Scalar LTI differential equations are not the most general case. The most general case are the coupled LTI differential equations. Below is the three equations in three unknowns:
\begin{eqnarray}
D y_1(t) + y_1(t)+ 5 y_2(t) – 4 D^2 y_3(t) &=& u_1(t)+2u_2(t) \\
2 y_1(t) + 5 D y_2(t) – D y_3(t) &=& u_1(t) \\
D^3 y_1(t) – 4 D y_3(t) &=& -u_2(t)
\end{eqnarray}
The unknowns are $y_1(t)$, $y_2(t)$ and $y_3(t)$. $u_1(t)$ and $u_2(t)$ are input signals. Note that inputs can also be differentiated.
Couled LTI differential equations can always be converted into a matrix form by introducing new variables. Here, rather than stating our results in a theorem-proof format, it is better to proceed with examples.
Example: Convert the above system into a matrix form.
Solution: Note that y_1 is of degree 3, y_2 is of degree 1, y_3 is of degree 2, u_1 is of degree 0, and u_2 is of degree 2. We define the auxiliary variables
\begin{eqnarray}
y_4 &=& Dy_1 \\
y_5 &=& Dy_4 = D^2y_1\\
y_6 &=& Dy_3 \\
\end{eqnarray}
Then the equation becomes
\begin{eqnarray}
y_4 &=& Dy_1 \\
y_5 &=& Dy_4 \\
y_6 &=& Dy_3 \\
y_4 + y_1 + 5 y_2 – 4 Dy_6 &=& u_1+2u_2 \\
2 y_1 + 5 D y_2 – y_6 &=& u_1 \\
D y_5 – 4 y_6 &=& -u_2
\end{eqnarray}
Taking derivatives to the left hand side
\begin{eqnarray}
Dy_1 &=& y_4 \\
5 Dy_2 &=& -2y_1 + y_6 + u_1 \\
Dy_3 &=& y_6 \\
Dy_4 &=& y_5\\
Dy_5 &=& 4 y_6 -u_2\\
4 Dy_6 &=& y_1 + 5y_2 + y_4 +u_1 +2u_2
\end{eqnarray}
Example 1: The following system has eigenvalues -1, 3 and 1. Find its solution.
\begin{eqnarray}
\frac{d}{dt}
\begin{bmatrix}
y_1\\
y_2\\
y_3
\end{bmatrix} =
\begin{bmatrix}
-3 & 8 & -28 \\
-8 & 19 & -64 \\
-2 & 4 &-13
\end{bmatrix}
\begin{bmatrix}
y_1\\
y_2\\
y_3
\end{bmatrix}
\end{eqnarray}
Example 2: The following system has eigenvalues 1, 1 and 2. Find its solution.
\begin{eqnarray}
\frac{d}{dt}
\begin{bmatrix}
y_1\\
y_2\\
y_3
\end{bmatrix} =
\begin{bmatrix}
1 & -1 & -1 \\
0 & 3 & 2 \\
0 & -1 & 0
\end{bmatrix}
\begin{bmatrix}
y_1\\
y_2\\
y_3
\end{bmatrix}
\end{eqnarray}
Example 3: The following system has eigenvalues 2, 1 and -1. Find its solution.
\begin{eqnarray}
\frac{d}{dt}
\begin{bmatrix}
y_1\\
y_2\\
y_3
\end{bmatrix} =
\begin{bmatrix}
5 & 0 & 3 \\
-4 & 1 & -2 \\
-6 & 0 & -4
\end{bmatrix}
\begin{bmatrix}
y_1\\
y_2\\
y_3
\end{bmatrix}
\end{eqnarray}
Example 4: The following system has eigenvalues $i, -i$. Find its solution.
\begin{eqnarray}
\frac{d}{dt}
\begin{bmatrix}
y_1\\
y_2
\end{bmatrix} =
\begin{bmatrix}
0 & -1 \\
1 & 0 \\
\end{bmatrix}
\begin{bmatrix}
y_1\\
y_2
\end{bmatrix}
\end{eqnarray}
with eigenvectors
\begin{eqnarray}
\begin{bmatrix}
1\\
-i
\end{bmatrix} ,
\begin{bmatrix}
1\\
i
\end{bmatrix}
\end{eqnarray}
corresponding to eigenvalues $i, -i$ respectively.
Example 5: The following system has eigenvalues $i$, $-i$ and 2. Find its solution.
\begin{eqnarray}
\frac{d}{dt}
\begin{bmatrix}
y_1\\
y_2\\
y_3
\end{bmatrix} =
\begin{bmatrix}
1 & -1 & 0 \\
1 & 0& 0 \\
0 & 0 & 2
\end{bmatrix}
\begin{bmatrix}
y_1\\
y_2\\
y_3
\end{bmatrix}
\end{eqnarray}
eigenvectors corresponding to eigenvalues $i$, $-i$ and 2 are
\begin{eqnarray}
\begin{bmatrix}
1\\
-i\\
0
\end{bmatrix} ,
\begin{bmatrix}
1\\
i\\
0
\end{bmatrix} ,
\begin{bmatrix}
0\\
0\\
1
\end{bmatrix}
\end{eqnarray}
in that order.
Example 6: The following system has eigenvalues $1+3i$, $1-3i$ and 2. Find its solution.
\begin{eqnarray}
\frac{d}{dt}
\begin{bmatrix}
y_1\\
y_2\\
y_3
\end{bmatrix} =
\begin{bmatrix}
1 & -3 & 0 \\
3 & 1 & 0 \\
0 & 0 & 2
\end{bmatrix}
\begin{bmatrix}
y_1\\
y_2\\
y_3
\end{bmatrix}
\end{eqnarray}
eigenvectors corresponding to eigenvalues $i$, $-i$ and 2 are
\begin{eqnarray}
\begin{bmatrix}
1\\
-i\\
0
\end{bmatrix} ,
\begin{bmatrix}
1\\
i\\
0
\end{bmatrix} ,
\begin{bmatrix}
0\\
0\\
1
\end{bmatrix}
\end{eqnarray}
in that order.
Example 7: The following system has eigenvalues $1+3i$, $1-3i$ and 2. Find its solution.
\begin{eqnarray}
\frac{d}{dt}
\begin{bmatrix}
y_1\\
y_2\\
y_3
\end{bmatrix} =
\begin{bmatrix}
2 & -4 & 0 \\
1 & 3 & 0 \\
0 & 0 & 5
\end{bmatrix}
\begin{bmatrix}
y_1\\
y_2\\
y_3
\end{bmatrix}
\end{eqnarray}
eigenvectors corresponding to eigenvalues $\frac{5}{2}+\frac{\sqrt{15}}{2}i$, $\frac{5}{2}-\frac{\sqrt{15}}{2}i$ and 5 are
\begin{eqnarray}
\begin{bmatrix}
1\\
\frac{1}{2}+\frac{\sqrt{15}}{2}i\\
0
\end{bmatrix} ,
\begin{bmatrix}
1\\
\frac{1}{2}-\frac{\sqrt{15}}{2}i\\
0
\end{bmatrix} ,
\begin{bmatrix}
4\\
3\\
1
\end{bmatrix}
\end{eqnarray}
in that order.