
% extra credit ch7 2250, S2015
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{\large\sf Problem #1}. {\bf #2}\par}
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\begin{document}
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 {\bf Name~}\underline{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}\hfill
 {Class Time~}\underline{~~~~~~~~}\newline


\begin{center}
\bf Math 2250 Extra Credit Problems \\
\bf Chapter 7 \\
{\small\bf S2015}
\end{center}

{\bf Submitted work}. Please submit one stapled package with this sheet
on top. Kindly check-mark the problems submitted and
label the paper \fbox{\bf Extra Credit}. Label each solved problem with
its corresponding problem number, e.g., \fbox{Xc10.3-20}.

\Prob{Xc7.1-8}{(Transform to a first order system)}
 Use the position-velocity substitution $u_1=x(t)$, $u_2=x'(t)$,
 $u_3=y(t)$, $u_4=y'(t)$ to transform the system below into
 vector-matrix form $\.u'(t)=A\.u(t)$. Do not attempt to solve the
 system.

 $$ x''-2x'+5y=0,\quad y'' + 2y' -5x=0.$$

\Prob{Xc7.1-20a}{(Dynamical systems)}
 Prove this result for system
 \begin{equation}\label{ch7.xc.dynamical-system}
 \begin{array}{lcrcrcr}
  x' &=& ax &+& by, \\
  y' &=& cx &+& dy.
 \end{array}
 \end{equation}

\begin{quote}\sf
\hrule

 \BF{Theorem}.  Let $A=\twobytwo{a&b}{c&d}$ and define
 ${\bf trace}(A)=a+d$. Then $p_1=-{\bf trace}(A)$, $p_2=\det(A)$ are the coefficients
 in the determinant expansion
  $$\det(A-rI)=r^2+p_1r+p_2$$
 and $x(t)$ and $y(t)$ in equation (\ref{ch7.xc.dynamical-system}) are both solutions of the differential
 equation $u''+p_1u'+p_2u=0$.

 \hrule
 \end{quote}

\Prob{xC7.1-20b}{(Solve dynamical systems)}

 (a) Apply the previous problem to solve
 \begin{displaymath}
  \begin{array}{lcrcrcr}
  x' &=& 2x &-& y, \\
  y' &=& x &+& 2y.
   \end{array}
 \end{displaymath}

 (b) Use first order methods to solve the system
 \begin{displaymath}
  \begin{array}{lcrcrcr}
  x' &=& 2x &-& y, \\
  y' &=&    & & 2y.
   \end{array}
 \end{displaymath}


\Prob{Xc7.2-12}{(General solution answer check)}
 (a) Verify that $\.x_1(t)=e^{3t}\colvector{1\\ -1}$ and
 $\.x_2(t)=e^{2t}\colvector{1\\ -2}$ are solutions of $\.x'=A\.x$,
 where
 $$A=\twobytwo{4&1}{-2&1}.$$

 (b) Apply the Wronskian test $\det(\AUG(\.x_1,\.x2))\ne 0$ to
 verify that the two solutions are independent.

 (c) Display the general solution of $\.x'=A\.x$.


\vfill {\bf Extra credit problems chapter 7 continue on the next
page.} \eject

\Prob{Xc7.2-14}{(Particular solution)}

 (a) Find the constants $c_1$, $c_2$ in the general solution
 $$ \.x(t)=c_1e^{3t}\colvector{1\\ -1} + c_2e^{5t}\colvector{1\\
 -3}$$
 satisfying the initial conditions $x_1(0)=4$, $x_2(0)=-1$.

  (b) Find the matrix $A$ in the equation $\.x'=A\.x$. Use the
  formula $AP=PD$ and Fourier's model for $A$, which is given
  implicitly in (a) above.

\Prob{Xc7.3-8}{(Eigenanalysis method $2\times 2$)}

 (a) Find $\lambda_1$, $\lambda_2$, $\.v_1$, $\.v_2$ in Fourier's
 model
 $A\left(c_1\.v_1+c_2\.v_2\right)=c_1\lambda_1\.v_1+c_2\lambda_2\.v_2$
 for
 $$A=\twobytwo{3&-4}{4&3}.$$

 (b) Display the general solution of $\.x'=A\.x$.

\Prob{Xc7.3-20}{(Eigenanalysis method $3\times 3$)}

(a) Find $\lambda_1$, $\lambda_2$, $\lambda_3$, $\.v_1$, $\.v_2$,
$\.v_3$ in Fourier's
 model
 $A\left(c_1\.v_1+c_2\.v_2+c_3\.v_3\right)=c_1\lambda_1\.v_1+c_2\lambda_2\.v_2+c_3\lambda_3\.v_3$
 for
 $$A=\threebythree{2&1&-1}{-4&-3&-1}{4&4&2}.$$

 (b) Display the general solution of $\.x'=A\.x$.

\Prob{Xc7.3-30}{(Brine Tanks)}
 Consider two brine tanks satisfying the equations
 $$
 x_1'(t) = -k_1x_1+k_2x_2,\quad x_2'=k_1x_1-k_2x_2.
 $$
 Assume $r=10$ gallons per minute, $k_1=r/V_1$, $k_2=r/V_2$, $x_1(0)=30$ and $x_2(0)=0$. Let the tanks have volumes
 $V_1=50$ and $V_2=25$ gallons. Solve for $x_1(t)$ and $x_2(t)$.


\Prob{Xc7.3-40}{(Eigenanalysis method $4\times 4$)}
 Display (a) Fourier's model and (b) the general solution of
 $\.x'=A\.x$ for the $4\times 4$ matrix
 $$A=\left(\begin{array}{rrrr}
  2 & 0 & 0 & 0 \\
   -21 & -5 & -27 & -9 \\
    0 & 0 & 5 & 0 \\
     0 & 0 & -16 & -4
  \end{array}\right).
 $$
%A:=matrix([[ 2 , 0 , 0 , 0],
%    [-21 , -5 , -27 , -9],
%    [ 0 , 0 , 5 , 0],
%    [  0 , 0 , -16 , -4]]);
%
% [5, 1, {vector([0, -11/10, 1, -16/9])}],
% [2, 1, {vector([1, -3, 0, 0])}],
% [-5, 1, {vector([0, 1, 0, 0])}],
% [-4, 1, {vector([0, -9, 0, 1])}]

 \vfill {\bf End of extra credit problems chapter 7.}

\end{document}
