519 lines
17 KiB
TeX
519 lines
17 KiB
TeX
\documentclass[a4paper,french,12pt]{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~: $e^{\alpha x}\cos{x}$ avec $\alpha = 0$ $\implies \alpha$ non 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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\end{document}
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