Finish E5, equa diff 2nd ordre

analyse/exercices/main.tex

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