From 6c95ba69325b8de94b20be75cdb9018bdb1b32eb Mon Sep 17 00:00:00 2001
From: Timon Idema <t.idema@tudelft.nl>
Date: Tue, 21 Nov 2023 11:15:04 +0100
Subject: [PATCH] Fixed ref in differentialequations.md

---
 content/differentialequations.md | 2 +-
 content/linearalgebra.md         | 4 +++-
 2 files changed, 4 insertions(+), 2 deletions(-)

diff --git a/content/differentialequations.md b/content/differentialequations.md
index d68ae41..a47af0e 100644
--- a/content/differentialequations.md
+++ b/content/differentialequations.md
@@ -131,7 +131,7 @@ $$
 x(t) = A e^{\lambda_{+}t} + B e^{\lambda_{-}t},
 $$ (secondorderodesol2)
 
-where $A$ and $B$ are set by either initial or boundary conditions. Since the $\lambda_\pm$ may be complex, so may $A$ and $B$; it's their combination that should give a real number (as $x(t)$ is real), see problem&nbsp;{numref}`app:solvingde`.\ref{prob:rewritesolutionsecondorderode}.
+where $A$ and $B$ are set by either initial or boundary conditions. Since the $\lambda_\pm$ may be complex, so may $A$ and $B$; it's their combination that should give a real number (as $x(t)$ is real), see {numref}`pb:secondorderodesolutions`a.
 
 In the case that equation&nbsp;{eq}`secondorderodesol1` gives only one solution, the corresponding exponential function is still a solution of equation&nbsp;{eq}`secondorderodeconstcoeff`, but it is not the most general one, as we only can put a single undetermined constant in front of it. We therefore need a second, independent solution. To guess one, here's a third useful trick<sup>[^4]</sup>: take the derivative of our known solution, $e^{\lambda t}$, with respect to the parameter $\lambda$. This gives a second Ansatz: $t e^{\lambda t}$, where $\lambda = -b/2a$. Substituting this Ansatz into equation&nbsp;{eq}`secondorderodeconstcoeff` for the case that $c = b^2/2a$, we find:
 
diff --git a/content/linearalgebra.md b/content/linearalgebra.md
index 20c20e5..2f2f198 100644
--- a/content/linearalgebra.md
+++ b/content/linearalgebra.md
@@ -144,7 +144,7 @@ A = \begin{pmatrix}
 5 \\ 4 \\ 3
 \end{pmatrix}.
 ```
-Find $\bm{y}$.  
+Find $\bm{y}$.
 
 ---
 **Solution**
@@ -262,6 +262,7 @@ A = \begin{pmatrix}
 2 & 1 \\ 2 & 4
 \end{pmatrix}.
 ```
+
 ---
 **Solution**
 We write the matrix $A$ in combination with the identity matrix, then apply row-reduction until we've reduced $A$ itself to the identity matrix.
@@ -397,6 +398,7 @@ Find the eigenvalues and eigenvectors of the following matrices:
 ```{math}
 A = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}, \qquad B = \begin{pmatrix} 1 & 0 & 1 \\ 0 & 2 & 0 \\ 1 & 0 & 1 \end{pmatrix}.
 ```
+
 ---
 **Solution**
 For the matrix $A$, we can write down and solve the characteristic equation with ease:
-- 
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