diff --git a/documents/by-course/math-4b/course-notes/main.typ b/documents/by-course/math-4b/course-notes/main.typ
index f42ac28..96b368e 100644
--- a/documents/by-course/math-4b/course-notes/main.typ
+++ b/documents/by-course/math-4b/course-notes/main.typ
@@ -67,6 +67,15 @@ Attendance to discussion sections is mandatory.
   - $y' = y^2$
 ]
 
+#definition[
+  In general, a differential equation is called linear if and only if it can be
+  written in the form
+  $
+    a_n (t) (dif^n y) / (dif t^n) + a_(n-1) (t) (d^(n-1) y) / (d t^(n-1)) + dots + a_1 (t) (dif y) / (dif t) + a_0 (t) y = g(t)
+  $
+  where $a_k (t)$ and $g(t)$ are single variable functions of $t$.
+]
+
 #definition[
   *Equilibrium solutions* for the ODE
   $ y' = F(x,y) $
@@ -87,3 +96,71 @@ Attendance to discussion sections is mandatory.
   What are the equilibria of the equation
   $ y' = y(y - x) $
 ]
+
+== General solution of a first order linear ODE
+
+We start with the differential equation in standard form
+
+$ (dif y) / (dif t) + p(t) y = g(t) $
+
+where $p(t)$ and $g(t)$ are continuous single variable functions of $t$.
+
+Then let us assume the existence of an *integrating factor* $mu(t)$, such that
+
+$ mu(t) p(t) = mu'(t) $
+
+and then multiplying each term by $mu(t)$ to obtain
+
+$ mu(t) (dif y) / (dif t) + mu(t) p(t) y = mu(t) g(t) $
+
+Then
+
+$ mu(t) (dif y) / (dif t) + mu'(t) y = mu(t) g(t) $
+
+Then recognize that the left side of the equation is the product rule to obtain
+
+$
+  (mu(t) y(t))' &= mu(t) g(t) \
+  integral (mu(t) y(t))' dif t &= integral mu(t) g(t) dif t \
+  mu(t) y(t) + C &= integral mu(t) g(t) dif t \
+  y(t) &= (integral mu(t) g(t) dif t + C) / (mu(t))
+$
+
+Now we have a general solution but we need to determine $mu(t)$.
+
+$
+  mu(t) p(t) &= mu'(t) \
+  (mu'(t)) / (mu(t)) &= p(t) \
+  (ln mu(t))' &= p(t)
+$
+
+So now
+
+$
+  integral mu(t) + k &= integral p(t) dif t \
+  ln mu(t) &= integral p(t) dif t + k \
+  mu(t) = e^(integral p(t) dif t + k) &= k e^(integral p(t) dif t)
+$
+
+Now substitute
+
+$
+  y(t) &= (integral k e^(integral p(t) dif t) g(t) dif t + C) / (k e^(integral p(t))) \
+  &= (k integral e^(integral p(t) dif t) g(t) dif t + C) / (k e^(integral p(t)))
+$
+
+Now do some cursed constant manipulation to obtain a final solution with only one arbitrary constant
+
+$
+  y(t) = (integral e^(integral p(t) dif t) g(t) dif t + C) / (e^(integral p(t)))
+$
+
+#remark[
+  The most useful result to us is
+  $
+    y(t) &= (integral mu(t) g(t) dif t + C) / (mu(t)) \
+    mu(t) &= e^(integral p(t) dif t)
+  $
+  We can easily obtain a solution form for any first order linear ODE simply by
+  identifying $p(t)$ and $g(t)$.
+]
diff --git a/documents/by-course/math-6a/course-notes/main.typ b/documents/by-course/math-6a/course-notes/main.typ
index 8fa6777..8c28c0f 100644
--- a/documents/by-course/math-6a/course-notes/main.typ
+++ b/documents/by-course/math-6a/course-notes/main.typ
@@ -59,3 +59,16 @@ point or vector.
   How far is the point $x_1, x_2, x_3$ from the origin? \
   Answer: $x_1^2 + x_2^2 + x_3^2$
 ]
+
+#definition[
+  For vectors $u$ and $v$, where
+  $ v = vec(v_1, v_2, dots.v, n), u = vec(u_1, u_2, dots.v, n) $
+  The dot product is defined as
+  $ sum_(i=1)^n v_i dot u_i $
+]
+
+#proposition[
+  The dot product of two vectors is the product of their magnitudes and the cosine of the angle between.
+
+  $ arrow(v) dot arrow(w) = ||arrow(v)|| dot ||arrow(w)|| cos theta $
+]