Skip to content
Snippets Groups Projects
Commit 6c95ba69 authored by Timon Idema's avatar Timon Idema
Browse files

Fixed ref in differentialequations.md

parent 86de7ebb
No related branches found
No related tags found
No related merge requests found
......@@ -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 {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:
......
......@@ -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:
......
0% Loading or .
You are about to add 0 people to the discussion. Proceed with caution.
Finish editing this message first!
Please register or to comment