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# 4.2: Linear Systems of Differential Equations

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A first order system of differential equations that can be written in the form

\label{eq:4.2.1}
\begin{array}{ccl}
y'_1&=&a_{11}(t)y_1+a_{12}(t)y_2+\cdots+a_{1n}(t)y_n+f_1(t)\\
y'_2&=&a_{21}(t)y_1+a_{22}(t)y_2+\cdots+a_{2n}(t)y_n+f_2(t)\\
&\vdots\\
y'_n &=& a_{n1}(t)y_1+a_{n2}(t)y_2+\cdots+a_{nn}(t)y_n+f_n(t)\end{array}

is called a $$\textcolor{blue}{\mbox{linear system}}$$.

The linear system \eqref{eq:4.2.1} can be written in matrix form as

\begin{eqnarray*}
{y'}_n = A_{nn} y_n + f_n,
\end{eqnarray*}

or more briefly as

\label{eq:4.2.2}
{\bf y}'=A(t){\bf y}+{\bf f}(t),

where

\begin{eqnarray*}
\end{eqnarray*}

We call $$A$$ the $$\textcolor{blue}{\mbox{coefficient matrix}}$$ of \eqref{eq:4.2.2} and $${\bf f}$$ the $$\textcolor{blue}{\mbox{forcing function}}$$. We'll say that $$A$$ and $${\bf f}$$ are $$\textcolor{blue}{\mbox{continuous}}$$ if their entries are continuous. If $${\bf f}={\bf 0}$$, then \eqref{eq:4.2.2} is $$\textcolor{blue}{\mbox{homogeneous}}$$; otherwise, \eqref{eq:4.2.2} is $$\textcolor{blue}{\mbox{nonhomogeneous}}$$.

An initial value problem for \eqref{eq:4.2.2} consists of finding a solution of \eqref{eq:4.2.2} that equals a given constant vector

\begin{eqnarray*}
{\bf k} = k_n.
\end{eqnarray*}

at some initial point $$t_0$$. We write this initial value problem as

\begin{eqnarray*}
{\bf y}' = A(t){\bf y} + {\bf f}(t), \quad {\bf y}(t_0) - {\bf k}.
\end{eqnarray*}

The next theorem gives sufficient conditions for the existence of solutions of initial value problems for \eqref{eq:4.2.2}. We omit the proof.

## Theorem $$\PageIndex{1}$$

Suppose the coefficient matrix $$A$$ and the forcing function $${\bf f}$$ are continuous on $$(a,b)$$, let $$t_0$$ be in $$(a,b)$$, and let $${\bf k}$$ be an arbitrary constant $$n$$-vector. Then the initial value problem

\begin{eqnarray*}
{\bf y}' = A(t) {\bf y} + {\bf f}(t), \quad {\bf y} (t_0) = {\bf k}
\end{eqnarray*}

has a unique solution on $$(a,b)$$.

Proof

Add proof here and it will automatically be hidden if you have a "AutoNum" template active on the page.

## Example $$\PageIndex{1}$$

(a) Write the system

\label{eq:4.2.3}
\begin{array}{rcl}
y_1'&=&\phantom{2}y_1+2y_2+2e^{4t} \\
y_2'&=&2y_1+\phantom{2}y_2+\phantom{2}e^{4t}
\end{array}

in matrix form and conclude from Theorem $$(4.2.1)$$ that every initial value problem for \eqref{eq:4.2.3} has a unique solution on $$(-\infty,\infty)$$.

(b) Verify that

\label{eq:4.2.4}
{\bf y} = {1\over 5}\left[ \begin{array} \\ 8 \\ 7e^{4t} \end{array} \right] + c_1 \left[ \begin{array} \\ 11 \\ e^{3t} \end{array} \right] + c_2 \left[ \begin{array} \\ 1 \\ {-1} e^{-t} \end{array} \right]

is a solution of \eqref{eq:4.2.3} for all values of the constants $$c_1$$ and $$c_2$$.

(c) Find the solution of the initial value problem

\label{eq:4.2.5}
{\bf y}' = \left[ \begin{array} \\ 1 & 2 \\ 2 & 1 \end{array} \right] {\bf y} + \left[ \begin{array} \\ 21 \\ e^{4t} \end{array} \right], \quad {\bf y}(0)= {1 \over 5} \left[ \begin{array} \\ 3 \\ 22 \end{array} \right].

(a) The system \eqref{eq:4.2.3} can be written in matrix form as

\begin{eqnarray*}
{\bf y}' = \left[\begin{array} \\ 1 & 2 \\ 2 & 1 \end{array} \right] {\bf y} + \left[ \begin{array} \\ 21 \\ e^{4t} \end{array} \right].
\end{eqnarray*}

An initial value problem for \eqref{eq:4.2.3} can be written as

\begin{eqnarray*}
{\bf y}' = \left[ \begin{array} \\ 1 & 2 \\ 2 & 1 \end{array} \right] {\bf y} + \left[ \begin{array} \\ 21 \\ e^{4t} \end{array} \right], \quad y(t_0) = \left[ \begin{array} \\ {k_1} \\ {k_2} \end{array} \right].
\end{eqnarray*}

Since the coefficient matrix and the forcing function are both continuous on $$(-\infty,\infty)$$, Theorem $$(4.2.1)$$ implies that this problem has a unique solution on $$(-\infty,\infty)$$.

(b) If $${\bf y}$$ is given by \eqref{eq:4.2.4}, then

\begin{eqnarray*}
A{\bf y}+{\bf f}&=&
{1\over5} \left[ \begin{array} \\ 1 & 2 \\ 2 & 1 \end{array} \right] \left[ \begin{array} \\ 8 \\ 7e^{4t} \end{array} \right] +
c_1 \left[ \begin{array} \\ 1 & 2 \\ 2 & 1 \end{array} \right] \left[ \begin{array} \\11 \\e^{3t} \end{array} \right] \\
&&+c_2 \left[ \begin{array} \\ 1 & 2 \\ 2 & 1 \end{array} \right] \left[ \begin{array} \\ 1 \\ {-1}e^{-t} \end{array} \right]
+ \left[ \begin{array} \\ 2 \\ 1e^{4t} \end{array} \right] \\
&=&{1\over5} \left[ \begin{array} \\ {22} \\ {23}e^{4t} \end{array} \right] +c_1 \left[ \begin{array} \\ 3 \\ 3e^{3t} \end{array} \right] + c_2 \left[ \begin{array} \\ {-1} \\ 1e^{-t} \end{array} \right]
+ \left[ \begin{array} \\ 2 \\ 1e^{4t} \end{array} \right] \\
&=&{1\over5} \left[ \begin{array} \\ {32} \\ {28}e^{4t} \end{array} \right] + 3c_1 \left[ \begin{array} \\ 11 \\ e^{3t} \end{array} \right] - c_2 \left[ \begin{array} \\ 1 \\ {-1}e^{-t} \end{array} \right]
={\bf y}'.
\end{eqnarray*}

(c) We must choose $$c_1$$ and $$c_2$$ in \eqref{eq:4.2.4} so that

\begin{eqnarray*}
\displaystyle\frac{1}{5} \left[ \begin{array} \\ 8 \\ 7 \end{array} \right] + c_1 \left[ \begin{array} \\ 1 \\ 1 \end{array} \right] + c_2 \left[ \begin{array} \\ 1 \\ -1 \end{array} \right] = \displaystyle\frac{1}{5} \left[ \begin{array} \\ 3 \\ {22} \end{array} \right],
\end{eqnarray*}

which is equivalent to

\begin{eqnarray*}
\left[ \begin{array} \\ 1 & 1 \\ 1 & {-1} \end{array} \right] \left[ \begin{array} \\ {c_1} \\ {c_2} \end{array} \right] = \left[ \begin{array} \\ {-1} \\ 3 \end{array} \right].
\end{eqnarray*}

Solving this system yields $$c_1=1$$, $$c_2=-2$$, so

\begin{eqnarray*}
{\bf y} = \displaystyle\frac{1}{5} \left[ \begin{array} \\ 8 \\ 7 e^{4t} \end{array} \right] + \left[ \begin{array} \\ 11 \\ e^{3t}-2 \end{array} \right] \left[ \begin{array} \\ 1 \\ {-1} e^{-t} \end{array} \right]
\end{eqnarray*}

is the solution of \eqref{eq:4.2.5}.

The theory of $$n\times n$$ linear systems of differential equations is analogous to the theory of the scalar $$n$$th order equation

\label{eq:4.2.6}
P_0(t)y^{(n)}+P_1(t)y^{(n-1)}+\cdots+P_n(t)y=F(t),

as developed in Sections 3.1. For example, by rewriting \eqref{eq:4.2.6} as an equivalent linear system it can be shown that Theorem $$(4.2.1)$$ implies Theorem $$(3.1.1)$$ (Exercise $$(4.2E.12)$$.