Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
[ "article:topic", "license:ccbyncsa", "showtoc:no", "authorname:wtrench" ]
Mathematics LibreTexts

10.4E: Constant Coefficient Homogeneous Systems I (Exercises)

  • Page ID
    18300
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \( \newcommand{\place}{\bigskip\hrule\bigskip\noindent} \newcommand{\threecol}[3]{\left[\begin{array}{r}#1\\#2\\#3\end{array}\right]} \newcommand{\threecolj}[3]{\left[\begin{array}{r}#1\\[1\jot]#2\\[1\jot]#3\end{array}\right]} \newcommand{\lims}[2]{\,\bigg|_{#1}^{#2}} \newcommand{\twocol}[2]{\left[\begin{array}{l}#1\\#2\end{array}\right]} \newcommand{\ctwocol}[2]{\left[\begin{array}{c}#1\\#2\end{array}\right]} \newcommand{\cthreecol}[3]{\left[\begin{array}{c}#1\\#2\\#3\end{array}\right]} \newcommand{\eqline}[1]{\centerline{\hfill$\displaystyle#1$\hfill}} \newcommand{\twochar}[4]{\left|\begin{array}{cc} #1-\lambda\\#3-\lambda\end{array}\right|} \newcommand{\twobytwo}[4]{\left[\begin{array}{rr} #1\\#3\end{array}\right]} \newcommand{\threechar}[9]{\left[\begin{array}{ccc} #1-\lambda\\#4-\lambda\\#7 -\lambda\end{array}\right]} \newcommand{\threebythree}[9]{\left[\begin{array}{rrr} #1\\#4\\#7 \end{array}\right]} \newcommand{\solutionpart}[1]{\vskip10pt\noindent\underbar{\color{blue}\sc Solution({\bf #1})\ }} \newcommand{\Cex}{\fbox{\textcolor{red}{C}}\, } \newcommand{\CGex}{\fbox{\textcolor{red}{C/G}}\, } \newcommand{\Lex}{\fbox{\textcolor{red}{L}}\, } \newcommand{\matfunc}[3]{\left[\begin{array}{cccc}#1_{11}(t)_{12}(t)&\cdots _{1#3}(t)\\#1_{21}(t)_{22}(t)&\cdots_{2#3}(t)\\\vdots& \vdots&\ddots&\vdots\\#1_{#21}(t)_{#22}(t)&\cdots_{#2#3}(t) \end{array}\right]} \newcommand{\col}[2]{\left[\begin{array}{c}#1_1\\#1_2\\\vdots\\#1_#2\end{array}\right]} \newcommand{\colfunc}[2]{\left[\begin{array}{c}#1_1(t)\\#1_2(t)\\\vdots\\#1_#2(t)\end{array}\right]} \newcommand{\cthreebythree}[9]{\left[\begin{array}{ccc} #1\\#4\\#7 \end{array}\right]} 1 \ newcommand {\ dy} {\ ,\ mathrm {d}y} \ newcommand {\ dx} {\ ,\ mathrm {d}x} \ newcommand {\ dyx} {\ ,\ frac {\ mathrm {d}y}{\ mathrm {d}x}} \ newcommand {\ ds} {\ ,\ mathrm {d}s} \ newcommand {\ dt }{\ ,\ mathrm {d}t} \ newcommand {\dst} {\ ,\ frac {\ mathrm {d}s}{\ mathrm {d}t}} \)

    In Exercises [exer:10.4.1}– [exer:10.4.15} find the general solution.

    [exer:10.4.1] \(\{{\bf y}'= \left[\begin{array}{rr} 1&2\\2&1\end{array}\right]{\bf y}}\)&

    [exer:10.4.2] \(\{{\bf y}'= {1\over4}\left[\begin{array}{rr}-5&3 \\3&-5\end{array}\right]{\bf y}}\)

    [exer:10.4.3] \(\{{\bf y}'= {1\over5}\left[\begin{array}{rr}-4&3\\ -2&-11\end{array}\right]{\bf y}}\)&

    [exer:10.4.4] \(\{{\bf y}'= \left[\begin{array}{rr}-1&-4\\-1&-1\end{array}\right]{\bf y}}\)

    [exer:10.4.5] \(\{{\bf y}'= \left[\begin{array}{rr} 2&-4\\-1&-1\end{array}\right]{\bf y}}\)&

    [exer:10.4.6] \(\{{\bf y}'= \left[\begin{array}{rr} 4&-3\\2&-1\end{array}\right]{\bf y}}\)

    [exer:10.4.7] \(\{{\bf y}'= \left[\begin{array}{rr}-6&-3\\1&-2\end{array}\right]{\bf y}}\)&

    [exer:10.4.8] \(\{{\bf y}'= \left[\begin{array}{rrr} 1&-1&-2\\1&-2&-3\\-4&1&-1\end{array}\right] {\bf y}}\)

    [exer:10.4.9] \(\{{\bf y}'= \left[\begin{array}{rrr} -6&-4&-8\\-4&0&-4\\-8&-4&-6\end{array}\right]{\bf y}}\)&

    [exer:10.4.10] \(\{{\bf y}'= \left[\begin{array}{rrr}3&5&8\\1&-1& -2\\-1&-1&-1\end{array}\right]{\bf y}}\)

    [exer:10.4.11] \(\{{\bf y}'= \left[\begin{array}{rrr} 1&-1&2\\12&-4 & 10\\-6&1&-7 \end{array}\right]{\bf y}}\)&

    [exer:10.4.12] \(\{{\bf y}'= \left[\begin{array}{rrr} 4&-1&-4\\4&-3&-2\\1&-1&-1\end{array}\right]{\bf y}}\)

    [exer:10.4.13] \(\{{\bf y}'= \left[\begin{array}{rrr}-2&2&-6\\2&6&2\\-2&-2& 2\end{array}\right]{\bf y}}\)&

    [exer:10.4.14] \(\{{\bf y}'= \left[\begin{array}{rrr}3&2&-2\\-2&7&-2\\ -10&10&-5\end{array}\right]{\bf y}}\)

    [exer:10.4.15] \(\{{\bf y}'= \left[\begin{array}{rrr}3&1&-1\\3&5&1\\-6&2&4\end{array} \right]{\bf y}}\)

    [exer:10.4.16] \(\{{\bf y}'=\twobytwo{-7}4{-6}7{\bf y},\quad{\bf y}(0)=\twocol2{-4}}\)

    [exer:10.4.17] \(\{{\bf y}'={1\over6}\twobytwo72{-2}2{\bf y},\quad{\bf y}(0)=\twocol0{-3}}\)

    [exer:10.4.18] \(\{{\bf y}'=\twobytwo{21}{-12}{24}{-15}{\bf y},\quad{\bf y}(0)=\twocol53}\)

    [exer:10.4.19] \(\{{\bf y}'=\twobytwo{-7}4{-6}7{\bf y},\quad{\bf y}(0)=\twocol{-1}7}\)

    [exer:10.4.20] \(\{{\bf y}'={1\over6}\threebythree1204{-1}0003{\bf y},\quad{\bf y}(0)=\threecol471}\)

    [exer:10.4.21] \(\{{\bf y}'={1\over3}\threebythree2{-2}3{-4}43210{\bf y},\quad{\bf y}(0)=\threecol115}\)

    [exer:10.4.22] \(\{{\bf y}'=\threebythree6{-3}{-8}21{-2}3{-3}{-5}{\bf y},\quad{\bf y}(0)=\threecol0{-1}{-1}}\)

    [exer:10.4.23] \(\{{\bf y}'={1\over3}\threebythree24{-7}15{-5}{-4}4{-1}{\bf y},\quad{\bf y}(0)=\threecol413}\)

    [exer:10.4.24] \(\{ {\bf y}'=\threebythree301{11}{-2}7103{\bf y},\quad {\bf y}(0)=\threecol276}\)

    [exer:10.4.25] \(\{ {\bf y}'=\threebythree{-2}{-5}{-1}{-4}{-1}145{3}{\bf y},\quad {\bf y}(0)=\threecol8{-10}{-4}}\)

    [exer:10.4.26] \(\{ {\bf y}'=\threebythree3{-1}04{-2}04{-4}2{\bf y},\quad {\bf y}(0)=\threecol7{10}2}\)

    [exer:10.4.27] \(\{{\bf y}'= \left[\begin{array}{rrr}-2&2&6\\2&6&2\\-2&-2& 2\end{array}\right]{\bf y}},\quad{\bf y}(0)=\threecol6{-10}7\)

    [exer:10.4.28] Let \(A\) be an \(n\times n\) constant matrix. Then Theorem [thmtype:10.2.1} implies that the solutions of \[{\bf y}'=A{\bf y} \eqno{\rm(A)}\]

    are all defined on \((-\infty,\infty)\).

    Use Theorem [thmtype:10.2.1} to show that the only solution of (A) that can ever equal the zero vector is \({\bf y}\equiv{\bf0}\).

    Suppose \({\bf y}_1\) is a solution of (A) and \({\bf y}_2\) is defined by \({\bf y}_2(t)={\bf y}_1(t-\tau)\), where \(\tau\) is an arbitrary real number. Show that \({\bf y}_2\) is also a solution of (A).

    Suppose \({\bf y}_1\) and \({\bf y}_2\) are solutions of (A) and there are real numbers \(t_1\) and \(t_2\) such that \({\bf y}_1(t_1)={\bf y}_2(t_2)\). Show that \({\bf y}_2(t)={\bf y}_1(t-\tau)\) for all \(t\), where \(\tau=t_2-t_1\).

    [exer:10.4.29] \({\bf y}'=\{\twobytwo111{-1}}{\bf y}\)&

    [exer:10.4.30] \({\bf y}'=\{\twobytwo{-4}3{-2}{-11}}{\bf y}\)

    [exer:10.4.31] \({\bf y}'=\{\twobytwo9{-3}{-1}{11}}{\bf y}\)&

    [exer:10.4.32] \({\bf y}'=\{\twobytwo{-1}{-10}{-5}4}{\bf y}\)

    [exer:10.4.33] \({\bf y}'=\{\twobytwo5{-4}1{10}}{\bf y}\)&

    [exer:10.4.34] \({\bf y}'=\{\twobytwo{-7}13{-5}}{\bf y}\)

    [exer:10.4.35] Suppose the eigenvalues of the \(2\times 2\) matrix \(A\) are \(\lambda=0\) and \(\mu\ne0\), with corresponding eigenvectors \({\bf x}_1\) and \({\bf x}_2\). Let \(L_1\) be the line through the origin parallel to \({\bf x}_1\).

    Show that every point on \(L_1\) is the trajectory of a constant solution of \({\bf y}'=A{\bf y}\).

    Show that the trajectories of nonconstant solutions of \({\bf y}'=A{\bf y}\) are half-lines parallel to \({\bf x}_2\) and on either side of \(L_1\), and that the direction of motion along these trajectories is away from \(L_1\) if \(\mu>0\), or toward \(L_1\) if \(\mu<0\).

    [exer:10.4.36] \({\bf y}'=\{\twobytwo{-1}11{-1}}{\bf y}\)&

    [exer:10.4.37] \({\bf y}'=\{\twobytwo{-1}{-3}26}{\bf y}\)

    [exer:10.4.38] \({\bf y}'=\{\twobytwo1{-3}{-1}3}{\bf y}\)&

    [exer:10.4.39] \({\bf y}'=\{\twobytwo1{-2}{-1}2}{\bf y}\)

    [exer:10.4.40] \({\bf y}'=\{\twobytwo{-4}{-4}11}{\bf y}\)&

    [exer:10.4.41] \({\bf y}'=\{\twobytwo3{-1}{-3}1}{\bf y}\)

    [exer:10.4.42] Let \(P=P(t)\) and \(Q=Q(t)\) be the populations of two species at time \(t\), and assume that each population would grow exponentially if the other didn’t exist; that is, in the absence of competition,

    \[P'=aP \mbox{\quad and \quad}Q'=bQ, \eqno{\rm(A)}\]

    where \(a\) and \(b\) are positive constants. One way to model the effect of competition is to assume that the growth rate per individual of each population is reduced by an amount proportional to the other population, so (A) is replaced by

    \[\begin{aligned} P'&=&\phantom{-}aP-\alpha Q\\ Q'&=&-\beta P+bQ,\end{aligned}\]

    where \(\alpha\) and \(\beta\) are positive constants. (Since negative population doesn’t make sense, this system holds only while \(P\) and \(Q\) are both positive.) Now suppose \(P(0)=P_0>0\) and \(Q(0)=Q_0>0\).

    For several choices of \(a\), \(b\), \(\alpha\), and \(\beta\), verify experimentally (by graphing trajectories of (A) in the \(P\)-\(Q\) plane) that there’s a constant \(\rho>0\) (depending upon \(a\), \(b\), \(\alpha\), and \(\beta\)) with the following properties:

    If \(Q_0>\rho P_0\), then \(P\) decreases monotonically to zero in finite time, during which \(Q\) remains positive.

    If \(Q_0<\rho P_0\), then \(Q\) decreases monotonically to zero in finite time, during which \(P\) remains positive.

    Conclude from (a) that exactly one of the species becomes extinct in finite time if \(Q_0\ne\rho P_0\). Determine experimentally what happens if \(Q_0=\rho P_0\).

    Confirm your experimental results and determine \(\gamma\) by expressing the eigenvalues and associated eigenvectors of

    \[A=\twobytwo a{-\alpha}{-\beta}b\]

    in terms of \(a\), \(b\), \(\alpha\), and \(\beta\), and applying the geometric arguments developed at the end of this section.