Solving coupled linear systems exactly Problem set B Answers

1. (a) \(\mathbf M=\begin{pmatrix}3&-5\\6&2\end{pmatrix}\)
2. (b) \(\mathbf M=\begin{pmatrix}-4&1\\2&0\end{pmatrix}\)

1. (a) \(\lambda=2,\ \mathbf v=\begin{pmatrix}1\\-2\end{pmatrix}\ \text{and}\ \lambda=5,\ \mathbf v=\begin{pmatrix}1\\1\end{pmatrix}\)
2. (b) \(\lambda=-3,\ \mathbf v=\begin{pmatrix}1\\-4\end{pmatrix}\ \text{and}\ \lambda=2,\ \mathbf v=\begin{pmatrix}1\\1\end{pmatrix}\)

\(\begin{pmatrix}x\\y\end{pmatrix}=Ae^{-3t}\begin{pmatrix}2\\-1\end{pmatrix}+Be^{2t}\begin{pmatrix}1\\2\end{pmatrix}\)

\(x=e^{-2t}+5e^{t},\ y=-2e^{-2t}+5e^{t}\)

\(x=2e^{-2t}+2e^{3t}\)

1. (a) \(\mathbf M=\begin{pmatrix}-3&1\\4&-3\end{pmatrix}\)
2. (b) \(\lambda=-5,\ \mathbf v=\begin{pmatrix}1\\-2\end{pmatrix}\ \text{and}\ \lambda=-1,\ \mathbf v=\begin{pmatrix}1\\2\end{pmatrix}\)
3. (c) \(x=3e^{-5t}+5e^{-t},\ y=-6e^{-5t}+10e^{-t}\)
4. (d) 1.86 mg; both tend to 0 because both eigenvalues are negative

1. (a) \(\lambda=1,\ \mathbf v=\begin{pmatrix}1\\-3\end{pmatrix}\ \text{and}\ \lambda=5,\ \mathbf v=\begin{pmatrix}1\\1\end{pmatrix}\)
2. (b) \(x=e^{t}+5e^{5t},\ y=-3e^{t}+5e^{5t}\)
3. (c) \(x(0.5)=62.6,\ y(0.5)=56.0\)