Edit problem 1 and add problem 3

latex/problems/problem1.tex

@@ -1,8 +1,17 @@
 \section*{Problem 1}
 
+First, we rearrange the equation.
+
+\begin{align*}
+    - \frac{d^2u}{dx^2} &= 100 e^{-10x} \\
+    \frac{d^2u}{dx^2} &= -100 e^{-10x} \\
+\end{align*}
+
+Now we find $u(x)$.
+
 % Do the double integral
 \begin{align*}
-    u(x) &= \int \int \frac{d^2 u}{dx^2} dx^2\\
+    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 \\
@@ -10,7 +19,7 @@
          &= -e^{-10x} + c_1 x + c_2
 \end{align*}
 
-Using the boundary conditions, we can find $c_1$ and $c_2$ as shown below:
+Using the boundary conditions, we can find $c_1$ and $c_2$
 
 \begin{align*}
     u(0) &= 0 \\

latex/problems/problem3.tex

@@ -2,7 +2,7 @@
 \section*{Problem 3}
 
 To derive the discretized version of the Poisson equation, we first need
-the taylor expansion for $u(x)$ around $x + h$ and $x - h$.
+the Taylor expansion for $u(x)$ around $x$ for $x + h$ and $x - h$.
 
 \begin{align*}
     u(x+h) &= u(x) + u'(x) h + \frac{1}{2} u''(x) h^2 + \frac{1}{6} u'''(x) h^3 + \mathcal{O}(h^4)
@@ -24,8 +24,8 @@ If we add the equations above, we get this new equation:
 We can then replace $\frac{d^2u}{dx^2}$ with the RHS (right-hand side) of the equation:
 
 \begin{align*}
-    - \frac{d^2u}{dx^2} &= 100 e^{-10x} \\
+    - \frac{d^2u}{dx^2} &= f(x) \\
-    \frac{ - u_{i+1} + 2 u_i - u_{i-1}}{h^2} + \mathcal{O}(h^2) &= 100 e^{-10x} \\
+    \frac{ - u_{i+1} + 2 u_i - u_{i-1}}{h^2} + \mathcal{O}(h^2) &= f_i \\
 \end{align*}
 
 And lastly, we leave out $\mathcal{O}(h^2)$ and change $u_i$ to $v_i$ to 
@@ -33,5 +33,5 @@ differentiate between the exact solution and the approximate solution,
 and get the discretized version of the equation:
 
 \begin{align*}
-align*    \frac{ - u_{i+1} + 2 u_i - u_{i-1}}{h^2} &= 100 e^{-10x} \\
+    \frac{ - v_{i+1} + 2 v_i - v_{i-1}}{h^2} &= 100 e^{-10x_i} \\
 \end{align*}