% 2019 Differential Equations 2280-1
\documentclass[12pt,leqno]{article}
\usepackage{linalg}
\setlength{\oddsidemargin}{0in} \setlength{\evensidemargin}{0in}
\setlength{\textwidth}{6.4in} \setlength{\textheight}{9.5in}
\setlength{\parskip}{5pt} \setlength{\topmargin}{-90pt}
\setlength{\parindent}{0pt}
%\newenvironment{exercises}{\begin{list}{}{}}{\end{list}}
\newenvironment{solution}[1]{\protect\par{\bf Solution to Problem #1}.~\small\rm}{\protect\par\normalsize\rm}
%\long\def\answer#1{\begin{quote}\sf {\large\sf Answer:}\par #1\end{quote}}
\def\SeparatedPages#1{#1}
%\def\SeparatedPages#1{}
\def\prob#1#2{\item[{\normalsize\bf #1}] {\bf #2}}
\def\L{{\cal L}}
\def\dd{\displaystyle}
\def\BF#1{{\large\bf #1}}
\def\.#1{\vec{\mbox{\bf #1}}}
\def\minicolvector#1{\left(\begin{array}{@{}r@{}}#1\end{array}\right)}
\def\minicolvectorC#1{\left(\begin{array}{@{}c@{}}#1\end{array}\right)}
\newif\ifsolutions
\solutionstrue
%\solutionsfalse
\ifsolutions
%\def\Answer#1{\rule{0.2\linewidth}{0.05em}\\ {\bf Answer}: {\sf #1}}
\def\Answer#1{\relax}
%\par\rule{\linewidth}{0.1em}}
\def\SeparatedPages#1{\relax}
\def\myspace#1{\relax}
\def\mynewpage{\relax}
\else
\def\Answer#1{\relax}
\def\myspace#1{\vspace*{#1}}
\def\mynewpage{\newpage}
\fi

\begin{document}
\thispagestyle{empty}
%{\large\bf Name}
%\underline{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}\hfill
%{{\bf Time of your class}~\underline{~~~~~~~~~~} }
\def\EJECT{\SeparatedPages
{\vfill\newpage
\thispagestyle{empty} \hfill  2280 Exam 2 S2019
%\newline{\large\bf Name}.
%\underline{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}\hfill
}
}
%%% {\bf Time of your class}~\underline{~~~~~~~~~~} }
\def\EJECTEND{\SeparatedPages{\vfill %Use this page to start your solution.
%Attach extra pages as needed, then staple.
\eject}}
%\def\EJECT{}\def\EJECTEND{}

\begin{center}\Large\bf
Differential Equations 2280 \\
\ifsolutions
\small Midterm Exam 2 with Solutions\\
\else
\small Midterm Exam 2\\
\fi
\small Exam Date: 5 April 2019 at 7:30am
\end{center}

{\bf Instructions}: This in-class exam is 80 minutes. No
calculators, notes, tables or books. No answer check is expected.
Details count 3/4, answers count 1/4. 

\section*{Laplace Theory: Chapter 7}
{\bf Laplace Theory Background}. Expected without notes or books is the $4$-entry Forward Laplace Table and these rules: Linearity, Parts, Shift, $s$-differentiation, Lerch's Theorem, Final Value Theorem, Existence theory.

\begin{exercises}
\prob{1.}{(Laplace Theory: Differential Equations)} ~\newline
{\bf Solve differential equations (1a), (1b) by Laplace's method}.

%\BF{(1a)} [30\%] Second order
%$x'' + 4x' + 4x = e^{-t}$, $x(0)=x'(0)=0$.

%de1:=diff(x(t),t,t)+4*diff(x(t),t)+4*x(t)=exp(-t);
%ic1:=x(0)=0,D(x)(0)=0;dsolve([de1,ic1],x(t));

 \BF{(1a)} [40\%] Third order $x'''+x'=0$, $x(0)=1$, $x'(0)=0$, $x''(0)=0$.

% de2:=diff(x(t),t,t,t)+4*diff(x(t),t)=8;
% ic2:=x(0)=1,D(x)(0)=0,D(D(x))(0)=0;dsolve([de2,ic2],x(t));
\mynewpage

  \BF{(1b)} [60\%] Dynamical system $x'=x+y$, $y'=2x+6$, $x(0)=0$, $y(0)=0$.


% de3:=diff(x(t),t)=x(t)+y(t),diff(y(t),t)=2*x(t)+6;
% ic3:=x(0)=0,y(0)=0;dsolve([de3,ic3],[x(t),y(t)]);


\Answer{~\newline
In progress
}
\end{exercises}
\EJECT

\begin{exercises}
\prob{2.}{(Laplace Theory: Backward Table)} ~\\
{\bf Solve for $f(t)$} in {\bf (2a), (2b), (2c), (2d)}.
\begin{itemize}\sf
\item[] {\large\bf Assumptions}. Below, $f(t)$ is of piecewise continuous of exponential order. Expression $u(t)$ denotes the unit step function.

\item[] {\large\bf Credit}. Document all steps, e.g., if you cancel $\L$ then cite Lerch's Theorem. The answer is 25\%. The documented steps are 75\%. Partial fraction
coefficients are expected to be evaluated last.
\end{itemize}


\BF{(2a)} [20\%] $\L(f(t))=\dd\frac{100}{(s^2+25)(s^2+4)}$

\myspace{25pc}

 \BF{(2b)} [20\%] $\L(f(t))= \dd \frac{s+1}{s^2+4s+29}$.

\mynewpage

\BF{(2c)} [30\%] $\L(f(t))=\dd\frac{1}{(s^2+2s)(s^2-3s)}$

\myspace{25pc}

\BF{(2d)} [30\%] $\dd\left.\frac{d}{ds}\L(f(t))\right|_{s \to (s-3)}=\dd\frac{d}{ds}\L(u(t-\pi)e^t\cos 25t)$.\\


\Answer{~\newline
In progress
}
\end{exercises}
\EJECT
\begin{exercises}
\prob{3.}{(Laplace Theory: Forward Table)} ~\newline
{\bf Compute the Laplace transform $\L(f(t))$}.

  \BF{(3a)} [20\%] $f(t)=(-t)e^{2t}\sin(3t)$.

\myspace{15pc}

  \BF{(3b)} [30\%] $f(t))=\dd e^{-\pi t}g(t)$ and
  $g(t)=\dd \frac{e^{2t}-e^{-2t}}{t}$.

\mynewpage

  \BF{(3c)} [20\%] $\dd\frac{d}{ds}\L(f(t))=\dd\frac{1}{s^{2019}} + \frac{1}{s^2+1}$.\\
   %{\small\sf The final value theorem or $s$-differentiation applies}.

\myspace{25pc}\
  \BF{(3d)} [30\%] Define $f(t)=e^{-2t}g(t))$ where
  $g(t)=\dd\frac{e^{2t}-e^{-2t}}{t}$.

\Answer{~\newline
In progress
}
\end{exercises}
\EJECT


\section*{Systems of Differential Equations: Chapters 4 and 5}


\BF{Background}.
\begin{itemize}
\item[] The \BF{Eigenanalysis Method} for a real $3\times 3$ matrix $A$ assumes eigenpairs
 $(\lambda_1,\.v_1)$,  $(\lambda_2,\.v_2)$,  $(\lambda_3,\.v_3)$.
 It says that the $3\times 3$ system
$\.x' = A\.x$ has general solution
$\.x(t)=c_1\.v_1e^{\lambda_1
t}+c_2\.v_2e^{\lambda_2 t}+c_3\.v_3e^{\lambda_3 t}$.

\item[]  The  \BF{Cayley-Hamilton-Ziebur method} is based upon this result:
\begin{quote}\sf
Let $A$ be an $n\times n$ real matrix. The components of solution $\.u$ of $\frac{d}{dt}\,\.u(t)=A\.u(t)$ are linear combinations of Euler solution atoms obtained from the roots of the characteristic equation $|A-\lambda I|=0$. Alternatively, $\.u(t)$ is a vector linear combination of the Euler solution atoms: $\.u(t)=\sum_{k=1}^n (\mbox{\em atom}_k)\.d_k$.
\end{quote}
\item[]  A
 \BF{Fundamental Matrix} is an $n\times n$ matrix $\Phi(t)$ with columns consisting of independent solutions of $\.x'(t)=A\.x(t)$, where $A$ is an $n\times n$ real matrix. The general solution of $\.x'(t)=A\.x(t)$ is $\.x(t)=\Phi(t)\.c$, where $\.c$ is a column vector of arbitrary constants $c_1, \ldots, c_n$. An alternate and widely used definition of fundamental matrix is  $\Phi'(t) = A\Phi(t)$, with $|\Phi(0)| \ne 0$ required to establish independence of the columns of $\Phi$.

\item[]
The \BF{Exponential Matrix}, denoted $e^{At}$, is the unique fundamental matrix $\Psi(t)$ such that $\Psi(0)=I$.  Matrix $A$ is an $n\times n$ real matrix. It is known that $e^{At}=\Phi(t)\Phi(0)^{-1}$ for {\em any} fundamental matrix $\Phi(t)$. Consequently, $\frac{d}{dt}\left(e^{At}\right)=A\,e^{At}$ and $\left. e^{At}\right|_{t=0}=I$.
\end{itemize}
\EJECT
\begin{exercises}
\prob{4.}{(Systems: Eigenanalysis Method)} ~\newline
Complete parts {\bf ((4a), (4b), (4c)}.

 \BF{(4a)} [40\%] Let
$A=\left(\begin{array}{rrr}
 5 & 1 &  1  \\
 1 & 5 &  1  \\
 0 & 0 &  7
 \end{array}\right)$.
 Display the linear algebra details for computing the three eigenpairs of $A$.

 \mynewpage


 \BF{(4b)} [30\%]
%with(LinearAlgebra):
%B:=Matrix([[4,1,-1,0],[1,4,-2,1],[0,0,2,0],[0,0,2,4]]);
%Eigenvalues(B);
Matrix $A=\left(\begin{array}{rrrr}
 4 & 1 & -1 & 0 \\
 1 & 4 & -2 & 1 \\
 0 & 0 &  2 & 0 \\
 0 & 0 &  2 & 4
 \end{array}\right)$ has eigenpairs
$\left(2,\minicolvector{ 1\\ -5\\ -3 \\ 3}\right)$,
$\left(3,\minicolvector{-1\\ 1\\ 0 \\ 0}\right)$,
$\left(4,\minicolvector{-1\\ 0\\ 0 \\ 1}\right)$,
$\left(5,\minicolvector{1 \\ 1 \\ 0 \\ 0}\right)$.
Display the general solution of $\dd\frac{d}{dt}\,\.u(t)=A\.u(t)$.

\myspace{10pc}


\BF{(4c)} [40\%] Let $\.u(t)=\.c\, e^{rt}$, the vector Euler substitution.
Assume real $n\times n$ matrix $A$ has a real eigenpair
$(\lambda,\.v)$. Prove that the Euler substitution with $r=\lambda$ and
$\.c = \.v$
applied to $\dd\frac{d}{dt}\,\.u(t)=A\.u(t)$ finds a nonzero solution of $\dd\frac{d}{dt}\,\.u(t)=A\.u(t)$.

\Answer{~\newline
In progress
}
\end{exercises}
\EJECT

\begin{exercises}
\prob{5.}{(Systems: First Order Cayley-Hamilton-Ziebur)}
\newline
\BF{(5a)} [30\%]  The eigenvalues are $2$, $6$ for the matrix
$A=\left(\begin{array}{rrrr}
 4 & 2   \\
 2 & 4
 \end{array}\right)$.
%with(LinearAlgebra):
%C:=Matrix([[4,2],[2,4]]); Eigenvectors(C);

Display the general solution of $\dd\frac{d}{dt}\,\.u(t)=A\.u(t)$ according to the Cayley-Hamilton-Ziebur shortcut (textbook chapters 4,5).

\mynewpage

\BF{(5b)} [40\%]  The $3\times 3$ triangular matrix
$A=\left(\begin{array}{rrr}
 2 & 1 &  1  \\
 0 & 5 &  1  \\
 0 & 0 &  -2
 \end{array}\right)$
 represents a linear cascade, such as found in brine tank models.
 Apply the {\bf linear integrating factor method} and related shortcuts (Ch 1 in the textbook) to find the components $x_1,x_2,x_3$ of the vector general solution $\.x(t)$ of $\dd\frac{d}{dt}\,\.x(t)=A\.x(t)$.
 
\mynewpage

   \BF{(5c)}  [30\%] The Cayley-Hamilton-Ziebur shortcut  applies to the system
   $$ x' = 3x + 2y,\quad y' = -2x + 3y,$$
   which has complex eigenvalues $\lambda = 3\pm 2i$. Find a fundamental matrix $\Phi(t)$ for this system, documenting all details of the computation.
   
% A:=<3,2|-2,3>^+;with(LinearAlgebra):Eigenvectors(A);   
\mynewpage


   
\Answer{~\newline
In progress
}  
 \end{exercises}
 \EJECT
 
 \begin{exercises}
\prob{6.}{(Systems: Second Order Cayley-Hamilton-Ziebur)}
\newline
% [[-13, -6], [6, -28]]
Assume below that real $2\times 2$ matrix $A=\twobytwo{-13&-6}{6&-28}$ has eigenpairs $\left(-25,\minicolvector{1 \\ 2} \right)$, $\left( -16,\minicolvector{2\\ 1}\right)$. Textbook theorems applied to $\dd\frac{d^2}{dt^2}\,\.u(t)=A\.u(t)$ report general solution
$$\.u(t) = (c_1\cos(5t)+c_2\sin(5t)\minicolvector{1\\ 2}
+ (c_3\cos(4t)+c_4\sin(4t))\minicolvector{2\\ 1}.$$

\BF{(6a)} [30\%] Derive the characteristic equation $|A-r^2I|=0$ for the second order equation $\dd\frac{d^2}{dt^2}\,\.u(t)=A\.u(t)$ from Euler's vector substitution $\.u(t)=\.c\,e^{rt}$. A proof is expected with details.\footnote{
{\bf Reminder}: Linear algebra writes eigenpair equation $A\.x=\lambda \.x$ equivalently as $A\.x=\lambda I \.x$ and then converts it to the homogeneous system of linear algebraic equations $(A-\lambda I)\.x=\.0$. The proof you write should apply without edits to $n\times n$ real matrices $A$.}

\mynewpage

\BF{(6b)} [40\%]
Substitute $\.u(t)=\.d\,\cos(5t)$ into $\dd\frac{d^2}{dt^2}\.u(t)=A\.u(t)$ to determine vector $\.d$ in terms of eigenpairs of $A$. Repeat for $\.u(t)=\.d\,\sin(5t)$ and report the answer.

\mynewpage

\BF{(6c)} [30\%]
The Euler solution atoms $\cos(5t),\sin(5t),\cos(4t),\sin(4t)$ are linearly independent on $(-\infty,\infty)$. Substitute $\.u(t)=\.d_1\cos(5t)+\.d_2\sin(5t)+\.d_3\cos(4t)+\.d_4\sin(4t)$ into $\dd\frac{d^2}{dt^2}\,\.u(t)=A\.u(t)$ and use independence (vector coefficients of atoms match) to determine the vectors $\.d_1,\.d_2,\.d_3,\.d_4$ in terms of eigenpairs of $A$.

\Answer{~\newline
In progress
}
\end{exercises}
\EJECTEND

\end{document}
\endinput
