diff --git a/latex/contents.tex b/latex/contents.tex
index d149977..1a37de7 100644
--- a/latex/contents.tex
+++ b/latex/contents.tex
@@ -35,15 +35,81 @@ Blabla
 \end{equation*}
 
 \begin{equation}
-	\Rightarrow\left\{\begin{aligned}
+	\Rightarrow\boxed{\left\{\begin{aligned}
 		\dot{x_{1}} &= \frac{\partial^2 e_2}{\partial\zeta^2} \\
 		\dot{x_{2}} &= -\frac{\partial^2 e_1}{\partial\zeta^2}-q(\zeta,t)
-	\end{aligned}\right.
+	\end{aligned}\right. }
 \end{equation}
 
 
 \subsection{Exercice 2}
 
+\begin{align*}
+		\left\{\begin{aligned}
+			x_1 &\approx \phi^Tx_{1d}(t) \\
+			x_2 &\approx \phi^Tx_{2d}(t)
+		\end{aligned}\right. & &
+		\left\{\begin{aligned}
+			e_1 &\approx \phi^Te_{1d}(t) \\
+			e_2 &\approx \phi^Te_{2d}(t)
+		\end{aligned}\right.
+\end{align*}
+
+En utilisant la première ligne de l'équation 1, on trouve :
+
+\begin{equation*}
+	\int\phi(\zeta)d\zeta\times\phi^T\dot{x_{1d}} = \int\phi(\zeta)d\zeta\times\frac{\partial^2}{\partial\zeta^2} = \ddot{\phi}(\zeta)^Te_{2*d}
+\end{equation*}
+
+\begin{equation}
+	\underbrace{\int_{0}^{L}\phi(\zeta)\phi(\zeta)^Td\zeta}_\text{E}\times\dot{x_{1d}} = \left(\int_{0}^{L}\phi(\zeta)\ddot{\phi}(\zeta)d\zeta\right)e_{2d}
+\end{equation}
+
+Intégration par partie (IPP) :
+
+\begin{equation*}
+	\int_{0}^{L}u\cdot \dot{v} d\zeta = [u\cdot v]_{0}^{L} - \int_{0}^{L}\dot{u}\cdot v d\zeta
+\end{equation*}
+
+Dans notre cas :
+
+\begin{align*}
+	\left\{\begin{aligned}
+		u&=\phi(\zeta) \\
+		\dot{v} &= \ddot{\phi}(\zeta)^T
+	\end{aligned}\right. &
+	\Rightarrow\left\{\begin{aligned}
+		\dot{u} &= \dot{\phi}(\zeta) \\
+		v &= \dot{\phi}(\zeta)^T
+	\end{aligned}\right.
+\end{align*}
+
+On applique plusieurs fois :
+
+\begin{equation*}
+	\begin{aligned} 
+		\int_{0}^{L}\phi(\zeta)\ddot{\phi}(\zeta)^T d\zeta &= [\phi(\zeta)\dot{\phi}(\zeta)^T]_{0}^{L} - \int_{0}^{L}\dot{\phi}(\zeta)\dot{\phi}(\zeta)^T d\zeta \\
+		&= \phi(L)\dot{\phi}(L)^T - \underbrace{\phi(0)\dot{\phi}(0)^T }_\text{=0} - \int_{0}^{L}\dot{\phi}(\zeta)\dot{\phi}(\zeta)^T d\zeta \\
+		&= \phi(L)\dot{\phi}(L)^T - \dot{\phi}(L)\phi(L)^T + \underbrace{\dot{\phi(0)}\phi(0)^T }_\text{=0} + \int_{0}^{L} \ddot{\phi}(\zeta)\phi(\zeta)^T d\zeta \\
+		&= \phi(L)\dot{\phi}(L)^T - \dot{\phi}(L)\phi(L)^T + \int_{0}^{L} \ddot{\phi}(\zeta)\phi(\zeta)^T d\zeta = D
+	\end{aligned}
+\end{equation*}
+
+On continue avec ça en remplaçant dans l'équation 2 :
+
+\begin{equation*}
+	\Rightarrow \boxed{E\dot{x}_{1d} = De_{2d}}
+\end{equation*}
+
+Puis on a dans la deuxième ligne de l'équation 1 : 
+
+\begin{equation*}
+	\begin{aligned}
+			\phi^T(\zeta)\dot{x}_{2d} &= - \frac{\partial^2}{\partial\zeta^2} (\phi(\zeta)^T e_{1d}) - q(\zeta, t) \\
+			\int_{0}^{L} \phi^T(\zeta)d\zeta\times\dot{x}_{2d} &= -e_{1d}\int_{0}^{L}\phi(\zeta)\ddot{\phi}(\zeta)^T d\zeta - \underbrace{\int_{0}^{L}\phi(\zeta)q(\zeta,t)d\zeta}_{=F_{ext}}
+	\end{aligned}
+\end{equation*}
+
 
 \subsection{Exercice 3}
 
diff --git a/latex/main.pdf b/latex/main.pdf
index dc3d030..8498ee6 100644
Binary files a/latex/main.pdf and b/latex/main.pdf differ
diff --git a/latex/main.tex b/latex/main.tex
index 5c19780..040d853 100644
--- a/latex/main.tex
+++ b/latex/main.tex
@@ -3,7 +3,7 @@
 
 \renewcommand{\familydefault}{\sfdefault} % Si vous voulez passer en Arial le texte
 
-\newcommand{\titre}{Mini-Projet Automatique}
+\newcommand{\titre}{BE Commande Numérique}
 \newcommand{\imagecouverture}{example-image}
 \newcommand{\firstcouverture}{
     \parbox{\textwidth}{
diff --git a/latex/template/table_des_matieres.tex b/latex/template/table_des_matieres.tex
index 1ee15fb..a5d101e 100644
--- a/latex/template/table_des_matieres.tex
+++ b/latex/template/table_des_matieres.tex
@@ -4,4 +4,4 @@
 \newpage
 
 \pagenumbering{arabic} % Numérotation en chiffres romains (i, ii, iii, ...)
-\setcounter{page}{1}
\ No newline at end of file
+\setcounter{page}{1}1
\ No newline at end of file