\documentclass[a4paper,french,12pt]{article}

\title{Analyse --- Exercices}
\author{}
\date{Dernière compilation~: \today{} à \currenttime}

\usepackage{../../cours}
\usepackage{enumitem}
\usepackage{xfrac}

\begin{document}

\maketitle

\section{Équations différentielles d'ordre 1}

    \paragraph{$(E_1)$}
        $y' - 2y = x^2$

        \begin{enumerate}[label=\alph*)]

            \item Solution homogène
                \begin{align*}
                    r = 2 \implies y_0 = \lambda e^{2x}
                \end{align*}

            \item Solution particulière
                \begin{align*}
                    \left\{
                    \begin{array}{l}
                        y_1 = ax^2 + bx + c \\
                        y_1' = 2ax + b \\
                    \end{array}
                    \right.
                \end{align*}
                \begin{align*}
                    \text{Dans } (E_1)
                    &\implies
                    2ax + b - 2(ax^2 + bx + c) = x^2 \\
                    &\iff
                    \left\{
                    \begin{array}{l}
                        -2a = 1 \\
                        2a - 2b = 0 \\
                        b - 2c = 0 \\
                    \end{array}
                    \right.
                    \iff
                    \left\{
                    \begin{array}{l}
                        a = \sfrac{-1}{2} \\
                        b = \sfrac{-1}{2} \\
                        c = \sfrac{-1}{4} \\
                    \end{array}
                    \right.
                \end{align*}
                \begin{align*}
                    \implies y_1 = -\frac{1}{2}x^2 -\frac{1}{2}x -\frac{1}{4}
                \end{align*}

            \item Solution générale
                \begin{equation*}
                    y = y_0 + y_1 = \boxed{-\frac{1}{2}x^2 -\frac{1}{2}x -\frac{1}{4} + \lambda e^{2x}}
                \end{equation*}

        \end{enumerate}

    \paragraph{$(E_2)$}
        $3y' - 9y = 7e^{3x}$

        \begin{enumerate}[label=\alph*)]

            \item Solution homogène
                \begin{align*}
                    r = 3 \implies y_0 = \lambda e^{3x}
                \end{align*}

            \item Solution particulière
                \begin{align*}
                    \left\{
                    \begin{array}{l}
                        y_1 = ax e^{3x} \\
                        y_1' = ae^{3x} + 3axe^{3x} = (3ax + a)e^{3x} \\
                    \end{array}
                    \right.
                \end{align*}
                \begin{align*}
                    \text{Dans } (E_2)
                    &\implies 3(3ax + a)e^{3x} - 9axe^{3x} = 7e^{3x} \\
                    &\iff (9ax + 3a)e^{3x} - 9axe^{3x} = 7e^{3x} \\
                    &\iff 9ax + 3a - 9ax = 7 \\
                    &\iff 3a = 7 \\
                    &\iff a = \frac{7}{3}
                \end{align*}
                \begin{align*}
                    \implies y_1 = \frac{7}{3}xe^{3x}
                \end{align*}

            \item Solution générale
                \begin{equation*}
                    y = y_0 + y_1 = \lambda e^{3x} + \frac{7}{3}xe^{3x} = \boxed{(\frac{7}{3}x + \lambda)e^{3x}}
                \end{equation*}

        \end{enumerate}

    \paragraph{$(E_3)$}
        $2y' - 4y = 5e^{3x}$

        \begin{enumerate}[label=\alph*)]

            \item Solution homogène
                \begin{align*}
                    r = 2 \implies y_0 = \lambda e^{2x}
                \end{align*}

            \item Solution particulière
                \begin{align*}
                    \left\{
                    \begin{array}{l}
                        y_1 = ae^{3x} \\
                        y_1' = 3ae^{3x} \\
                    \end{array}
                    \right.
                \end{align*}
                \begin{align*}
                    \text{Dans } (E_3)
                    &\iff 6ae^{3x} - 4ae^{3x} = 5e^{3x} \\
                    &\iff 2a = 5
                    \iff a = \frac{5}{2}
                \end{align*}
                \begin{align*}
                    \implies y_1 = \frac{5}{2}e^{3x}
                \end{align*}

            \item Solution générale
                \begin{equation*}
                    y = y_0 + y_1 = \boxed{\lambda e^{2x} + \frac{5}{2}e^{3x}}
                \end{equation*}

        \end{enumerate}

    \paragraph{$(E_4)$}
        $y' + 4y = 3\cos(2x)$

        \begin{enumerate}[label=\alph*)]

            \item Solution homogène
                \begin{align*}
                    r = -4 \implies y_0 = \lambda e^{-4x}
                \end{align*}

            \item Solution particulière
                \begin{align*}
                    \left\{
                    \begin{array}{l}
                        y_1 = a\cos(2x) + b\sin(2x) \\
                        y_1' = -2a\sin(2x) + 2b\cos(2x) \\
                    \end{array}
                    \right.
                \end{align*}
                \begin{align*}
                    \text{Dans } (E_4)
                    &\implies -2a\sin(2x) + 2b\cos(2x) + 4(a\cos(2x) + b\sin(2x)) = 3\cos(2x) \\
                    &\iff
                    \left\{
                    \begin{array}{l}
                        4a + 2b = 3 \\
                        4b - 2a = 0 \\
                    \end{array}
                    \right.
                    \iff
                    \left\{
                    \begin{array}{l}
                        4a + 2b = 3 \\
                        4a - 8b = 0 \\
                    \end{array}
                    \right.
                    \iff
                    \left\{
                    \begin{array}{l}
                        10b = 3 \\
                        4b - 2a = 0 \\
                    \end{array}
                    \right. \\
                    &\iff
                    \left\{
                    \begin{array}{l}
                        b = \sfrac{3}{10} \\
                        a = \sfrac{12}{20} = \sfrac{3}{5} \\
                    \end{array}
                    \right.
                \end{align*}
                \begin{align*}
                    \implies y_1 = \frac{3}{5}\cos(2x) + \frac{3}{10}\sin(2x)
                \end{align*}

            \item Solution générale
                \begin{equation*}
                    y = y_0 + y_1 = \boxed{\lambda e^{-4x} + \frac{3}{5}\cos(2x) + \frac{3}{10}\sin(2x)}
                \end{equation*}

        \end{enumerate}

\section{Équations différentielles d'ordre 2}

    \paragraph{$(E_5)$}
        $y'' - 7y' + 10 y = (x + 3) e^{2x}$

        \begin{enumerate}[label=\alph*)]

            \item Solution homogène
                \begin{align*}
                    r^2 - 7r + 10 = 0
                    \implies \Delta = 9
                    \implies
                    \left\{
                    \begin{array}{l}
                        r_1 = \frac{7 - 3}{2} = 2 \\\\
                        r_2 = \frac{7 + 3}{2} = 5 \\
                    \end{array}
                    \right.
                \end{align*}
                \begin{align*}
                    \implies
                    y_0 = \lambda e^{2x} + \mu e^{5x}
                \end{align*}

            \item Solution particulière

                second membre~: $(x + 3)e^{\alpha x}$ avec $\alpha = 2$ $\implies \alpha$ racine de l'équation caractéristique.
                \begin{align*}
                    &\left\{
                    \begin{array}{l}
                        y_1 = xe^{2x} (ax + b) \\
                        y_1' = (2x + 1)e^{2x}(ax + b) + axe^{2x} \\
                        y_1'' = (4ax + 2a + 2b)e^{2x} + (4ax^2 + 4ax + 4bx + 2b)e^{2x} \\
                    \end{array}
                    \right. \\
                    \implies
                    &\left\{
                    \begin{array}{l}
                        y_1 = xe^{2x} (ax + b) \\
                        y_1' = (2ax^2 + 2ax + 2bx + b)e^{2x} \\
                        y_1'' = (4ax^2 + 8ax + 4bx + 2a + 4b)e^{2x} \\
                    \end{array}
                    \right.
                \end{align*}

                Dans $(E_5)$~:

                $(4ax^2 + 8ax + 4bx + 2a + 4b)e^{2x} - 7(2ax^2 + 2ax + 2bx + b)e^{2x} + 10(ax + b)xe^{2x} = (x + 3) e^{2x}$
                \begin{align*}
                    \implies
                    -6ax + 2a - 3b = x + 3
                    \implies
                    \left\{
                    \begin{array}{l}
                        -6a = 1 \\
                        2a - 3b = 3 \\
                    \end{array}
                    \right.
                    \implies
                    \left\{
                    \begin{array}{l}
                        a = \frac{-1}{6} \\\\
                        b = \frac{-10}{9} \\
                    \end{array}
                    \right.
                \end{align*}
                \begin{align*}
                    \implies
                    y_1 = xe^{2x}(-\frac{1}{6}x - \frac{10}{9})
                \end{align*}

            \item Solution générale
                \begin{equation*}
                    y = y_0 + y_1 = \boxed{\lambda e^{2x} + \mu e^{5x} + xe^{2x}(-\frac{1}{6}x - \frac{10}{9})}
                \end{equation*}

        \end{enumerate}

    \paragraph{$(E_6)$}
        $y'' - y = x^3$

        \begin{enumerate}[label=\alph*)]

            \item Solution homogène
                \begin{align*}
                    r^2 - 1 = 0
                    \implies \Delta = 4
                    \implies
                    \left\{
                    \begin{array}{l}
                        r_1 = \frac{0 - 2}{2} = -1 \\\\
                        r_2 = \frac{0 + 2}{2} = 1 \\
                    \end{array}
                    \right.
                \end{align*}
                \begin{align*}
                    \implies
                    y_0 = \lambda e^{-x} + \mu e^{x}
                \end{align*}

            \item Solution particulière

                second membre~: $x^3e^{\alpha x}$ avec $\alpha = 0$ $\implies \alpha$ non racine de l'équation caractéristique.
                \begin{align*}
                    &\left\{
                    \begin{array}{l}
                        y_1 = ax^3 + bx^2 + cx + d \\
                        y_1' = 3ax^2 + 2bx + c \\
                        y_1'' = 6ax + 2b \\
                    \end{array}
                    \right.
                \end{align*}

                Dans $(E_6)$~:

                $6ax + 2b - ax^3 - bx^2 - cx - d = x^3$
                \begin{align*}
                    \implies
                    -ax^3 - bx^2 + 6ax - cx + 2b - d = x^3
                    \implies
                    \left\{
                    \begin{array}{l}
                        -a = 1 \\
                        -b = 0 \\
                        6a - c = 0 \\
                        2b - d = 0 \\
                    \end{array}
                    \right.
                    \implies
                    \left\{
                    \begin{array}{l}
                        a = -1 \\
                        b = 0 \\
                        c = -6 \\
                        d = 0 \\
                    \end{array}
                    \right.
                \end{align*}
                \begin{align*}
                    \implies
                    y_1 = -x^3 - 6x
                \end{align*}

            \item Solution générale
                \begin{equation*}
                    y = y_0 + y_1 = \boxed{\lambda e^{-x} + \mu e^{x} - x^3 - 6x}
                \end{equation*}

        \end{enumerate}

    \paragraph{$(E_7)$}
        $y'' + y = \cos{x}$

        \begin{enumerate}[label=\alph*)]

            \item Solution homogène
                \begin{align*}
                    r^2 + 1 = 0
                    \implies \Delta = -4
                    \implies
                    \left\{
                    \begin{array}{l}
                        r_1 = \frac{0 - 2i}{2} \\\\
                        r_2 = \frac{0 + 2i}{2} \\
                    \end{array}
                    \right.
                    \implies
                    \left\{
                    \begin{array}{l}
                        \alpha = 0 \\
                        \beta = 1
                    \end{array}
                    \right.
                \end{align*}
                \begin{align*}
                    \implies
                    y_0 = e^{0}(\lambda \cos{x} + \mu \sin{x})
                    = \lambda \cos{x} + \mu \sin{x}
                \end{align*}

            \item Solution particulière

                second membre~: $e^{\alpha x}\cos{x}$ avec $\alpha = 0$ $\implies \alpha$ non racine de l'équation caractéristique.
                \begin{align*}
                    y_1 &= xe^{\alpha x}(a\cos{\beta x} + b\sin{\beta x}) \\
                        &= x(a\cos{x} + b\sin{x}) \\
                    \implies
                    &\left\{
                    \begin{array}{l}
                        y_1 = x(a\cos{x} + b\sin{x}) \\
                        y_1' = a\cos{x} + b\sin{x} - x(a\sin{x} - b\cos{x}) \\
                        y_1'' = -a\sin{x} + b\cos{x} - a\sin{x} + b\cos{x} - x(a\cos{x} + b\sin{x}) \\
                    \end{array}
                    \right. \\
                    \implies
                    &\left\{
                    \begin{array}{l}
                        y_1 = x(a\cos{x} + b\sin{x}) \\
                        y_1' = a\cos{x} + b\sin{x} - x(a\sin{x} - b\cos{x}) \\
                        y_1'' = -2a\sin{x} + 2b\cos{x} - x(a\cos{x} + b\sin{x}) \\
                    \end{array}
                    \right.
                \end{align*}

                Dans $(E_7)$~:
                \begin{align*}
                    -2a\sin{x} + 2b\cos{x} - x(a\cos{x} + b\sin{x}) + x(a\cos{x} + b\sin{x}) = \cos{x} \\
                    \iff
                    -2a\sin{x} + 2b\cos{x} = \cos{x} \\
                    \iff
                    2b\cos{x} - 2a\sin{x} = \cos{x} \\
                    \implies
                    \left\{
                    \begin{array}{l}
                        2b = 1 \\
                        -a = 0 \\
                    \end{array}
                    \right.
                    \implies
                    \left\{
                    \begin{array}{l}
                        a = 0 \\
                        b = \frac{1}{2} \\
                    \end{array}
                    \right.
                \end{align*}
                \begin{align*}
                    \implies
                    y_1 = \frac{x\sin{x}}{2} \\
                \end{align*}

            \item Solution générale
                \begin{equation*}
                    y = y_0 + y_1 = \boxed{\lambda \cos{x} + \mu \sin{x} + \frac{x\sin{x}}{2}}
                \end{equation*}

        \end{enumerate}

    \paragraph{$(E_8)$}
        $y'' - 4y = (-4x + 3) e^{2x}$

        \begin{enumerate}[label=\alph*)]

            \item Solution homogène
                \begin{align*}
                    r^2 - 4 = 0
                    \implies \Delta = 16
                    \implies
                    \left\{
                    \begin{array}{l}
                        r_1 = \frac{0 - 4}{2} = -2 \\\\
                        r_2 = \frac{0 + 4}{2} = 2 \\
                    \end{array}
                    \right.
                \end{align*}
                \begin{align*}
                    \implies
                    y_0 = \lambda e^{-2x} + \mu e^{2x}
                \end{align*}

            \item Solution particulière

                second membre~: $(-4x + 3)e^{\alpha x}$ avec $\alpha = 2$ $\implies \alpha$ racine de l'équation caractéristique.
                \begin{align*}
                    &\left\{
                    \begin{array}{l}
                        y_1 = xe^{2x} (ax + b) \\
                        y_1' = (2x + 1)e^{2x}(ax + b) + axe^{2x} \\
                        y_1'' = (4ax + 2a + 2b)e^{2x} + (4ax^2 + 4ax + 4bx + 2b)e^{2x} \\
                    \end{array}
                    \right. \\
                    \implies
                    &\left\{
                    \begin{array}{l}
                        y_1 = xe^{2x} (ax + b) \\
                        y_1' = (2ax^2 + 2ax + 2bx + b)e^{2x} \\
                        y_1'' = (4ax^2 + 8ax + 4bx + 2a + 4b)e^{2x} \\
                    \end{array}
                    \right.
                \end{align*}

                Dans $(E_8)$~:

                $(4ax^2 + 8ax + 4bx + 2a + 4b)e^{2x} - 4x(ax + b)e^{2x} = (-4x + 3) e^{2x}$

                \begin{align*}
                    \implies
                    8ax + 2a + 4b = -4x + 3
                    \implies
                    \left\{
                    \begin{array}{l}
                        8a = -4 \\
                        2a + 4b = 3 \\
                    \end{array}
                    \right.
                    \implies
                    \left\{
                    \begin{array}{l}
                        a = \frac{-1}{2} \\
                        b = 1 \\
                    \end{array}
                    \right.
                \end{align*}
                \begin{align*}
                    \implies
                    y_1 = xe^{2x} (\frac{-x}{2} + 1)
                \end{align*}

            \item Solution générale
                \begin{equation*}
                    y = y_0 + y_1 = \boxed{\lambda e^{-2x} + \mu e^{2x} + xe^{2x} (\frac{-x}{2} + 1)}
                \end{equation*}

        \end{enumerate}

\end{document}