diff --git a/latex/assignment_1.pdf b/latex/assignment_1.pdf
new file mode 100644
index 0000000..1f17a59
Binary files /dev/null and b/latex/assignment_1.pdf differ
diff --git a/latex/assignment_1.tex b/latex/assignment_1.tex
index b4f8fa9..37e4f28 100644
--- a/latex/assignment_1.tex
+++ b/latex/assignment_1.tex
@@ -80,7 +80,7 @@
 \begin{document}
 
 \title{Project 1}      % self-explanatory
-\author{Cory Balaton \& Janita Willumsen}          % self-explanatory
+\author{Cory Alexander Balaton \& Janita Ovidie Sandtrøen Willumsen}          % self-explanatory
 \date{\today}                             % self-explanatory
 \noaffiliation                            % ignore this, but keep it.
 
@@ -107,4 +107,6 @@
 
 \input{problems/problem9}
 
+\input{problems/problem10}
+
 \end{document}
diff --git a/latex/problems/problem1.tex b/latex/problems/problem1.tex
index 8f53c11..cae7156 100644
--- a/latex/problems/problem1.tex
+++ b/latex/problems/problem1.tex
@@ -40,5 +40,5 @@ 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}
+         &= 1 - (1 - e^{-10})x - e^{-10x}
 \end{align*}
diff --git a/latex/problems/problem5.tex b/latex/problems/problem5.tex
index 5d32e6f..abc9654 100644
--- a/latex/problems/problem5.tex
+++ b/latex/problems/problem5.tex
@@ -1,6 +1,6 @@
 
 \section*{Problem 5}
 
-\subsection*{a)}
+\subsection*{a \& b)}
 
-\subsection*{b)}
+$n = m - 2$ since when solving for $\vec{v}$, we are finding the solutions for all the points that are in between the boundaries and not the boundaries themselves. $\vec{v}^*$ on the other hand includes the boundary points.
diff --git a/latex/problems/problem6.tex b/latex/problems/problem6.tex
index 36efa4f..18c9b60 100644
--- a/latex/problems/problem6.tex
+++ b/latex/problems/problem6.tex
@@ -44,4 +44,4 @@ Following Thomas algorithm for gaussian elimination, we first perform a forward
 Counting the number of FLOPs for the general algorithm by looking at one procedure at a time. 
 For every iteration of i in forward sweep we have 1 division, 2 multiplications, and 2 subtractions, resulting in $5(n-1)$ FLOPs. 
 For backward sweep we have 1 division, and for every iteration of i we have 1 subtraction, 1 multiplication, and 1 division, resulting in $3(n-1)+1$ FLOPs. 
-Total FLOPs for the general algorithm is $8(n-1)+1$.
\ No newline at end of file
+Total FLOPs for the general algorithm is $8(n-1)+1$.
diff --git a/src/problem2.cpp b/src/problem2.cpp
deleted file mode 100644
index e69de29..0000000