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--- a/latex/assignment_1.tex
+++ b/latex/assignment_1.tex
@@ -74,7 +74,7 @@
 
 \begin{document}
 
-\title{Title of the document}      % self-explanatory
+\title{Project 1}      % self-explanatory
 \author{Cory Balaton \& Janita}          % self-explanatory
 \date{\today}                             % self-explanatory
 \noaffiliation                            % ignore this, but keep it.
@@ -87,6 +87,38 @@
 \section*{Problem 1}
 
 % Do the double integral
+\begin{align*}
+    u(x) &= \int \int \frac{d^2 u}{dx^2} dx^2\\
+         &= \int \int -100 e^{-10x} dx^2 \\
+         &= \int \frac{-100 e^{-10x}}{-10} + c_1 dx \\
+         &= \int 10 e^{-10x} + c_1 dx \\
+         &= \frac{10 e^{-10x}}{-10} + c_1 x + c_2 \\
+         &= -e^{-10x} + c_1 x + c_2
+\end{align*}
+
+Using the boundary conditions, we can find $c_1$ and $c_2$ as shown below:
+
+\begin{align*}
+    u(0) &= 0 \\
+    -e^{-10 \cdot 0} + c_1 \cdot 0 + c_2 &= 0 \\
+    -1 + c_2 &= 0 \\
+    c_2 &= 1
+\end{align*}
+
+\begin{align*}
+    u(1) &= 0 \\
+    -e^{-10 \cdot 1} + c_1 \cdot 1 + c_2 &= 0 \\
+    -e^{-10} + c_1 + c_2 &= 0 \\
+    c_1 &= e^{-10} - c_2\\
+    c_1 &= e^{-10} - 1\\
+\end{align*}
+
+Using the values that we found for $c_1$ and $c_2$, we get
+
+\begin{align*}
+    u(x) &= -e^{-10x} + (e^{-10} - 1) x + 1 \\
+         &= 1 - (1 - e^{-10}) - e^{-10x}
+\end{align*}
 
 \section*{Problem 2}
 
@@ -106,9 +138,9 @@
 
 \subsection*{b)}
 
-\section{Problem 6}
+\section*{Problem 6}
 
-\subsection{a)}
+\subsection*{a)}
 
 % Use Gaussian elimination, and then use backwards substitution to solve the equation