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Mathematics LibreTexts

10.6E: Constant Coefficient Homogeneous Systems III (Exercises)

  • Page ID
    18302
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    In Exercises [exer:10.6.1}– [exer:10.6.16} find the general solution.

    [exer:10.6.1] \(\{\bf y'=\twobytwo{-1}2{-5}5\bf y }\)

    [exer:10.6.2] \(\{\bf y'=\twobytwo{-11}4{-26}9\bf y }\)

    [exer:10.6.3] \(\{\bf y'=\twobytwo12{-4}5\bf y}\)

    [exer:10.6.4] \(\{\bf y'=\twobytwo5{-6}3{-1}\bf y}\)

    [exer:10.6.5] \(\{\bf y'=\threebythree3{-3}1022511\bf y}\)

    [exer:10.6.6] \(\{\bf y'=\threebythree{-3}311{-5}{-3}{-3}73\bf y}\)

    [exer:10.6.7] \(\{\bf y'=\threebythree21{-1}011101\bf y}\)

    [exer:10.6.8] \(\{\bf y'=\threebythree{-3}1{-3}4{-1}24{-2}3\bf y}\)

    [exer:10.6.9] \(\{\bf y'=\twobytwo5{-4}{10}1\bf y}\)

    [exer:10.6.10] \(\{\bf y'={1\over3}\twobytwo7{-5}25\bf y}\)

    [exer:10.6.11] \(\{\bf y'=\twobytwo32{-5}1\bf y}\)

    [exer:10.6.12] \(\{\bf y'=\twobytwo{34}{52}{-20}{-30}\bf y}\)

    [exer:10.6.13] \(\{\bf y' =\threebythree11210{-1}{-1}{-2}{-1}\bf y}\)

    [exer:10.6.14] \(\{\bf y' =\threebythree3{-4}{-2}{-5}7{-8}{-10}{13}{-8}\bf y}\)

    [exer:10.6.15] \(\{\bf y'=\threebythree60{-3}{-3}331{-2}6\bf y'}\)

    [exer:10.6.16] \(\{\bf y'=\threebythree12{-2}02{-1}100\bf y'}\)

    [exer:10.6.17] \(\{\bf y'=\twobytwo4{-6}3{-2}\bf y,\quad \bf y(0)=\twocol52}\)

    [exer:10.6.18] \(\{\bf y'=\twobytwo7{15}{-3}1\bf y,\quad \bf y(0)=\twocol51}\) [exer:10.6.19] \(\{\bf y'=\twobytwo7{-15}3{-5}\bf y,\quad \bf y(0)=\twocol{17}7}\)

    [exer:10.6.20] \(\{\bf y'={1\over6}\twobytwo4{-2}52\bf y,\quad \bf y(0)=\twocol1{-1}}\)

    [exer:10.6.21] \(\{\bf y'=\threebythree52{-1}{-3}22132\bf y,\quad \bf y(0)=\threecol406}\)

    [exer:10.6.22] \(\{\bf y'=\threebythree4408{10}{-20}23{-2}\bf y,\quad \bf y(0)=\threecol865}\)

    [exer:10.6.23] \(\{\bf y'=\threebythree1{15}{-15}{-6}{18}{-22}{-3}{11}{-15}\bf y,\quad \bf y(0)=\threecol{15}{17}{10}}\)

    [exer:10.6.24] \(\{\bf y'=\threebythree4{-4}4{-10}3{15}2{-3}1\bf y,\quad \bf y(0)=\threecol{16}{14}6}\)

    [exer:10.6.25] Suppose an \(n\times n\) matrix \(A\) with real entries has a complex eigenvalue \(\lambda=\alpha+i\beta\) (\(\beta\ne0\)) with associated eigenvector \({\bf x}={\bf u}+i{\bf v}\), where \({\bf u}\) and \({\bf v}\) have real components. Show that \({\bf u}\) and \({\bf v}\) are both nonzero.

    [exer:10.6.26] Verify that

    \[\bf y_1=e^{\alpha t}({\bf u}\cos\beta t-{\bf v}\sin\beta t) \mbox{\quad and\quad} \bf y_2=e^{\alpha t}({\bf u}\sin\beta t+{\bf v}\cos\beta t),\]

    are the real and imaginary parts of

    \[e^{\alpha t}(\cos\beta t+i\sin\beta t)({\bf u}+i{\bf v}).\]

    [exer:10.6.27] Show that if the vectors \({\bf u}\) and \({\bf v}\) are not both \({\bf 0}\) and \(\beta\ne0\) then the vector functions

    \[\bf y_1=e^{\alpha t}({\bf u}\cos\beta t-{\bf v}\sin\beta t)\quad \mbox{ and }\quad \bf y_2=e^{\alpha t}({\bf u}\sin\beta t+{\bf v}\cos\beta t)\]

    are linearly independent on every interval.

    [exer:10.6.28] Suppose \({\bf u}=\{\twocol{u_1}{u_2}}\) and \({\bf v}=\{\twocol{v_1}{v_2}}\) are not orthogonal; that is, \(({\bf u},{\bf v})\ne0\).

    Show that the quadratic equation

    \[({\bf u},{\bf v})k^2+(\|{\bf v}\|^2-\|{\bf u}\|^2)k-({\bf u},{\bf v})=0\]

    has a positive root \(k_1\) and a negative root \(k_2=-1/k_1\).

    Let \({\bf u}_1^{(1)}={\bf u}-k_1{\bf v}\), \({\bf v}_1^{(1)}={\bf v}+k_1{\bf u}\), \({\bf u}_1^{(2)}={\bf u}-k_2{\bf v}\), and \({\bf v}_1^{(2)}={\bf v}+k_2{\bf u}\), so that \(({\bf u}_1^{(1)},{\bf v}_1^{(1)}) =({\bf u}_1^{(2)},{\bf v}_1^{(2)})=0\), from the discussion given above. Show that

    \[{\bf u}_1^{(2)}={{\bf v}_1^{(1)}\over k_1} \mbox{\quad and \quad} {\bf v}_1^{(2)}=-{{\bf u}_1^{(1)}\over k_1}.\]

    Let \({\bf U}_1\), \({\bf V}_1\), \({\bf U}_2\), and \({\bf V}_2\) be unit vectors in the directions of \({\bf u}_1^{(1)}\), \({\bf v}_1^{(1)}\), \({\bf u}_1^{(2)}\), and \({\bf v}_1^{(2)}\), respectively. Conclude from (a) that \({\bf U}_2={\bf V}_1\) and \({\bf V}_2=-{\bf U}_1\), and that therefore the counterclockwise angles from \({\bf U}_1\) to \({\bf V}_1\) and from \({\bf U}_2\) to \({\bf V}_2\) are both \(\pi/2\) or both \(-\pi/2\).

    [exer:10.6.29]   \(\{\bf y'=\twobytwo3{-5}5{-3}\bf y}\) [exer:10.6.30]   \(\{\bf y'=\twobytwo{-15}{10}{-25}{15}\bf y}\)
    [exer:10.6.31]   \(\{\bf y'=\twobytwo{-4}8{-4}4\bf y}\) [exer:10.6.32]   \(\{\bf y'=\twobytwo{-3}{-15}33\bf y}\)
    [exer:10.6.33]   \(\{\bf y'=\twobytwo{-5}6{-12}7\bf y}\) [exer:10.6.34]   \(\{\bf y'=\twobytwo5{-12}6{-7}\bf y}\)
    [exer:10.6.35]   \(\{\bf y'=\twobytwo4{-5}9{-2}\bf y}\) [exer:10.6.36]   \(\{\bf y'=\twobytwo{-4}9{-5}2\bf y}\)
    [exer:10.6.37]   \(\{\bf y'=\twobytwo{-1}{10}{-10}{-1}\bf y}\) [exer:10.6.38]   \(\{\bf y'=\twobytwo{-1}{-5}{20}{-1}\bf y}\)
    [exer:10.6.39]   \(\{\bf y'=\twobytwo{-7}{10}{-10}9\bf y}\) [exer:10.6.40]   \(\{\bf y'=\twobytwo{-7}6{-12}5\bf y}\)