(Solved): Please help on number 20 X=[5553]X,X(0)=[21] soL.uTIoN: Eligenvalues: det(AI ...

Please help on number 20

${\vec{X}}^{\prime }=\left[\begin{array}{ll}5& -5\\ 5& -3\end{array}\right]\vec{X},\vec{X}(0)=\left[\begin{array}{l}2\\ 1\end{array}\right]$ soL.uTIoN: Eligenvalues: $\begin{array}{c}\det (A-\vec{\lambda }I)=\left|\begin{array}{cc}5-\lambda & -5\\ 5& -3-\lambda \end{array}\right|=(5-\lambda )(-3-\lambda )+25={\lambda }^2-2\lambda +10.\\ \lambda =\frac{2\pm \sqrt{4-40}}{2}=\frac{2\pm 6i}{2}=1\pm 3i.\end{array}$ Eigenvectors: We only find an eigenvector for $\lambda =1+3i$. $(A-(1+3i)I)\vec{v}=0\Rightarrow \left[\begin{array}{cc}4-3i& -5\\ 5& -4-3i\end{array}\right]\left[\begin{array}{l}x\\ y\end{array}\right]=\left[\begin{array}{l}0\\ 0\end{array}\right]$ The first row above gives $(4-3i)x-5y=0$, or $5y=(4-3i)x$, or $y=\frac{1}{5}(4-3i)x$. To avold fractions we can choose $x=5$. Then $y=4-3i$. $\vec{v}=\left[\begin{array}{c}5\\ 4-3i\end{array}\right]=\left[\begin{array}{l}5\\ 4\end{array}\right]+\left[\begin{array}{c}0\\ -3\end{array}\right]i$ Recall that if $\vec{v}=\vec{a}+\vec{b}i$ is an eigenvector for $\lambda =\alpha +\beta i$, then the general solution is $\begin{array}{rl}\vec{X}(t)& =c_1{e}^{at}(\cos (\beta t)\vec{a}-\sin (\beta t)\vec{b})\\ & +c_2{e}^{at}(\sin (\beta t)\vec{a}+\cos (\beta t)\vec{b}).\end{array}$ So in our case, $\begin{array}{rl}\vec{X}(t)& =c_1{e}^t\left(\cos (3t)\left[\begin{array}{c}5\\ 4\end{array}\right]-\sin (3t)\left[\begin{array}{c}0\\ -3\end{array}\right]\right)+c_2{e}^t\left(\sin (3t)\left[\begin{array}{c}5\\ 4\end{array}\right]+\cos (3t)\left[\begin{array}{c}0\\ -3\end{array}\right]\right)\\ & =\left[\begin{array}{c}5c_1{e}^t\cos (3t)+5c_2{e}^t\sin (3t)\\ \left(4c_1-3c_2\right)\cos (3t)+\left(3c_1+4c_2\right)e^t\sin (3t)\end{array}\right].\end{array}$ The initial conditions then give $5c_1=2$ and $4c_1-3c_2=1$. So $c_1=\frac{2}{5},c_2=\frac{1}{5}$. Plugging these back into the above gives $\vec{X}(t)=\left[\begin{array}{l}2e^t\cos (3t)+e^t\sin (3t)\\ e^t\cos (3t)+2e^t\sin (3t)\end{array}\right]$ 22. $\begin{array}{rl}{\vec{X}}^{\prime }& =\left[\begin{array}{cc}3& -1\\ 2& 1\end{array}\right]\vec{X},\vec{X}(0)=\left[\begin{array}{c}1\\ -2\end{array}\right].\\ {\vec{X}}^{\prime }& =\left[\begin{array}{cc}1& -2\\ 5& -1\end{array}\right]\vec{X},\vec{X}(0)=\left[\begin{array}{c}1\\ -1\end{array}\right].\end{array}$ $X^{\prime }=\left[\begin{array}{cc}-2& 5\\ -1& 2\end{array}\right]X,X(0)=\left[\begin{array}{c}-1\\ 1\end{array}\right]$ 23. ${\vec{X}}^{\prime }=\left[\begin{array}{cc}0& 5\\ -1& 2\end{array}\right]\vec{X},\vec{X}(0)=\left[\begin{array}{c}-3\\ 1\end{array}\right]$ 24.