Two tanks that exchange brine, predators and prey, two masses joined by springs: many models involve several unknown functions that change together. Writing them as one vector equation lets the tools of linear algebra, starting with the matrix equation $A\mathbf{x} = \mathbf{b}$, do the work.
Key idea
A linear system $x_1' = a_{11}x_1 + a_{12}x_2$, $x_2' = a_{21}x_1 + a_{22}x_2$ is $\mathbf{x}' = A\mathbf{x}$ with $A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}$. Any $n$th-order equation becomes a first-order system: for $y'' + by' + cy = 0$, set $x_1 = y$, $x_2 = y'$ to get $$\mathbf{x}' = \begin{bmatrix} 0 & 1 \\ -c & -b \end{bmatrix}\mathbf{x}.$$
Why it works
With $x_1 = y$ and $x_2 = y'$, the first equation $x_1' = x_2$ is just the definition of $x_2$, and the second comes from solving the original equation for the highest derivative: $x_2' = y'' = -cy - by'$ $= -cx_1 - bx_2$. The state $(y, y')$, position and velocity, is exactly what a second-order equation needs to know at each instant to decide what happens next. That is why every higher-order equation fits this first-order mold, and why numerical solvers and the phase plane work with systems.
Worked example: Reading a matrix
Write $x' = 3y - x$, $y' = 2x$ as $\mathbf{x}' = A\mathbf{x}$.
- Keep the columns in the order $x$, $y$. Row 1 (for $x'$): coefficient of $x$ is $-1$, of $y$ is 3.
- Row 2 (for $y'$): coefficient of $x$ is 2, of $y$ is 0.
Answer: $A = \begin{bmatrix} -1 & 3 \\ 2 & 0 \end{bmatrix}$
Worked example: A second-order equation
Write $2y'' + 6y' + 10y = 0$ as a system.
- Divide by 2: $y'' = -5y - 3y'$.
- With $x_1 = y$, $x_2 = y'$: $x_1' = x_2$ and $x_2' = -5x_1 - 3x_2$.
Answer: $A = \begin{bmatrix} 0 & 1 \\ -5 & -3 \end{bmatrix}$
Worked example: Third order
Write $y''' - 2y'' + y' - 4y = 0$ as a system.
- Let $x_1 = y$, $x_2 = y'$, $x_3 = y''$, so $x_1' = x_2$ and $x_2' = x_3$.
- $y''' = 4y - y' + 2y''$, so $x_3' = 4x_1 - x_2 + 2x_3$.
Answer: $A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 4 & -1 & 2 \end{bmatrix}$
A vector solution can be checked like any other: if $\mathbf{x}(t) = e^{2t}\begin{bmatrix} 1 \\ k \end{bmatrix}$, then $\mathbf{x}' = 2\mathbf{x}$, so $\mathbf{x}' = A\mathbf{x}$ forces $A\begin{bmatrix} 1 \\ k \end{bmatrix} = 2\begin{bmatrix} 1 \\ k \end{bmatrix}$. That is an eigenvector equation, the key to solving systems.
Common mistake
Forgetting to divide by the leading coefficient before reading off the last row, or putting the coefficients in the wrong columns when a term is written first or missing. Every row lists the coefficient of $x_1$, then $x_2$ (then $x_3$); a missing term is a 0.