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Finish E5, equa diff 2nd ordre

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flyingscorpio@arch-desktop 2021-09-20 09:57:35 +02:00
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analyse/exercices/main.tex
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@ -206,6 +206,79 @@
\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{$(E_6)$}
$y'' - y = x^3$