# Homogeneous linear first-order ODEs with constant coefficients

Systems of homogeneous linear first-order ODEs with constant coefficients can
be written in the form:

\begin{equation}
\vv{y}' = \vv{A} \vv{y}
\end{equation}

Let's guess a solution of the form:

\begin{align}
\vv{y} &= e^{\lambda t} \vv{x} \\
\vv{y}' &= \lambda e^{\lambda t} \vv{x}
\end{align}

Substituting in the ODE:

\begin{align}
\lambda e^{\lambda t} \vv{x} &= \vv{A} e^{\lambda t}\\
\lambda \vv{x} &= \vv{A} \vv{x}
\end{align}

We find that our solution must satisfy an eigenvalue problem! $\lambda$ is an
eigenvalue and **x** is an eigenvector of **A**. We know an *n* x *n* matrix
has *n* eigenvalues. Let's suppose the corresponding eigenvectors are all
linearly independent. What do we do?

```{topic} Superposition principle
For a linear ODE, if $\vv{y}_1$ and $\vv{y}_2$ are both solutions, then:

\begin{equation}
\vv{y} = c_1\vv{y_1} + c_2\vv{y_2}
\end{equation}

is also a solution.
```

In practice, this means that

\begin{equation}
\vv{y} = \sum_{i=1}^n c_i e^{\lambda_i t} \vv{x}_i
  = c_1 e^{\lambda_1 t}\vv{x}_1 + \cdots + c_n e^{\lambda_n t}\vv{x}_n
\end{equation}

(There are some cases this fails, but we will not cover them.) To apply the
initial condition $\vv{y}(0) = \vv{y}_0$, substitute and rewrite as a linear
system:

\begin{align}
\vv{y}(0) = c_1 \vv{x}_1 + \cdots c_n \vv{x}_n &= \vv{y}_0 \\
\vv{X} \vv{c} &= \vv{y}_0
\end{align}

where $\vv{X} = [\vv{x}_1 \cdots \vv{x}_n]$ is the matrix whose columns are the
eigenvectors of **A** and **c** is the column vector of unknown coefficients.

```{example} Reaction network (again)
We previously analyzed the concentration of three species (A, B, and C)
undergoing a sequence of first-order reactions, $A \to B \to C$, where the first
reaction has rate constant $k_1$ and the second reaction has rate constant $k_2$.
It gave rise to a system of first-order ODEs:

\begin{equation}
\vv{c}' = \vv{A} \vv{c}, \quad
  \vv{c}(0) = \begin{bmatrix} c_{{\rm A},0} \\ 0 \\ 0 \end{bmatrix}
\end{equation}

where

\begin{equation}
\vv{c} = \begin{bmatrix} c_{\rm A} \\ c_{\rm B} \\ c_{\rm C} \end{bmatrix} \qquad
\vv{A} = \begin{bmatrix} -k_1 & 0 & 0 \\ k_1 & -k_2 & 0 \\ 0 & k_2 & 0 \end{bmatrix}
\end{equation}

Solve again using eigenvalues and eigenvectors.

---

**A** is lower triangular, so its eigenvalues are: $\lambda_1 = -k_1$
$\lambda_2 = -k_2$, and $\lambda_3 = 0$. We need to find the corresponding
eigenvectors. Starting with $\lambda_1 = -k_1$:

\begin{equation}
\vv{A} - \lambda_1 \vv{I} =
\begin{bmatrix} 0 & 0 & 0 \\ k_1 & -k_2 + k_1 & 0 \\ 0 & k_2 & k_1 \end{bmatrix}
\end{equation}

from which we obtain two equations:

\begin{align}
k_1 x_1 - (k_2 - k_1) x_2 &= 0 \\
k_2 x_2 + k_1 x_3 &= 0 \\
\end{align}

with $x_3$ free. Taking $x_3 = 1$ gives

\begin{equation}
\vv{x}_1 = \begin{bmatrix} \dfrac{-k_2 - k_1}{k_2} \\ \dfrac{-k_1}{k_2} \\ 1 \end{bmatrix}
\end{equation}

Next, we will solve with $\lambda_2 = -k_2$, which gives:

\begin{equation}
\vv{A} - \lambda_2 \vv{I} =
\begin{bmatrix} k_2-k_1 & 0 & 0 \\ k_1 & 0 & 0 \\ 0 & k_2 & k_2 \end{bmatrix}
\end{equation}

from which we obtain:

\begin{equation}
\vv{x}_2 = \begin{bmatrix} 0 \\
-1 \\
1 \end{bmatrix}
\end{equation}

Finally, we solve with $\lambda_3 = 0$, which gives:

\begin{equation}
\vv{A} - \lambda_3 \vv{I} =
\begin{bmatrix} -k_1 & 0 & 0 \\ k_1 & -k_2 & 0 \\ 0 & k_2 & 0 \end{bmatrix}
\end{equation}

from which we obtain:

\begin{equation}
\vv{x}_3 = \begin{bmatrix} 0\\
0 \\
1 \end{bmatrix}
\end{equation}

Hence, the general solution is

\begin{equation}
\vv{c} = a_1 e^{-k_1t} \begin{bmatrix}
\dfrac{-k_2 - k_1}{k_2} \\ \dfrac{-k_1}{k_2} \\ 1 \end{bmatrix}  +
a_2 e^{-k_2t} \begin{bmatrix} 0 \\ -1 \\ 1 \end{bmatrix} +
a_3 e^{0 t} \begin{bmatrix} 0\\ 0 \\ 1 \end{bmatrix}
\end{equation}

Now, we apply the initial condition:

\begin{align}
\begin{bmatrix}
\dfrac{-k_2 - k_1}{k_2} & 0 & 0 \\
\dfrac{-k_1}{k_2} & -1 & 0 \\
1 & 1 & 1
\end{bmatrix}
\begin{bmatrix}
a_1 \\ a_2 \\ a_3
\end{bmatrix} =
\begin{bmatrix} c_{{\rm A},0}\\ 0 \\ 0 \end{bmatrix}
\end{align}

Thus,

\begin{align}
a_1 &= -c_{{\rm A},0}\frac{k_2}{k_2-k_1}  \\
a_2 &= -\frac{k_1}{k_2}a_1 = c_{{\rm A},0}\frac{k_1}{k_2-k_1} \\
a_3 &= -a_1 - a_2 = c_{{\rm A},0}
\end{align}

Finally, substitute these constants into the general solution and add across
the rows to obtain solutions for each concentration:

\begin{align}
c_{\rm A}(t) &= a_1e^{-k_1t}\frac{-k_2-k_1}{k_2} = c_{{\rm A},0}e^{-k_1t}  \\
c_{\rm B}(t) &= a_1e^{-k_1t}\frac{-k_1}{k_2} - a_2e^{-k_2t} \\
  &= c_{{\rm A},0}\frac{k_1}{k_2 - k_1}(e^{-k_1t} - e^{-k_2t}) \\
  c_{\rm C}(t) &= a_1e^{-k_1t} + a_2e^{-k_2t} + a_3 \\
&= c_{{\rm A},0}\Biggl(1 - \frac{k_2}{k_2 - k_1} e^{-k_1t} +
  \frac{k_1}{k_2 - k_1} e^{-k_2t}\Biggr)
\end{align}

This matches our old answer!
```

````{example} Diffusion cell
Consider a diffusion cell consisting of two compartments with solute
concentrations $c_1$ and $c_2$ separated by a membrane.

```{image} ./_images/diffusioncell.png
:alt: Diffusion cell
:align: center
:width: 300px
```

Based on unsteady mole balances, the concentrations change according to:

\begin{align}
c_1' &= -0.1c_1 + 0.1c_2 \\
c_2' &= 0.1c_1 - 0.1c_2
\end{align}

Solve for $c_1$ and $c_2$ as a function of time if they are initially 1 M and
0 M, respectively.

---

Write in matrix form $\vv{c}' = \vv{A} \vv{c}$, where

\begin{equation}
\vv{c} = \begin{bmatrix}c_1 \\ c_2\end{bmatrix}
\qquad \vv{A} = \begin{bmatrix} -0.1 & 0.1 \\ 0.1 & -0.1 \end{bmatrix}
\end{equation}

Since **A** is symmetric, it has real eigenvalues:

\begin{align}
|\vv{A} - \lambda \vv{I}|
&= \begin{vmatrix} -0.1 - \lambda & 0.1 \\ 0.1 & -0.1 - \lambda \end{vmatrix}\\
&= (\lambda + 0.1)^2 - 0.1^2 \\
&= \lambda^2 + 0.2\lambda \\
& = \lambda(\lambda + 0.2) = 0
\end{align}

So, the eigenvalues are $\lambda_1 = 0$ and $\lambda_2 = -0.2$. Next, we find
the eigenvectors:

\begin{align}
\vv{A}-\lambda_1 \vv{I} &= \begin{bmatrix} -0.1 & 0.1 \\ 0.1 & -0.1 \end{bmatrix}
\to \vv{x}_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \\
\vv{A} - \lambda_2 \vv{I} &= \begin{bmatrix} 0.1 & 0.1 \\ 0.1 & 0.1 \end{bmatrix}
\to \vv{x}_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}
\end{align}

The general solution is:

\begin{align}
\vv{c}(t) &= a_1 e^{0t} \begin{bmatrix} 1 \\ 1 \end{bmatrix} +
a_2 e^{-0.2t} \begin{bmatrix} 1 \\ -1 \end{bmatrix} \\
&= a_1 \begin{bmatrix} 1 \\ 1 \end{bmatrix} + a_2 e^{-0.2t}
\begin{bmatrix} 1 \\ -1 \end{bmatrix}
\end{align}

Apply the initial conditions by solving $\vv{X}\vv{a} = \vv{c}(0)$ using
Gauss-Jordan elimination:

\begin{align}
\begin{bmatrix} 1 & 1 & 1\\ 1 & -1 & 0\end{bmatrix}
\begin{matrix}\vphantom{R_1} \\ -R_1 \end{matrix}
&\to \begin{bmatrix} 1 & 1 & 1\\ 0 & -2 & -1\end{bmatrix}
\begin{matrix}\vphantom{R_1} \\ \div -2 \end{matrix} \\
&\to \begin{bmatrix} 1 & 1 & 1\\ 0 & 1 & 1/2\end{bmatrix}
\begin{matrix} -R_2 \\ \vphantom{R_2} \end{matrix} \\
&\to \begin{bmatrix} 1 & 0 & 1/2 \\ 0 & 1 & 1/2\end{bmatrix}
\end{align}

so $a_1 = a_2 = 1/2$. Substituting back, the general solution for
$\vv{c}(t)$ becomes:

\begin{equation}
\vv{c}(t) = \frac{1}{2} \begin{bmatrix} 1 \\ 1 \end{bmatrix} +
  \frac{1}{2} e^{-0.2t} \begin{bmatrix} 1 \\ -1 \end{bmatrix}
\end{equation}

or equivalently

\begin{align}
c_1(t) &= \frac{1}{2} \left(1 + e^{-0.2 t}\right) \\
c_2(t) &= \frac{1}{2} \left(1 - e^{-0.2 t}\right)
\end{align}

````

## Types of critical points

We can use the eigenvalues and eigenvectors to anticipate what solutions
around *critical points* ($\vv{y}' = \vv{0}$, or steady states) look like. We
will focus our discussion on only 2 x 2 systems.

- If the eigenvalues are real and distinct, $\lambda_1 \ne \lambda_2$:

  ::::{grid}
  :::{grid-item-card} Unstable improper node

  $$
  \lambda_1 > \lambda_2 > 0
  $$

  ```{image} ./_images/unstable_improper_node.png
  :alt: Unstable improper node
  :width: 150px
  :align: center
  ```

  :::
  :::{grid-item-card} Stable improper node

  $$
  \lambda_1 < \lambda_2 < 0
  $$

  ```{image} ./_images/stable_improper_node.png
  :alt: Unstable improper node
  :width: 150px
  :align: center
  ```

  :::
  :::{grid-item-card} Saddle

  $$
  \lambda_1 > 0, \quad \lambda_2 < 0
  $$

  ```{image} ./_images/saddle.png
  :alt: Saddle
  :width: 150px
  :align: center
  ```

  :::
  ::::

- If the eigenvalues are complex, $\lambda_{1,2} = \alpha \pm  i\omega$:

  ::::{grid}
  :::{grid-item-card} Unstable spiral

  $$
  \alpha > 0
  $$

  ```{image} ./_images/unstable_spiral.png
  :alt: Unstable spiral
  :width: 150px
  :align: center
  ```

  :::
  :::{grid-item-card} Stable spiral

  $$
  \alpha < 0
  $$

  ```{image} ./_images/stable_spiral.png
  :alt: Stable spiral
  :width: 150px
  :align: center
  ```

  :::
  :::{grid-item-card} Center

  $$
  \alpha = 0
  $$

  ```{image} ./_images/center.png
  :alt: Center
  :width: 150px
  :align: center
  ```

  :::
  ::::

  The direction of the orbit depends on the matrix and can be checked for
  some point. If

  \begin{equation}
  \vv{A} = \begin{bmatrix}a & b \\ c & d\end{bmatrix}
  \end{equation}

  The orbit is clockwise if $b > c$ and counterclockwise if $c < b$.

- If the eigenvalues are real and repeated, $\lambda_1 = \lambda_2$:

  ::::{grid}
  :::{grid-item-card} Proper node / star

  **A** is a multiple of **I**.

  ```{image} ./_images/proper_node.png
  :alt: Proper node
  :width: 150px
  :align: center
  ```

  :::
  :::{grid-item-card} Degenerate node

  ```{image} ./_images/degenerate_node.png
  :alt: Degenerate node
  :width: 150px
  :align: center
  ```

  :::
  ::::

## Skill builder problems

Solve the inital value problem $ \vv{y}' = \vv{A}\vv{y}$ with
$\vv{y}(0) = [1 \quad 0]^{\rm T}$ for the following matrices. Also classify the
type of critical point using the eigenvalues.

1. For:

   \begin{equation}
   \vv{A} = \begin{bmatrix} -2 & 0  \\ 0 & -2  \end{bmatrix}
   \end{equation}

   ```{solution}

   **A** is upper triangular, so its eigenvalues are on its diagonal,
   $\lambda_1 = \lambda_2 = -2$. Its eigenvectors are:

   \begin{equation}
   \vv{x}_1 = \begin{bmatrix} 1 \\ 0  \end{bmatrix} \qquad
   \vv{x}_2 = \begin{bmatrix} 0  \\  1  \end{bmatrix}
   \end{equation}

   since $\vv{A} - \lambda_{1,2} \vv{I} = 0$. The general solution is:

   \begin{equation}
   \vv{y} = c_1 e^{-2t} \begin{bmatrix} 1 \\ 0 \end{bmatrix} +
     c_2 e^{-2t} \begin{bmatrix} 0 \\ 1 \end{bmatrix}
   \end{equation}

   To solve for the initial condition, form the augmented matrix for
   $\vv{X}\vv{c} = \vv{y}(0)$,

   \begin{equation}
   \begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}
   \end{equation}

   so $c_1 = 1$ and $c_2 = 0$.

   Hence,
   \begin{equation}
   \vv{y} =e ^{-2t} \begin{bmatrix} 1 \\ 0 \end{bmatrix}
   \end{equation}

   or

   \begin{align}
   y_1 &= e^{-2t} \\
   y_2 &= 0
   \end{align}

   The critical point is a stable proper node because
   $\lambda_{1} = \lambda_{2} < 0$ and **A** is a multiple of **I**.
   ```

2. For:

   \begin{equation}
   \vv{A} = \begin{bmatrix} 0 & 3 \\ 12 & 0 \end{bmatrix}
   \end{equation}

   ```{solution}
   First, find the eigenvalues of **A**:

   \begin{align}
   |\vv{A}-\lambda \vv{I}|
   &= \begin{vmatrix}  -\lambda & 3  \\ 12 & -\lambda \end{vmatrix} \\
   &= \lambda^{2}-36=0 \\
   \lambda_{1,2} &= \pm 6
   \end{align}

   For $\lambda_1 = 6$,

   \begin{equation}
   \vv{A}-\lambda_1\vv{I} =
   \begin{bmatrix} -6 & 3 \\ 12 & -6 \end{bmatrix}
   \to \vv{x}_1 = \begin{bmatrix} 1 \\ 2 \end{bmatrix}
   \end{equation}

   For $\lambda_2 = -6$,

   \begin{equation}
   \vv{A}-\lambda_2\vv{I} =
   \begin{bmatrix} 6 & 3 \\ 12 & 6 \end{bmatrix}
   \to \vv{x}_2 = \begin{bmatrix} 1 \\ -2 \end{bmatrix}
   \end{equation}

   The general solution is:

   \begin{equation}
   \vv{y} = c_1 e^{6t} \begin{bmatrix} 1 \\ 2 \end{bmatrix} +
     c_2 e^{-6t} \begin{bmatrix} 1 \\ -2 \end{bmatrix}
   \end{equation}

   To solve for the initial condition, form the augmented matrix for
   $\vv{X}\vv{c} = \vv{y}(0)$,

   \begin{align}
   \begin{bmatrix} 1 & 2 & 1 \\ 2 & -2 & 0 \end{bmatrix}
   \begin{matrix} \vphantom{R_1} \\ -R_1 \end{matrix}
   & \to \begin{bmatrix} 1 & 1 & 1 \\ 0 & -4 & -2 \end{bmatrix}
   \begin{matrix} \vphantom{R_1} \\ \div -4 \end{matrix} \\
   & \to \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & \frac{1}{2} \end{bmatrix}
   \begin{matrix} -R_2 \\ \vphantom{R_2} \end{matrix} \\
   & \to \begin{bmatrix} 1 & 0 & \frac{1}{2} \\ 0 & 1 & \frac{1}{2}  \end{bmatrix}
   \end{align}

   so $c_1 = 1/2$ and $c_2 = 1/2$ and

   \begin{equation}
   \vv{y} = \frac{1}{2}e^{-6t} \begin{bmatrix} 1 \\ 2 \end{bmatrix} +
     \frac{1}{2}e^{-6t} \begin{bmatrix} 1 \\ -2 \end{bmatrix}
   \end{equation}

   or

   \begin{align}
   y_1 & =\frac{1}{2}(e^{6t}+e^{-6t}) \\
   y_2 &= e^{6t}+e^{-6t}
   \end{align}

   The critical point is a saddle because the eigenvalues are real and
   $\lambda_1 > 0$ but $\lambda_2 < 0$.
   ```

3. For:

   \begin{equation}
   \vv{A} = \begin{bmatrix} 0 & 4 \\ -4 & 0 \end{bmatrix}
   \end{equation}

   ``` {solution}
   First, find the eigenvalues of **A**:

   \begin{align}
   |\vv{A} - \lambda \vv{I}|
   &= \begin{vmatrix} -\lambda & 4 \\ -4 & -\lambda \end{vmatrix} \\
   &= \lambda^2 + 16 = 0 \\
   \lambda_{1,2} &= \pm 4i
   \end{align}

   For $\lambda_1 = 4i$,

   \begin{equation}
   \vv{A}-\lambda_1\vv{I} =
   \begin{bmatrix} -4i & 4 \\ -4 & -4i \end{bmatrix}
   \to \vv{x}_1 = \begin{bmatrix} -i \\ 1 \end{bmatrix}
   \end{equation}


   For $\lambda_2 = -4i$,

   \begin{equation}
   \vv{A}-\lambda_2\vv{I} =
   \begin{bmatrix} 4i & 4 \\ -4 & 4i \end{bmatrix}
   \to \vv{x}_2 = \begin{bmatrix} i \\ 1 \end{bmatrix}
   \end{equation}

   The general solution is:

   \begin{equation}
   \vv{y} = c_1 e^{4it} \begin{bmatrix} -i \\ 1 \end{bmatrix} +
   c_2 e^{-4it} \begin{bmatrix} i \\ 1 \end{bmatrix}
   \end{equation}

   To solve for the initial condition, form the augmented matrix for
   $\vv{X}\vv{c} = \vv{y}(0)$,

   \begin{align}
   \begin{bmatrix}
   -i & i & 1 \\
   1 & 1 & 0
   \end{bmatrix}
   \begin{matrix} {\rm swap} \\ \vphantom{R_2}\end{matrix}
   &\to
   \begin{bmatrix}
   1 & 1 & 0 \\
   -i & i & 1
   \end{bmatrix}
   \begin{matrix} \vphantom{R_1} \\ -R_1 \end{matrix} \\
   &\to
   \begin{bmatrix}
   1 & 1 & 0 \\
   0 & 2i & 1
   \end{bmatrix}
   \begin{matrix} \vphantom{R_1} \\ \div 2i \end{matrix} \\
   &\to
   \begin{bmatrix}
   1 & 1 & 0 \\
   0 & 1 & \frac{1}{2i}
   \end{bmatrix}
   \begin{matrix} -R_2 \\ \vphantom{R_2} \end{matrix} \\
   &\to
   \begin{bmatrix}
   1 & 0 & -\frac{1}{2i} \\
   0 & 1 & \frac{1}{2i}
   \end{bmatrix}
   \end{align}

   so $c_1 = -1/(2i) = i/2$ and $c_2 = 1/(2i) = -i/2$ and

   \begin{equation}
   \vv{y} = \frac{i}{2} e^{4it} \begin{bmatrix} -i \\ 1 \end{bmatrix}
   - \frac{i}{2} e^{-4it} \begin{bmatrix} i \\ 1 \end{bmatrix}
   \end{equation}

   This solution is acceptable, but we can simplify it further using Euler's
   identity:

   \begin{align}
   \vv{y} &= \frac{1}{2} \left( \cos 4t + i \sin 4t \right)
     \begin{bmatrix} 1 \\ i \end{bmatrix}
   - \frac{1}{2} \left( \cos 4t - i \sin 4t \right)
     \begin{bmatrix} -1 \\ i \end{bmatrix} \\
   &= \begin{bmatrix} \cos 4t \\ -\sin 4t \end{bmatrix}
   \end{align}

   so

   \begin{align}
   y_1 &= \cos 4 t \\
   y_2 &= -\sin 4t
   \end{align}

   The critial point is a center because the eigenvalues are purely imaginary.
   ```

4. For:

   \begin{equation} \vv{A} =
   \begin{bmatrix}
   2 & -2\\
   2 & 2
   \end{bmatrix}
   \end{equation}

   ```{solution}
   First, find the eigenvalues of **A**

   \begin{align}
   \vv{A} - \lambda\vv{I} &= \begin{vmatrix}
   2- \lambda & -2\\
   2 & 2-\lambda
   \end{vmatrix} \\
   &= (\lambda - 2)^2 + 4 = 0 \\
   \lambda_{1,2} &= 2 \pm 2i
   \end{align}

   For $\lambda_1 = 2+2i$,

   \begin{equation}
   \vv{A}-\lambda_1\vv{I} =
   \begin{bmatrix}
   -2 i & -2\\
   2 & -2i
   \end{bmatrix}
   \to \vv{x}_1 = \begin{bmatrix} i \\ 1 \end{bmatrix}
   \end{equation}

   For $\lambda_2 = 2-2i$,

   \begin{equation}
   \vv{A}-\lambda_2\vv{I} =
   \begin{bmatrix}
   2 i & -2\\
   2 & 2i
   \end{bmatrix}
   \to \vv{x}_2 = \begin{bmatrix} -i \\ 1\end{bmatrix}
   \end{equation}

   The general solution is:

   \begin{align}
   \vv{y} &= c_1 e^{(2+2i)t} \begin{bmatrix} i\\ 1 \end{bmatrix} +
     c_2 e^{(2-2i)t} \begin{bmatrix} -i\\ 1 \end{bmatrix} \\
   &= e^{2t} \Biggl( c_1 e^{2it} \begin{bmatrix} i\\ 1 \end{bmatrix} +
   + c_2 e^{-2it} \begin{bmatrix} -i\\ 1 \end{bmatrix}\Biggr)
   \end{align}

   To solve for the initial condition, form the augmented matrix for
   $\vv{X}\vv{c} = \vv{y}(0)$,

   \begin{align}
   \begin{bmatrix}
   i & -i & 1\\
   1 & 1 & 0
   \end{bmatrix}
   \begin{matrix}{\rm swap} \\ \vphantom{R_2}\end{matrix}
   &\to
   \begin{bmatrix}
   1 & 1 & 0\\
   i & -i & 1
   \end{bmatrix}
   \begin{matrix}\vphantom{R_1} \\ -i R_1 \end{matrix} \\
   &\to
   \begin{bmatrix}
   1 & 1 & 0\\
   0 & -2i & 1
   \end{bmatrix}
   \begin{matrix}\vphantom{R_1} \\ \div -2i \end{matrix} \\
   &\to
   \begin{bmatrix}
   1 & 1 & 0\\
   0 & 1 & \frac{-1}{2i}
   \end{bmatrix}
   \begin{matrix} -R_2 \\ \vphantom{R_2}\end{matrix} \\
   &\to
   \begin{bmatrix}
   1 & 0 & \frac{1}{2i}\\
   0 & 1 & - \frac{1}{2i}
   \end{bmatrix}
   \end{align}

   so $c_1 = 1/(2i) = -i/2$ and $c_2 = -1/(2i) = i/2$. Hence,

   \begin{equation}
   \vv{y} = e^{2t} \Biggl(
     - \frac{i}{2} e^{2it} \begin{bmatrix} i \\ 1 \end{bmatrix}
     + \frac{i}{2} e{-2it} \begin{bmatrix} -i\\ 1 \end{bmatrix} \Biggr)
   \end{equation}

   Simplifying using Euler's identity $e^{ix} = \cos x + i \sin x$ and adding
   across the rows gives

   \begin{align}
   y_1 &= e^{2t} \cos 2t\\
   y_2 &= e^{2t} \sin 2t
   \end{align}

   The critical point is an unstable spiral because the eigenvalues are complex,
   and the real part is positive.
   ```