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\begin{document}





 \problem 
If $r=\rho$ is a double eigenvalue of the 2 x 2 matrix A
with just one linearly independent eigenvector $\xi$ and $\eta$ is a vector satisfying $(A - \rho
I)\eta = \xi$, then the general solution of $x'=Ax$ is: 

\correct $c_{1} e^{\rho t} \xi + c_{2}[t e^{\rho t} \xi + e^{\rho t}\eta]$

\wrong $c_{1} e^{\rho t} \xi + c_{2}t e^{\rho t} \xi + \eta$ 
\wrong $c_{1} e^{\rho t} \xi + c_{2} e^{\rho t} \eta$ 
\wrong $c_{1} e^{\rho t} \xi + c_{2} e^{\rho t} \eta$ 
\wrong $c_{1} e^{\rho t} \xi + c_{2} e^{\rho t} \eta$ 

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 \problem 
 The general solution of $x' = \left( \hspace{-2mm}  \begin{array}{cc}
0 & 1 \\ -1 & 0 \\  \end{array} \hspace{-2mm} \right) x$ is 

\correct
 $c_{1} \left( \hspace{-2mm}  
\begin{array}{c} \cos\; t \\ - \sin\; t \\ \end{array} \hspace{-2mm} \right)$
+ $c_{2}\left( \hspace{-2mm}  
\begin{array}{c}\sin\; t\\ \cos\; t\\\end{array} \hspace{-2mm} \right)$

\wrong
 $c_{1}e^{t} \left( \hspace{-2mm}  
\begin{array}{c} 1 \\ 0\\ \end{array} \hspace{-2mm} \right)$ +
$c_{2}e^{-t}\left( \hspace{-2mm}  
\begin{array}{c}-1\\ 0 \\ \end{array} \hspace{-2mm} \right)$

\wrong
 $c_{1} \left( \hspace{-2mm}  
\begin{array}{c} \cos\; t \\ \cos\; t \\ \end{array} \hspace{-2mm} \right)$ +
$c_{2}\left( \hspace{-2mm}  
\begin{array}{c}\sin\; t\\ \sin\; t \\ \end{array} \hspace{-2mm} \right)$

\wrong
 $c_{1} \left( \hspace{-2mm} 
 \begin{array}{c} e^{t} \cos\; t \\ -e^{t}\sin\; t \\ \end{array}
\hspace{-2mm} \right)$ + $c_{2}\left( \hspace{-2mm} 
 \begin{array}{clcr}e^{t}\cos\; t\\ e^{t} \sin\;t \\ \end{array}
\hspace{-2mm} \right)$

\wrong
 $c_{1}e^{t} \left( \hspace{-2mm}  
\begin{array}{c} 1 \\ 1 \\ \end{array} \hspace{-2mm} \right)$ +
$c_{2}e^{-t}\left( \hspace{-2mm}  
\begin{array}{c}0\\ 1\end{array} \hspace{-2mm} \right)$


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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 
 \problem 

Assume x' = Ax has a fundamental matrix
$\Psi(t) = \left( \hspace{-2mm}  \begin{array}{cc} (1 + t) & (2 - t) \\ (t - 2) & (1 +
t) \\ \end{array} \hspace{-2mm} \right).$  \, \, Note that 
$\Psi(0) \neq I.$ Assume $x(t)$ is the
solution of the initial value problem
 $x' = Ax$ and $x(0) = \left( \hspace{-2mm}  \begin{array}{c} 2 \\ 3 \\ \end{array}
\hspace{-2mm} \right).$ Calculate $x(t)$ at $t=2$.

\correct 
$x(2) = \frac {3}{5} \left( \hspace{-2mm}  \begin{array}{c} -4 \\ 7\\ \end{array} 
\hspace{-2mm} \right) $

\wrong $x(2) = \left( \hspace{-2mm} 
 \begin{array}{c} 6 \\ 9\\ \end{array} \hspace{-2mm} \right)$


\wrong $x(2) = \left( \hspace{-2mm}  
\begin{array}{c} 3 \\ 0\\ \end{array} \hspace{-2mm} \right)$

\wrong $x(2) = \left( \hspace{-2mm}  
\begin{array}{c} 8 \\ -1\\ \end{array} \hspace{-2mm} \right)$

\wrong $x(2) = \frac {3}{7} \left( \hspace{-2mm}  
\begin{array}{c} 2 \\ 3\\ \end{array} \hspace{-2mm} \right)$

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\problem
The matrix $A = \left( \hspace{-2mm} \begin {array}{cc} 1 & 2 \\
0 & -3 \\ \end{array} \hspace{-2mm} \right)$ has eigenvalues $1$ and $-3$ with corresponding
eigenvectors $\xi^{1} = \left( \hspace{-2mm}\begin{array}{c} 1 \\ 0\\ \end{array} \hspace{-2mm}
\right)$
 and
$\xi^{2} = \left( \hspace{-2mm}  
\begin{array}{c} 1 \\ -2 \\ \end{array} \hspace{-2mm} \right).$

Now consider the nonhomogeneous equation $x' = Ax + g(t)$ where $A$ is the above matrix and $g(t) = \left( \hspace{-2mm}  
\begin{array}{c} 0 \\ -2e^{-2t} \end{array} \hspace{-2mm} \right).$ 
 
Using the above information and the method of variation of parameters with $X_p = \Psi U$, the
equations satisfied by the components $u_1'$ and $u_2'$ of $U'$ are

\correct 
$\begin{array}{ccc} e^{t}u_1' + e^{-3t}u_2' &=& 0 \\  
e^{-3t}u_2' &=& e^{-2t}\\ \end{array}$

\wrong
$\begin{array}{ccc}e^{-t}u_1' + e^{-3t}u_2' &=& 0 \\
e^{-3t}u_2' &=& -2e^{-2t}\\ \end{array}$ 

\wrong
$\begin{array}{ccc}e^{t}u_1'+ e^{-3t}u_2' &=& 0 \\ 
3e^{-3t}u_2' &=& 0\\ \end{array}$   


\wrong
$\begin{array}{ccc}u_1' + e^{2t}u_2' &=& 0 \\
e^{-2t}u_2' &=& -2e^{2t'}\\ \end{array}$


\wrong
$\begin{array}{ccc}u_1' + u_2' &=& e^t \\
e^{-3t}u_2' &=& -2e^{-2t}\\ \end{array}$


 \end{document}



