diff --git a/latex/problems/problem3.tex b/latex/problems/problem3.tex
index 03fc02f..2d7ab8b 100644
--- a/latex/problems/problem3.tex
+++ b/latex/problems/problem3.tex
@@ -1,4 +1,37 @@
 
 \section*{Problem 3}
 
-% Show how it's derived and where we found the derivation.
+To derive the discretized version of the Poisson equation, we first need
+the taylor expansion for $u(x)$ around $x + h$ 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)
+\end{align*}
+
+\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)
+\end{align*}
+
+If we add the equations above, we get this new equation:
+
+\begin{align*}
+    u(x+h) + u(x-h) &= 2 u(x) + u''(x) h^2 + \mathcal{O}(h^4) \\
+    u(x+h) - 2 u(x) + u(x-h) + \mathcal{O}(h^4) &= u''(x) h^2 \\
+    u''(x) &= \frac{u(x+h) - 2 u(x) + u(x-h)}{h^2} + \mathcal{O}(h^2) \\
+    u_i''(x) &= \frac{u_{i+1} - 2 u_i + u_{i-1}}{h^2} + \mathcal{O}(h^2) \\
+\end{align*}
+
+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{ - u_{i+1} + 2 u_i - u_{i-1}}{h^2} + \mathcal{O}(h^2) &= 100 e^{-10x} \\
+\end{align*}
+
+And lastly, we leave out $\mathcal{O}(h^2)$ and change $u_i$ to $v_i$ to 
+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} \\
+\end{align*}