diff --git a/documents/by-course/math-4b/course-notes/main.typ b/documents/by-course/math-4b/course-notes/main.typ
index 320d1fc..eabf3a0 100644
--- a/documents/by-course/math-4b/course-notes/main.typ
+++ b/documents/by-course/math-4b/course-notes/main.typ
@@ -338,7 +338,7 @@ Consider the IVP
 $ y' = 1 / 2 cos(y) + t, y(0) = -1 $
 
 The righthand side is nice, by @eutheorem, it has a unique solution. We can't
-find an explicit solution.
+find an explicit solution. So use Euler's method.
 
 = Lecture #datetime(day: 23, month: 1, year: 2025).display()
 
@@ -387,4 +387,146 @@ has a unique solution $y(t)$ defined on $I$. Same as first order case. Always wr
 
 == 2nd order linear homogenous ODEs
 
+How to find solutions to 2nd order linear ODE? Recall
 
+$
+  y'' = p(t)y' + q(t)y = 0
+$
+
+Consider 2nd order linear homogenous ODEs with constant coefficients $a,b,c, a!= 0$.
+
+$
+  a y'' + b y' + c y = 0
+$
+
+Judging from this equation it seems $y$ should be an exponential since we want
+a function whose derivatives can cancel each other out (with constants
+applied).
+
+Trying $y=e^(r t)$,
+
+$
+  y' = r e^(r t) \
+  y'' = r^2 e^(r t) \
+  a r^2 e^(r t) + b r e^(r t) + c e^(r t) = 0
+$
+
+$e^(r t)$ is never 0 (in $RR$) so we can divide without losing or gaining solutions.
+
+$
+  e^(r t) (a r^2 + b r + c) = 0
+$
+
+Then we simply just need to solve a quadratic for $r$.
+
+$
+  a r^2 + b r + c = 0
+$
+
+We may have two distinct real solutions, one repeated solution, or complex
+solutions (no solutions in $RR$). For now let us consider the distinct real
+solutions.
+
+#fact[
+  We conclude $y(t) = e^(r t)$ is a solution of
+  $
+    a y'' + b y' + c y = 0
+  $
+  provided $r$ satisfies
+  $
+    a r^2 + b r + c = 0
+  $
+  This is the _characteristic equation_.
+]
+
+#example[
+  $
+    y'' + 5y' + 6y = 0
+  $
+  It is homogenous, so we find the characteristic equation.
+
+  $
+    r^2 + 5 r + 6 = 0 \
+    r = -2, -3 \
+    y_1 = e^(-2t), y_2 = e^(-3t)
+  $
+]<particular-solutions>
+
+#example[
+  $ y'' - 2y' + y = 0 $
+  The characteristic equation is
+  $
+    r^2 - 2r + 1 = 0 \
+    r = 1 \
+    y_1 = e^t
+  $
+]
+
+== Superposition principle for 2nd order linear homogenous ODE
+
+We can find a few particular solutions to our ODE, but how can we find all of the solutions?
+
+#fact[
+  Suppose $y_1(t), y_2(t)$ are a pair of solutions of
+  $
+    y'' + p(t) y' + q(t) y = 0
+  $
+  Then the linear combination
+  $
+    c_1 y_1(t) + c_2 y_2(t)
+  $
+  is also a solution.
+]<superposition-principle>
+
+#proof[
+  $
+    (c_1 y_1(t) + c_2 y_2(t))'' + p(t) (c_1 y_1(t) + c_2 y_2(t))' + q(t)(c_1 y_1(t) + c_2 y_2(t)) \
+    = c_1 y_1'' + c_2 y_2'' + p(t)(c_1 y_1' + c_2 y_2') + g(t)(c_1 y_1 + c_2 y_2) \
+    = c_1 (y''_1 + p(t)y'_1 + q(t) y_1) + c_2 (y''_2 + p(t) y'_2 + q(t) y_2) \
+    = c_1 (0) + c_2(0) = 0
+  $
+]
+
+#example[Revisiting @particular-solutions][
+  $
+    y'' + 5y' + 6y = 0
+  $
+  It is homogenous, so we find the characteristic equation.
+
+  $
+    r^2 + 5 r + 6 = 0 \
+    r = -2, -3 \
+    y_1 = e^(-2t), y_2 = e^(-3t)
+  $
+
+  So by @superposition-principle,
+  $
+    y(t) = c_1 e^(-2t) + c_2 e^(-3t)
+  $
+  is a general solution.
+]
+
+#example[
+  Find the general solution of
+  $
+    y'' + 2y' - 3y = 0 \
+    y(1) = 1, y'(1) = -1
+  $
+  and then find the solution satisfying the initial conditions
+  $
+    y(1) = 1, y'(1) = -1
+  $
+  General solution is
+  $
+    y(t) &= c_1 e^(-3t) + c_2 e^t \
+    y'(t) &= -3 c_1 e^(-3t) + c_2 e^t
+  $
+  Now we need the particular solution.
+  $
+    1 &= c_1 e^(-3) + c_2 e \
+    -1 &= -3 c_1 e^(-3) + c_2 e \
+  $
+
+  Now we have a system and we can solve it using standard linear algebra
+  techniques.
+]