218 lines
6.8 KiB
TeX
218 lines
6.8 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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y_1' - 2 y_1 = x^2
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&\Leftrightarrow
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2ax + b - 2(ax^2 + bx + c) = x^2 \\
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&\Leftrightarrow
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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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\Leftrightarrow
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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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3y_1' - 9y_1 = 7e^{3x}
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&\Leftrightarrow 3(3ax + a)e^{3x} - 9axe^{3x} = 7e^{3x} \\
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&\Leftrightarrow (9ax + 3a)e^{3x} - 9axe^{3x} = 7e^{3x} \\
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&\Leftrightarrow 9ax + 3a - 9ax = 7 \\
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&\Leftrightarrow 3a = 7 \\
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&\Leftrightarrow 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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2y_1' - 4y_1 = 5e^{3x}
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&\Leftrightarrow 6ae^{3x} - 4ae^{3x} = 5e^{3x} \\
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&\Leftrightarrow 2a = 5 \\
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&\Leftrightarrow 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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y_1' + 4y_1 = 3\cos(2x)
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&\Leftrightarrow -2a\sin(2x) + 2b\cos(2x) + 4(a\cos(2x) + b\sin(2x)) = 3\cos(2x) \\
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&\Leftrightarrow
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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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\Leftrightarrow
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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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\Leftrightarrow
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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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&\Leftrightarrow
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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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\paragraph{$(E_6)$}
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$y'' - y = x^3$
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\paragraph{$(E_7)$}
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$y'' + y = \cos{x}$
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\paragraph{$(E_8)$}
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$y'' - 4y = (-4x + 3) e^{2x}$
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\end{document}
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