Exam 3 Answers

10 b.

\[\left[\begin{array}{c} \dot{x} \\ \dot{y} \end{array}\right] = \left[\begin{array}{cc} 1 & -4 \\ 2 & -3 \end{array}\right] \left[\begin{array}{c} x \\ y \end{array}\right]\]

The characteristic polynomial for the coefficient matrix is \(\lambda^2 +2\lambda + 5\) . The roots of this are \(\lambda = -1 \pm 2i\).  These are the eigenvalues of the coefficient matrix.

To find the eigenvectors, solve

\[ \left[\begin{array}{cc} 1-\lambda & -4 \\ 2 & -3-\lambda \end{array}\right] \left[\begin{array}{c} a_1 \\ a_2 \end{array}\right] = \left[\begin{array}{c} 0 \\ 0 \end{array}\right]\]

with \(\lambda=-1+2i\).

We get \((2-2i)a_1-4a_2=0\).  So, if \(a_1=4\) then \(a_2=2-2i\).  Hence an eigenvector is \(\left[\begin{array}{c} 4 \\ 2-2i \end{array}\right]\)

So a complex solution is

\[\left[\begin{array}{c} x \\ y \end{array}\right]=e^{-t}\left[\begin{array}{c} 4 \\ 2-2i \end{array}\right](\cos(2t)+i\cdot\sin(2t))\]


\[\left[\begin{array}{c} x \\ y \end{array}\right]=e^{-t}\left[\begin{array}{c} 4\cos(2t)+i\cdot4\sin(2t) \\ (2-2i)(\cos(2t)+i\cdot\sin(2t)) \end{array}\right]\]


\[\left[\begin{array}{c} x \\ y \end{array}\right]=e^{-t}\left[\begin{array}{c} 4\cos(2t)+i\cdot4\sin(2t) \\ 2\cos(2t)+2\sin(2t) +i(-2\cos(2t)+2\sin(2t)) \end{array}\right]\]

So
\(x(t)=4e^{-t}(c_1\cos(2t)+c_2\sin(2t))\)
and
\(y(t)=2e^{-t}\big(c_1(\cos(2t)+\sin(2t))+c_2(-\cos(2t)+\sin(2t))\big)\)
\(y(t)\) could also be written as
\(y(t)=2e^{-t}\big((c_1-c_2)\cos(2t)+(c_1+c_2)\sin(2t)\big)\)

Here is a slope field with a solution curve through the point \((0,1)\) at time \(t=0\) with a counter-clockwise direction:

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