diff --git a/exercise7/main.pdf b/exercise7/main.pdf
index be11ac8..c86342c 100644
Binary files a/exercise7/main.pdf and b/exercise7/main.pdf differ
diff --git a/exercise7/main.typ b/exercise7/main.typ
index eb22182..5ed9a5f 100644
--- a/exercise7/main.typ
+++ b/exercise7/main.typ
@@ -37,3 +37,7 @@ this document was created using
 = problem 2
 
 #include "problem2.typ"
+
+= problem 4
+
+#include "problem4.typ"
diff --git a/exercise7/problem4.typ b/exercise7/problem4.typ
new file mode 100644
index 0000000..eae10d1
--- /dev/null
+++ b/exercise7/problem4.typ
@@ -0,0 +1,70 @@
+#import "@preview/physica:0.9.6": *
+
+== a)
+
+given the explicit midpoint method
+$
+  y_(n+1) = y_n + h f(t_n + 0.5h, y_n + 0.5h f(t_n, y_n)),
+$
+notice $k_1 = f(t_n, y_n)$ appears at depth 1 meaning the funcion is only nested
+twice, such that our number of stages $s = 2$. furthermore, we have $c_2 = 0.5$
+from the time step and $a_21 = 0.5$ from the function step. we can also observe
+that $b_1 = 1$ and $b_2 = 0$. this can be visualized in a mnemonic butcher
+tableau.
+
+#let butcher(s, nums) = {
+  let nums = nums.map(x => $#x$)
+  table(
+    stroke: (x, y) => if x == 0 and y == s {
+      (right: 0.7pt + black, top: 0.7pt + black)
+    } else if x == 0 {
+      (right: 0.7pt + black)
+    } else if y == s {
+      (top: 0.7pt + black)
+    },
+    align: (x, y) => (
+      if x > 0 { center } else { left }
+    ),
+    columns: s + 1,
+    $0$, ..range(s).map(x => none), // first row
+    ..range(2, s + 1)
+      .map(i => {
+        let p(i) = calc.floor(i * i / 2 + i / 2 - 1)
+        (
+          nums.slice(p(i - 1), p(i - 1) + i),
+          range(i, s + 1).map(x => none),
+        ).flatten()
+      })
+      .flatten(),
+    none, ..nums.rev().slice(0, s).rev() // last row
+  )
+}
+
+#align(center)[#butcher(2, (0.5, 0.5, 1, 0))]
+
+#pagebreak()
+
+== b)
+
+I don't wanna write python right now, so I'm gonna solve this in typst. yes, I'm
+coding this in the typesetting language I use to write this text.
+
+#let code = ```typc
+  let h = 0.01;
+  let f(t, y) = calc.exp(-y * y);
+  let y_1(t_0, y_0) = y_0
+      + h * f(t_0 + 0.5 * h,
+              y_0 + 0.5 * h * f(t_0, y_0));
+  let t = 0;
+  let y = 1;
+  while t < 1 {
+    y = y_1(t, y);
+    t += h;
+  };
+  [the value for $y$ at $t = 1$ is
+  #calc.round(y, digits: 4)]
+```
+
+#eval(code.text.split("\n").join())
+
+#code