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Problems #1 on pg.409 and #3 on pg.420 are graded.
#1,pg.409: Find the general solution of $\vec{y}'= \begin{bmatrix} 3 & -2 \\ 4 & -1 \end{bmatrix} \vec{y}$.
Solution: Assume that $\vec{y}=\vec{k}e^{\lambda t}$. This leads us to the eigenvalue problem $\begin{bmatrix} 3 & -2 \\ 4 & -1 \end{bmatrix} \vec{k} = \lambda \vec{k}.$ The eigenvalues and eigenvectors of this matrix can be found by routine calculation and are $\lambda_1=1+2i$ with associated eigenvector $\vec{k}^{(1)}=\begin{bmatrix} \frac{1}{2}+\frac{1}{2}i \\ 1 \end{bmatrix}$ and $\lambda_2=1-2i$ with associated eigenvector $\vec{k}^{(2)}=\begin{bmatrix} \frac{1}{2} - \frac{1}{2}i \\ 1 \end{bmatrix}$. With this information we see that
$$\begin{array}{ll}
\vec{y}^{(1)} &= \begin{bmatrix} \frac{1}{2}+\frac{1}{2}i \\ 1 \end{bmatrix} e^{(1+2i)t} \\
&= \left( \begin{bmatrix} \frac{1}{2} \\ 1 \end{bmatrix} + i \begin{bmatrix} \frac{1}{2} \\ 0 \end{bmatrix} \right) \left( e^t \cos(2t) + i e^t \sin(2t) \right) \\
&= \left( \begin{bmatrix} \frac{1}{2} \\ 1 \end{bmatrix} e^t \cos(2t) - \begin{bmatrix} \frac{1}{2} \\ 0 \end{bmatrix} e^t \sin(2t) \right) + i \left( \begin{bmatrix} \frac{1}{2} \\ 0 \end{bmatrix} e^t \cos(2t) + \begin{bmatrix} \frac{1}{2} \\ 1 \end{bmatrix} e^t \sin(2t) \right).
\end{array}$$
We will write real-valued solutions by taking the real and imaginary parts of the solution $\vec{y}^{(1)}$. From our calculation above it is clear that we can take
$$\tilde{\vec{y}}^{(1)} = \mathrm{Re} \left( \vec{y}^{(1)} \right) = \begin{bmatrix} \frac{1}{2} \\ 1 \end{bmatrix} e^t \cos(2t) - \begin{bmatrix} \frac{1}{2} \\ 0 \end{bmatrix} e^t \sin(2t)$$
and
$$\tilde{\vec{y}}^{(2)} = \mathrm{Im} \left( \vec{y}^{(1)} \right) = \begin{bmatrix} \frac{1}{2} \\ 0 \end{bmatrix} e^t \cos(2t) + \begin{bmatrix} \frac{1}{2} \\ 1 \end{bmatrix} e^t \sin(2t).$$
Therefore the general solution is
$$\vec{y}(t) = c_1 \begin{bmatrix} \frac{1}{2}e^t \cos(2t) - \frac{1}{2} e^t \sin(2t) \\ e^t \cos(2t) \end{bmatrix} + c_2 \begin{bmatrix} \frac{1}{2}e^t \cos(2t) + \frac{1}{2}e^t \sin(2t) \\ e^t \sin(2t) \end{bmatrix}.$$
NOTE: your answer may look very different from this -- you could have also chosen an eigenvalue with the imaginary terms in the 2nd row which would lead to an eigenvector like $\vec{k}^{(1)}=\begin{bmatrix}1 \\ 1-i \end{bmatrix}$.
#3,pg.420: First find a fundamental matrix for the system $\vec{y}'=\begin{bmatrix} 2 & -1 \\ 3& -2 \end{bmatrix}\vec{y}$. After that find the fundamental matrix that satisfies the initial value $\Phi(0)=I$.
Solution: To find the fundamental matrix we must solve the problem. The eigenvalues and associated eigevectors of the matrix are $\lambda_1 = -1$ with eigenvector $\vec{k}^{(1)} = \begin{bmatrix} 1 \\ 3 \end{bmatrix}$ and $\lambda_2=1$ with eigenvector $\vec{k}^{(2)}=\begin{bmatrix} 1 \\ 1 \end{bmatrix}$. Hence the general solution is
$$\vec{y}(t) = c_1 \begin{bmatrix} e^{-t} \\ 3e^{-t} \end{bmatrix} + c_2 \begin{bmatrix} e^t \\ e^t \end{bmatrix}.$$
From this we see the fundamental matrix is $\psi(t) = \begin{bmatrix} e^{-t} & e^t \\ 3e^{-t} & e^t \end{bmatrix}$ and the general solution can be expressed as $\Phi(t)=\psi(t)\vec{c}$, where $\vec{c}=\begin{bmatrix} c_{11} & c_{12} \\ c_{21} & c_{22} \end{bmatrix}$. To find the initial condition that satisfies $\Phi(0)=I$ we compute
$$\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix} = I = \Phi(0) = \psi(0)\vec{c} = \begin{bmatrix} 1 & 1 \\ 3 & 1 \end{bmatrix} \begin{bmatrix} c_{11} & c_{12} \\ c_{21} & c_{22} \end{bmatrix}=\begin{bmatrix} c_{11} + c_{21} & c_{12} + c_{22} \\ 3c_{11} + c_{21} & 3c_{12} + c_{22} \end{bmatrix}.$$
Thus we have the system
$$\left\{ \begin{array}{ll}
c_{11}+c_{21} &= 1 \\
c_{12}+c_{22} &= 0 \\
3c_{11} + c_{21} &= 0 \\
3c_{12} + c_{22} &= 1,
\end{array} \right.$$
whose solution can be found by routine calculation to be
$$c_{11} = -\dfrac{1}{2}, c_{12} = \dfrac{1}{2}, c_{21} = \dfrac{3}{2}, c_{22} =-\dfrac{1}{2}.$$
Therefore the solution of the IVP is
$$\Phi(t) = \psi(t) \vec{c} = \begin{bmatrix} e^{-t} & e^t \\ 3e^{-t} & e^t \end{bmatrix} \begin{bmatrix} -\frac{1}{2} & \frac{1}{2} \\ \frac{3}{2} & -\frac{1}{2} \end{bmatrix} = \begin{bmatrix} -\frac{1}{2}e^{-t}+\frac{3}{2}e^t & \frac{1}{2}e^{-t}-\frac{1}{2}e^t \\ -\frac{3}{2}e^{-t}+\frac{3}{2}e^t & \frac{3}{2}e^{-t} - \frac{1}{2}e^t \end{bmatrix}.$$
