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Problems #15,32 on pg.398 are graded.

#15,pg.398: Solve the given initial value problem. Describe the behavior of the solution as $t \rightarrow \infty$: $$\vec{y}' = \begin{bmatrix} 5 & -1 \\ 3 & 1 \end{bmatrix}\vec{y}; \vec{y}(0) = \begin{bmatrix} 2 \\ -1 \end{bmatrix}.$$
Solution: Assume $\vec{y} = \vec{k}e^{\lambda t}$ so that $\vec{y}' = \lambda \vec{k} e^{\lambda t}$. Plug these into the system to get the eigenvalue problem $\lambda \vec{k}=\begin{bmatrix} 5 & -1 \\ 3 & 1 \end{bmatrix} \vec{k}$. To find the eigenvalues of $\begin{bmatrix} 5 & -1 \\ 3 & 1 \end{bmatrix}$ we must solve the equation $$0 = \det \left( \begin{bmatrix} 5 & -1 \\ 3 & 1 \end{bmatrix} - \lambda I \right) = \det \left( \begin{bmatrix} 5-\lambda & -1 \\ 3 & 1-\lambda \end{bmatrix} \right)=(5-\lambda)(1-\lambda) - (-1)(3) = \lambda^2 -6\lambda + 8.$$ This equation factors to $0=(\lambda - 2)(\lambda - 4)$ yielding eigenvalues $\lambda_1=2$ and $\lambda_2=4$. We must find the eigenvectors $\vec{k}^{(1)}$ associated with $\lambda_1$ and $\vec{k}^{(2)}$ associated with $\lambda_2$. To find $\vec{k}^{(1)}=\begin{bmatrix} k_1^{(1)} \\ k_2^{(1)} \end{bmatrix}$, we look at the eigenvalue problem we derived above to see $$\begin{bmatrix} 5 & -1 \\ 3 & 1 \end{bmatrix} \begin{bmatrix} k_1^{(1)} \\ k_2^{(1)} \end{bmatrix} = \lambda_1 \begin{bmatrix} k_1^{(1)} \\ k_2^{(1)} \end{bmatrix}.$$ Carry out the multiplications to get $$\begin{bmatrix} 5k_1^{(1)} - k_2^{(1)} \\ 3k_1^{(1)} + k_2^{(1)} \end{bmatrix}= \begin{bmatrix} 2k_1^{(1)} \\ 2k_2^{(1)} \end{bmatrix}$$ and subtract the right side to the left side to get $$\begin{bmatrix} 3k_1^{(1)} - k_2^{(1)} \\ 3k_1^{(1)} - k_2^{(1)} \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}.$$ Of course this equation encodes the following system of equations: $$\left\{ \begin{array}{ll} 3k_1^{(1)} - k_2^{(1)} = 0 \\ 3k_1^{(1)} - k_2^{(1)} = 0. \end{array} \right.$$ Of course these equations are the same and hence one of them is redundant. This leads us to conclude that $\vec{k}^{(1)}$ is determined only by the formula $3k_1^{(1)} = k_2^{(1)}$. We are free to choose any value for $k_1^{(1)}$ and so we will pick $k_1^{(1)}=1$ yielding $\vec{k}^{(1)} = \begin{bmatrix} 1 \\ 3 \end{bmatrix}$. Similarly we may find an eigenvector associated with $\lambda_2$ and get $\vec{k}^{(2)}=\begin{bmatrix} 1 \\ 1 \end{bmatrix}$. Hence we get the general solution $$\vec{y}(t) = c_1 \vec{k}^{(1)} e^{\lambda_1 t} + c_2 \vec{k}^{(2)} e^{\lambda_2 t}=c_1 \begin{bmatrix} 1 \\ 3 \end{bmatrix} e^{2t} + c_2 \begin{bmatrix} 1 \\ 1 \end{bmatrix} e^{4t}.$$ Now we apply the initial condition: $$\begin{bmatrix} 2 \\ -1 \end{bmatrix} = \vec{y}(0) = c_1 \begin{bmatrix} 1 \\ 3 \end{bmatrix} + c_2 \begin{bmatrix} 1 \\ 1 \end{bmatrix}.$$ This vector equation encodes the following system of equations: $$\left\{ \begin{array}{ll} c_1 + c_2 &= 2 \\ 3c_1 + c_2 &= -1. \end{array} \right.$$ The solution of this system is $c_1 = -\dfrac{3}{2}$ and $c_2=\dfrac{7}{2}.$ Therefore the solution of the IVP is $$\vec{y}(t) = -\dfrac{3}{2}\begin{bmatrix} 1 \\ 3 \end{bmatrix} e^{2t} + \dfrac{7}{2} \begin{bmatrix} 1 \\ 1 \end{bmatrix} e^{4t}.$$ As $t \rightarrow \infty$ the solution diverges.

#32,pg.398: Find the general solution of the equation $$\dfrac{d}{dt} \begin{bmatrix} I \\ V \end{bmatrix} = \begin{bmatrix} -\dfrac{R_1}{L} & -\dfrac{1}{L} \\ \dfrac{1}{C} & -\dfrac{1}{CR_2} \end{bmatrix} \begin{bmatrix} I \\ V \end{bmatrix},$$ where $R_1=1 \Omega, R_2 = \dfrac{3}{5} \Omega, L=2 H,$ and $C=\dfrac{2}{3} F$. Show that $I(t) \rightarrow 0$ and $V(t) \rightarrow 0$ as $t \rightarrow \infty$ regardless of the initial values of $I(0)$ and $V(0)$.
Solution: Plugging in the given values (not writing in the units) we get the equation $$\dfrac{d}{dt} \begin{bmatrix} I \\ V \end{bmatrix} = \begin{bmatrix} -\dfrac{1}{2} & -\dfrac{1}{2} \\ \dfrac{3}{2} & -\dfrac{15}{6} \end{bmatrix} \begin{bmatrix} I \\ V \end{bmatrix}.$$ The eigenvalues and associated eigenvectors of the matrix are $\lambda_1=-2$ with eigenvector $\vec{k}^{(1)}=\begin{bmatrix} \frac{1}{3} \\ 1 \end{bmatrix}$ and $\lambda_2=-1$ with eigenvector $\vec{k}^{(2)} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$. Hence the general solution is $$\begin{bmatrix} I \\ V \end{bmatrix} = c_1 \begin{bmatrix} \frac{1}{3} \\ 1 \end{bmatrix} e^{-2t} + c_2 \begin{bmatrix} 1 \\ 1 \end{bmatrix} e^{-t}.$$ This means that $I(t) = \dfrac{1}{3}e^{-2t} + c_2 e^{-t}$ from which it is clear that no matter the values of $c_1$ and $c_2$, $$\displaystyle\lim_{t \rightarrow \infty} I(t) = 0,$$ and $V(t) = c_1 e^{-2t} + c_2e^{-t}$ from which it is clear that no matter the values of $c_1$ and $c_2$, $$\displaystyle\lim_{t \rightarrow \infty} V(t) =0.$$