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Mathematics LibreTexts

10.2: Linear Systems of Differential Equations

  • Page ID
    9454
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    A first order system of differential equations that can be written in the form

    \[\label{eq:10.2.1} \begin{array}{ccl} y'_1&=&a_{11}(t)y_1+a_{12}(t)y_2+\cdots+a_{1n}(t)y_n+f_1(t)\\ y'_2&=&a_{21}(t)y_1+a_{22}(t)y_2+\cdots+a_{2n}(t)y_n+f_2(t)\\ &\vdots\\ y'_n& =&a_{n1}(t)y_1+a_{n2}(t)y_2+\cdots+a_{nn}(t)y_n+f_n(t)\end{array}\]

    is called a linear system. The linear system Equation \ref{eq:10.2.1} can be written in matrix form as

    \[\col{y'}n=\matfunc ann\col yn+\colfunc fn, \nonumber\]

    or more briefly as

    \[\label{eq:10.2.2} {\bf y}'=A(t){\bf y}+{\bf f}(t),\]

    where

    \[\bf y=\col yn,\quad A(t)=\matfunc ann,\quad \text{and} \quad{\bf f}(t)=\colfunc fn.\]

    We call \(A\) the coefficient matrix of Equation \ref{eq:10.2.2} and \({\bf f}\) the forcing function. We’ll say that \(A\) and \({\bf f}\) are continuous if their entries are continuous. If \(\bf f={\bf 0}\), then Equation \ref{eq:10.2.2} is homogeneous; otherwise, Equation \ref{eq:10.2.2} is nonhomogeneous.

    An initial value problem for Equation \ref{eq:10.2.2} consists of finding a solution of Equation \ref{eq:10.2.2} that equals a given constant vector

    \[\bf k =\col kn. \nonumber\]

    at some initial point \(t_0\). We write this initial value problem as

    \[\bf y'=A(t){\bf y}+{\bf f}(t), \quad {\bf y}(t_0)={\bf k}.\nonumber\]

    The next theorem gives sufficient conditions for the existence of solutions of initial value problems for Equation \ref{eq:10.2.2}. We omit the proof.

    Theorem \(\PageIndex{1}\): Existence

    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

    \[\bf y'=A(t){\bf y}+{\bf f}(t), \quad {\bf y}(t_0)= \bf k \nonumber\]

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

    Example \(\PageIndex{1}\)

    Write the system

    \[\label{eq:10.2.3} \begin{array}{rcl} y_1'&=&\phantom{2}y_1+2y_2+2e^{4t} \\[4pt] y_2'&=&2y_1+\phantom{2}y_2+\phantom{2}e^{4t} \end{array}\]

    in matrix form and conclude from Theorem \(\PageIndex{1}\) that every initial value problem for Equation \ref{eq:10.2.3} has a unique solution on \((-\infty,\infty)\).

    Verify that

    \[\label{eq:10.2.4} {\bf y}= {1\over5}\twocol87e^{4t}+c_1\twocol11e^{3t}+c_2\twocol1{-1}e^{-t}\]

    is a solution of Equation \ref{eq:10.2.3} for all values of the constants \(c_1\) and \(c_2\).

    Find the solution of the initial value problem

    \[\label{eq:10.2.5} {\bf y}'=\twobytwo1221{\bf y}+\twocol21e^{4t},\quad {\bf y}(0)={1\over5}\twocol3{22}.\]

    The system Equation \ref{eq:10.2.3} can be written in matrix form as

    \[{\bf y}'=\twobytwo1221{\bf y}+\twocol21e^{4t}.\nonumber\]

    An initial value problem for Equation \ref{eq:10.2.3} can be written as

    \[{\bf y}'=\twobytwo1221{\bf y}+\twocol21e^{4t}, \quad y(t_0)=\twocol{k_1}{k_2}.\]

    Since the coefficient matrix and the forcing function are both continuous on \((-\infty,\infty)\), Theorem \(\PageIndex{1}\) implies that this problem has a unique solution on \((-\infty,\infty)\).

    If \({\bf y}\) is given by Equation \ref{eq:10.2.4}, then

    \[\begin{aligned} A{\bf y}+{\bf f}&=& {1\over5}\twobytwo1221\twocol87e^{4t}+ c_1\twobytwo1221\twocol11e^{3t}\\[4pt] &&+c_2\twobytwo1221\twocol1{-1}e^{-t} +\twocol21e^{4t}\\[4pt] &=&{1\over5}\twocol{22}{23}e^{4t}+c_1\twocol33e^{3t}+c_2\twocol{-1}1e^{-t} +\twocol21e^{4t}\\[4pt] &=&{1\over5}\twocol{32}{28}e^{4t}+3c_1\twocol11e^{3t}-c_2\twocol1{-1}e^{-t} ={\bf y}'.\end{aligned}\]

    We must choose \(c_1\) and \(c_2\) in Equation \ref{eq:10.2.4} so that

    \[{1\over5}\twocol87+c_1\twocol11+c_2\twocol1{-1}={1\over5}\twocol3{22},\nonumber\]

    which is equivalent to

    \[\twobytwo111{-1}\twocol{c_1}{c_2}=\twocol{-1}3.\nonumber\]

    Solving this system yields \(c_1=1\), \(c_2=-2\), so

    \[{\bf y}={1\over5}\twocol87e^{4t}+\twocol11e^{3t}-2\twocol1{-1}e^{-t}\nonumber\]

    is the solution of Equation \ref{eq:10.2.5}.