end{array}
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\right.
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\end{align*}
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\begin{align*}
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\implies
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y_1 = \frac{x\sin{x}}{2} \\
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\end{align*}
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\item Solution générale
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\begin{equation*}
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y = y_0 + y_1 = \boxed{\lambda \cos{x} + \mu \sin{x} + \frac{x\sin{x}}{2}}
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\end{equation*}
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\end{enumerate}
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\paragraph{$(E_8)$}
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$y'' - 4y = (-4x + 3) e^{2x}$
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\begin{enumerate}[label=\alph*)]
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\item Solution homogène
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\begin{align*}
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r^2 - 4 = 0
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\implies \Delta = 16
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\implies
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\left\{
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\begin{array}{l}
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r_1 = \frac{0 - 4}{2} = -2 \\\\
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r_2 = \frac{0 + 4}{2} = 2 \\
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\end{array}
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\right.
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\end{align*}
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\begin{align*}
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\implies
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y_0 = \lambda e^{-2x} + \mu e^{2x}
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\end{align*}
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\item Solution particulière
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second membre~: $(-4x + 3)e^{\alpha x}$ avec $\alpha = 2$ $\implies \alpha$ racine de l'équation caractéristique.
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\begin{align*}
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&\left\{
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\begin{array}{l}
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y_1 = xe^{2x} (ax + b) \\
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y_1' = (2x + 1)e^{2x}(ax + b) + axe^{2x} \\
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y_1'' = (4ax + 2a + 2b)e^{2x} + (4ax^2 + 4ax + 4bx + 2b)e^{2x} \\
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\end{array}
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\right. \\
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\implies
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&\left\{
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\begin{array}{l}
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y_1 = xe^{2x} (ax + b) \\
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y_1' = (2ax^2 + 2ax + 2bx + b)e^{2x} \\
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y_1'' = (4ax^2 + 8ax + 4bx + 2a + 4b)e^{2x} \\
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\end{array}
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\right.
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\end{align*}
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Dans $(E_8)$~:
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$(4ax^2 + 8ax + 4bx + 2a + 4b)e^{2x} - 4x(ax + b)e^{2x} = (-4x + 3) e^{2x}$
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\begin{align*}
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\implies
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8ax + 2a + 4b = -4x + 3
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\implies
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\left\{
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\begin{array}{l}
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8a = -4 \\
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2a + 4b = 3 \\
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\end{array}
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\right.
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\implies
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\left\{
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\begin{array}{l}
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a = \frac{-1}{2} \\
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b = 1 \\
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\end{array}
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\right.
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\end{align*}
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\begin{align*}
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\implies
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y_1 = xe^{2x} (\frac{-x}{2} + 1)
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\end{align*}
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\item Solution générale
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\begin{equation*}
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y = y_0 + y_1 = \boxed{\lambda e^{-2x} + \mu e^{2x} + xe^{2x} (\frac{-x}{2} + 1)}
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\end{equation*}
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\end{enumerate}
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\section{Intégrales généralisées}
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\subsection{Exercice 1}
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Étudier la convergence des intégrales généralisées suivantes.
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\paragraph{$(I_1)$}
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$\int_1^{+\infty} \frac{2x + x^3}{x^3 + x^4} \dif x$
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\begin{align*}
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\forall x \in [1; +\infty[\quad \frac{2x + x^3}{x^3 + x^4} \geq 0 \\\\
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\frac{2x + x^3}{x^3 + x^4} = \frac{x^3(\frac{2}{x^2} + 1)}{x^4(\frac{1}{x} + 1)} \sim \frac{x^3}{x^4} \sim \frac{1}{x} \\
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\text{Or } \int_1^{+\infty}\frac{1}{x}\dif x \text{ diverge donc $I_1$ diverge par équivalence.} \\
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\end{align*}
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\paragraph{$(I_2)$}
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$\int_0^{+\infty} \frac{x\sqrt{x}}{x^2 \sqrt[3]{x} + 4} \dif x$
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\begin{align*}
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\forall x \in [0; +\infty[\quad \frac{x\sqrt{x}}{x^2 \sqrt[3]{x} + 4} \geq 0 \\\\
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\frac{x\sqrt{x}}{x^2 \sqrt[3]{x} + 4}
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= \frac{x^{1+\frac{1}{2}}}{x^{2+\frac{1}{3}}(1 + \frac{4}{x^{\frac{7}{3}}})}
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\sim \frac{x^{1+\frac{1}{2}}}{x^{\frac{7}{3}}}
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\sim x^{\frac{3}{2} - \frac{7}{3}}
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\sim x^{-\frac{5}{6}}
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\sim \frac{1}{x^{\frac{5}{6}}} \\
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\text{Or } \int_1^{+\infty}\frac{1}{x^{\frac{5}{6}}}\dif x \text{ diverge donc $I_2$ diverge par équivalence.} \\
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\end{align*}
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\paragraph{$(I_3)$}
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$\int_1^{+\infty} \frac{5x + x^2}{x^3 + x^3\sqrt{x}} \dif x$
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\begin{align*}
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\forall x \in [1; +\infty[\quad \frac{5x + x^2}{x^3 + x^3\sqrt{x}} \geq 0 \\\\
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\frac{5x + x^2}{x^3 + x^3\sqrt{x}}
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= \frac{x^2(\frac{5x}{x^2} + 1)}{x^3\sqrt{x}(\frac{1}{\sqrt{x}} + 1)}
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= \frac{x^2(\frac{5}{x} + 1)}{x^{3 + \frac{1}{2}}(\frac{1}{\sqrt{x}} + 1)}
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\sim \frac{x^2}{x^{\frac{7}{2}}}
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\sim \frac{1}{x^{\frac{3}{2}}} \\
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\text{Or } \int_1^{+\infty}\frac{1}{x^{\frac{3}{2}}}\dif x \text{ converge donc $I_3$ converge par équivalence.} \\
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\end{align*}
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|
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\paragraph{$(I_4)$}
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$\int_0^{+\infty} \frac{x \sqrt[3]{x}}{x^2\sqrt{x} + 5} e^{-x} \dif x$
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\begin{align*}
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\forall x \in [0; +\infty[ \quad \frac{x \sqrt[3]{x}}{x^2\sqrt{x} + 5} e^{-x} \geq 0 \\\\
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\frac{x \sqrt[3]{x}}{x^2\sqrt{x} + 5} e^{-x}
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= \frac{x^{\frac{4}{3}}}{x^2\sqrt{x}(1 + \frac{5}{x^2\sqrt{x}})} e^{-x}
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\sim \frac{x^{\frac{4}{3}}}{x^{\frac{5}{2}}} e^{-x}
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\sim x^{\frac{-7}{6}} e^{-x}
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\sim \frac{1}{x^{\frac{7}{6}}} e^{-x} \\
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\text{Or } \int_0^{+\infty}\frac{1}{x^{\frac{7}{6}}} e^{-x}\dif x \text{ converge (l'exponentielle l'emporte) donc $I_4$ converge par équivalence.} \\
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|
(\sim e^{-kx} \text{ avec } k = 1 > 0)
|
|
\end{align*}
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|
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\subsection{Exercice 2}
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|
|
|
Étudier la convergence absolue de l'intégrale généralisée suivante.
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|
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\paragraph{$(I_5)$}
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$\int_1^{+\infty} \frac{\sin x}{x^2} \dif x$
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|
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$(I_5)$ n'est pas strictement positif sur $[1;+\infty[$.
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|
Nous allons donc étudier sa valeur absolue.
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|
\begin{align*}
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|
-1 \leq \sin{x} \leq 1 \\
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0 \leq |\sin{x}| \leq 1 \\
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0 \leq \left|\frac{\sin{x}}{x^2}\right| \leq \frac{1}{x^2} \\
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\text{Or } \int_1^{+\infty}\frac{1}{x^2}\dif x \text{ converge donc } \left|\frac{\sin{x}}{x^2}\right| \text{ converge par majoration.} \\
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\implies I_5 \text{ converge absolument.} \\
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\end{align*}
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\section{Séries de Fourier}
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\subsection{Exercice 1}
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|
|
|
\begin{enumerate}
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|
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|
\item Déterminer le développement en série de Fourier de la fonction $f$, 2-périodique, définie par~: \\
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|
$f(x) = |x| \quad \forall x \in [-1;1[$
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|
$f(-x) = |-x| = |x| = f(x)$ \quad donc $f$ est paire $\implies b_n = 0$
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|
|
\begin{tabularx}{\linewidth}{XX}
|
|
|
|
{\begin{align*}
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a_0 &= \frac{1}{2} \int_{-1}^1 |x| \dif x \\
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|
&= \int_{0}^1 x \dif x \\
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|
&= \left[\frac{x^2}{2}\right]_0^1 \\
|
|
a_0 &= \frac{1}{2}
|
|
\end{align*}} &
|
|
{\begin{align*}
|
|
a_n &= \frac{2}{T} \int_{-T/2}^{T/2} |x| \cos{\frac{2\pi nx}{T}} \dif x \\
|
|
a_n &= \frac{2}{2} \int_{-1}^1 |x| \cos{\frac{2\pi nx}{2}} \dif x \\
|
|
&= 2\int_0^1 x \cos(n\pi x) \dif x \\
|
|
\text{IPP }
|
|
&\left\{
|
|
\begin{array}{ll}
|
|
u = x & u' = 1 \\
|
|
v' = \cos(n\pi x) & v = \frac{\sin{n\pi x}}{n\pi} \\
|
|
\end{array}
|
|
\right. \\
|
|
a_n &= 2 \left(\left[x\frac{\sin{n\pi x}}{n\pi}\right]_0^1 - \int_0^1 \frac{\sin{n\pi x}}{n\pi} \dif x \right) \\
|
|
&= 2 \left(\frac{\sin{n\pi}}{n\pi} - 0 + \left[\frac{\cos{n\pi x}}{(n\pi)^2}\right]_0^1 \right) \\
|
|
&= 2 \left(0 + \frac{\cos{n\pi}}{(n\pi)^2} - \frac{\cos{0}}{(n\pi)^2}\right) \\
|
|
a_n &= 2\frac{(-1)^n - 1}{(n\pi)^2}
|
|
\end{align*}} \\
|
|
|
|
\end{tabularx}
|
|
|
|
\begin{equation*}
|
|
\implies f(x) = \frac{1}{2} + \sum_{n=1}^{+\infty} 2\frac{(-1)^n - 1}{(n\pi)^2} \cos(n\pi x)
|
|
\end{equation*}
|
|
|
|
\item En déduire que \quad $\frac{\pi^2}{8} = \sum_{n=0}^{+\infty} \frac{1}{(2n + 1)^2}$
|
|
|
|
\begin{tabularx}{\linewidth}{XX}
|
|
|
|
{pour $x = 0$~:
|
|
\begin{align*}
|
|
f(0) &= \frac{1}{2} + \sum_{n=1}^{+\infty} 2\frac{(-1)^n - 1}{(n\pi)^2} \cos{0} \\
|
|
\iff 0 &= \frac{1}{2} + \sum_{n=1}^{+\infty} 2\frac{(-1)^n - 1}{(n\pi)^2} \\
|
|
\iff -\frac{1}{2} &= 2 \sum_{n=1}^{+\infty} \frac{(-1)^n - 1}{(n\pi)^2} \\
|
|
\iff -\frac{1}{4} &= \sum_{n=1}^{+\infty} \frac{(-1)^n - 1}{(n\pi)^2} \\
|
|
\iff \frac{1}{4} &= \sum_{n=1}^{+\infty} \frac{(-1)^{n+1} + 1}{(n\pi)^2} \\
|
|
\iff \frac{\pi^2}{4} &= \sum_{n=1}^{+\infty} \frac{(-1)^{n+1} + 1}{n^2} \\
|
|
\iff \frac{\pi^2}{8} &= \sum_{n=1}^{+\infty} \frac{(-1)^{n+1} + 1}{2n^2}
|
|
\end{align*}} &
|
|
|
|
{pour $x = -1$~:
|
|
\begin{align*}
|
|
f(-1) &= \frac{1}{2} + \sum_{n=1}^{+\infty} 2\frac{(-1)^n - 1}{(n\pi)^2} \cos(-n\pi) \\
|
|
\iff 1 &= \frac{1}{2} + 2\sum_{n=1}^{+\infty} \frac{(-1)^n - 1}{(n\pi)^2} \cos(n\pi) \\
|
|
\iff \frac{1}{2} &= 2\sum_{n=1}^{+\infty} \frac{(-1)^n - 1}{(n\pi)^2} (-1)^n \\
|
|
\iff \frac{1}{4} &= \sum_{n=1}^{+\infty} \frac{(-1)^{2n} - (-1)^n}{(n\pi)^2} \\
|
|
\iff \frac{1}{4} &= \sum_{n=1}^{+\infty} \frac{1 - (-1)^n}{(n\pi)^2} \\
|
|
\iff \frac{\pi^2}{4} &= \sum_{n=1}^{+\infty} \frac{1 + (-1)^{n+1}}{n^2} \\
|
|
\iff \frac{\pi^2}{8} &= \sum_{n=1}^{+\infty} \frac{(-1)^{n+1} + 1}{2n^2} \\
|
|
\end{align*}} \\
|
|
|
|
\end{tabularx}
|
|
|
|
\item Appliquer l'identité de Parseval-Bessel.
|
|
En déduire la valeur de la somme \quad $\sum_{n=0}^{+\infty}\frac{1}{(2n + 1)^4}$
|
|
|
|
\begin{align*}
|
|
a_0^2 + \frac{1}{2}\sum_{n=1}^{+\infty}a_n^2 = \frac{1}{2}\int_{-1}^{1}f^2(x) \dif x \\
|
|
\end{align*}
|
|
|
|
\end{enumerate}
|
|
|
|
\subsection{Exercice 2}
|
|
|
|
\begin{enumerate}
|
|
|
|
\item Déterminer la série de Fourier de la fonction $f$, $2\pi$-périodique, définie par~: \\
|
|
\begin{align*}
|
|
\left\{
|
|
\begin{array}{l}
|
|
f(x) = e^x \\
|
|
f(\pi) = \cosh{\pi}
|
|
\end{array}
|
|
\right.
|
|
\quad \forall x \in \; ]{-\pi}, \pi[
|
|
\end{align*}
|
|
|
|
$f$ est ni paire, ni impaire, ni continue.
|
|
|
|
La série de Fourier est donc de la forme~:
|
|
\begin{equation*}
|
|
S_f(x) = a_0 + \sum_{n=1}^{+\infty} ( a_n \cos(nx) + b_n \sin(nx) )
|
|
\end{equation*}
|
|
|
|
\begin{align*}
|
|
a_0 &= \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) \dif x \\
|
|
&= \frac{1}{2\pi} \int_{-\pi}^{\pi} e^x \dif x \\
|
|
&= \frac{1}{2\pi} [e^x]_{-\pi}^{\pi} \\
|
|
&= \frac{1}{2\pi} (e^{\pi} - e^{-\pi}) \\
|
|
&= \frac{1}{2\pi} \left(e^{\pi} - \frac{1}{e^{\pi}}\right) \\
|
|
&= \frac{1}{2\pi} \left(\frac{e^{2\pi}}{e^{\pi}} - \frac{1}{e^{\pi}}\right) \\
|
|
&= \frac{1}{2\pi} \times \frac{e^{2\pi} - 1}{e^{\pi}} \\
|
|
&= \frac{e^{2\pi} - 1}{2\pi e^{\pi}} \\
|
|
\end{align*}
|
|
\begin{align*}
|
|
a_n &= \frac{2}{2\pi} \int_{-\pi}^{\pi} f(x) \cos{\frac{2\pi nx}{2\pi}} \dif x \\
|
|
&= \frac{1}{\pi} \int_{-\pi}^{\pi} e^x \cos(nx) \dif x \\
|
|
&\text{IPP }
|
|
\left\{
|
|
\begin{array}{ll}
|
|
u = e^x & u' = e^x \\
|
|
v' = \cos(nx) & v = \frac{\sin(nx)}{n} \\
|
|
\end{array}
|
|
\right. \\
|
|
a_n &= \frac{1}{\pi} \left(\left[e^x \frac{\sin{nx}}{n} \right]_{-\pi}^{\pi} - \int_{-\pi}^{\pi} e^x \frac{\sin(nx)}{n} \dif x \right) \\
|
|
&= \frac{1}{\pi} \left(\left[e^x \frac{\sin{nx}}{n} \right]_{-\pi}^{\pi} - \frac{1}{n} \int_{-\pi}^{\pi} e^x \sin(nx) \dif x \right) \\
|
|
&\text{IPP }
|
|
\left\{
|
|
\begin{array}{ll}
|
|
u = e^x & u' = e^x \\
|
|
v' = \sin(nx) & v = \frac{-\cos(nx)}{n} \\
|
|
\end{array}
|
|
\right. \\
|
|
a_n &= \frac{1}{\pi} \left(\left[e^x \frac{\sin{nx}}{n} \right]_{-\pi}^{\pi} - \frac{1}{n} \left(\left[-e^x\frac{\cos(nx)}{n}\right]_{-\pi}^{\pi} - \int_{-\pi}^{\pi} -e^x \frac{\cos(nx)}{n} \dif x \right)\right) \\
|
|
&= \frac{1}{\pi} \left(\left[e^x \frac{\sin{nx}}{n} \right]_{-\pi}^{\pi} - \frac{1}{n} \left(\left[-e^x\frac{\cos(nx)}{n}\right]_{-\pi}^{\pi} + \int_{-\pi}^{\pi} e^x \frac{\cos(nx)}{n} \dif x \right)\right) \\
|
|
&= \frac{1}{\pi} \left(\left[e^x \frac{\sin{nx}}{n} \right]_{-\pi}^{\pi} + \frac{1}{n} \left(\left[e^x\frac{\cos(nx)}{n}\right]_{-\pi}^{\pi} - \frac{1}{n} \int_{-\pi}^{\pi} e^x \cos(nx) \dif x \right)\right) \\
|
|
\end{align*}
|
|
\begin{align*}
|
|
b_n &=
|
|
\end{align*}
|
|
|
|
\end{enumerate}
|
|
|
|
\end{document}
|