Solving coupled linear systems exactly Problem set A Answers

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

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

\(\begin{pmatrix}x\\y\end{pmatrix}=Ae^{t}\begin{pmatrix}2\\-1\end{pmatrix}+Be^{4t}\begin{pmatrix}1\\1\end{pmatrix}\)

\(x=2e^{-t}+3e^{2t},\ y=-2e^{-t}+6e^{2t}\)

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

1. (a) \(\mathbf M=\begin{pmatrix}-3&1\\2&-2\end{pmatrix}\)
2. (b) \(\lambda=-4,\ \mathbf v=\begin{pmatrix}1\\-1\end{pmatrix}\ \text{and}\ \lambda=-1,\ \mathbf v=\begin{pmatrix}1\\2\end{pmatrix}\)
3. (c) \(x=6e^{-4t}+3e^{-t},\ y=-6e^{-4t}+6e^{-t}\)
4. (d) 1.21 kg; both tend to 0 because both eigenvalues are negative

1. (a) \(\lambda=2,\ \mathbf v=\begin{pmatrix}1\\3\end{pmatrix}\ \text{and}\ \lambda=4,\ \mathbf v=\begin{pmatrix}1\\1\end{pmatrix}\)
2. (b) \(x=2e^{2t}+4e^{4t},\ y=6e^{2t}+4e^{4t}\)
3. (c) \(x(0.5)=35.0,\ y(0.5)=45.9\)