Question

Problem 5. (1 point) Consider the linear system a. Find the eigenvalues and eigenvectors for the coefficient matrix. and iz = b. Find the real-valued solution to the initial value problem - -3y - 2y2 Syı + 3y2 yı(0) = -7, (0) = 10 Use I as the independent variable in your answers. Y() = Note: You can earn partial credit on this problem.

Answer 1

(a) We have, the eigenvectors of the coefficient matrix are given by the roots of the characteristic equation, so

$\small \det (A-xI)= 0 \\ \\ \Rightarrow \begin{vmatrix} -3-x & -2\\ 5 & 3-x \end{vmatrix}=0 \Rightarrow x^2+1=0 \Rightarrow x=i, -i$

Thus, the eigenvalues of the coefficient matrix are $\small \lambda=i,-i$

An eigenvector corresponding to the eigenvalue i is given by

$\small \begin{bmatrix}-3+i\\ 5\end{bmatrix}$

and, an eigenvector corresponding to the eigenvalue -i is

$\small \begin{bmatrix}-3-i\\ 5\end{bmatrix}$

So, the answer to (a) is

$\small \lambda_1=i,\;\vec{v}_1=\begin{bmatrix}-3+i\\ 5\end{bmatrix},\text{ and }\lambda_2=-i,\; \vec{v}_2=\begin{bmatrix}-3-i\\ 5\end{bmatrix}$

(b) Now, we have

$\small \mathrm{For\:two\:distinct\:complex\:eigenvalues}\:\lambda_1\ne \lambda_2,\:\mathrm{where\:}\lambda_1=\alpha +i\beta ,\:\lambda_2=\alpha -i\beta$

$\small \mathrm{and\:corresponding\:eigenvectors\;}\eta_1\ne \eta_2,\:\mathrm{where}\:\eta_1=u+iv,\:\eta_2=u-iv$

$\small \mathrm{the\:general\:solution\:takes\:the\:form:}$

$\small x=c_1e^{\alpha t}\left(\cos \left(\beta t\right)u-\sin \left(\beta t\right)v\right)+c_2e^{\alpha t}\left(\cos \left(\beta t\right)v+\sin \left(\beta t\right)u\right)$

Now, in our case

$\small \lambda=0\pm i\Rightarrow \alpha=0;\; \beta=1$

and

$\small \eta=\begin{bmatrix}-3\pm i\\ 5\end{bmatrix}=\begin{bmatrix}-3\\ 5\end{bmatrix}\pm i\begin{bmatrix}1\\ 0\end{bmatrix};\;u=\begin{bmatrix}-3\\ 5\end{bmatrix},\; v=\begin{bmatrix}1\\ 0\end{bmatrix}$

So, the general solution of the given system is

$\small \vec y=c_1e^0\left(\cos \left(t\right)\begin{bmatrix}-3\\ 5\end{bmatrix}-\sin \left(t\right)\begin{bmatrix}1\\ 0\end{bmatrix}\right)+c_2e^0\left(\cos \left(t\right)\begin{bmatrix}1\\ 0\end{bmatrix}+\sin \left(t\right)\begin{bmatrix}-3\\ 5\end{bmatrix}\right)$

Simplifying we get

$\small \vec y=c_1\left(\cos \left(t\right)\begin{bmatrix}-3\\ 5\end{bmatrix}-\sin \left(t\right)\begin{bmatrix}1\\ 0\end{bmatrix}\right)+c_2\left(\cos \left(t\right)\begin{bmatrix}1\\ 0\end{bmatrix}+\sin \left(t\right)\begin{bmatrix}-3\\ 5\end{bmatrix}\right)$

Now, we have the initial condition,

$\small \vec y(0)=\begin{bmatrix} -7\\ 10 \end{bmatrix}$

Applying this initial condition we get

$\small c_1=2,\;c_2=-1$

so our particular solution is

$\small \vec y=2\left(\cos \left(t\right)\begin{bmatrix}-3\\ 5\end{bmatrix}-\sin \left(t\right)\begin{bmatrix}1\\ 0\end{bmatrix}\right)-\left(\cos \left(t\right)\begin{bmatrix}1\\ 0\end{bmatrix}+\sin \left(t\right)\begin{bmatrix}-3\\ 5\end{bmatrix}\right)$

Simplifying further we have

$\small \vec y=\cos (t) \begin{bmatrix} -7\\ 10 \end{bmatrix}+\sin (t) \begin{bmatrix} 1\\ -5 \end{bmatrix}$

Thus, writing down the functions separately, the solution to (b) is

$\small y_1(t)=-7\cos(t)+\sin(t)$

$\small y_2(t)=10\cos(t)-5\sin(t)$