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\begin{document}
\thispagestyle{plain}
\def\EJECT{\SeparatedPages{\vfill %Use this page to start your solution. Attach extra pages as needed, then staple.
\eject
\thispagestyle{plain}
%{\large\bf Name}. \underline{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}
\hfill{2280 Final Exam S2019}}}
\def\EJECTEND{\SeparatedPages{\vfill
%Use this page to start your solution. Attach extra pages as needed, then staple.
\eject}}
\noindent{\large\bf Name}.\underline{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~} \hfill{2280 Final Exam S2019}\\
\vspace*{-2pc}
\begin{center}\large\bf
Final Exam Differential Equations 2280\\
\small\sf Tuesday, 30 April 2019, 7:30am to 10:15am
\end{center}
\vspace*{-1ex}
\begin{minipage}{\textwidth}\small\sf
{\bf Instructions}: No
calculators, notes, tables or books. No answer check is expected.
A correct answer without details counts $25$\%.\end{minipage}

\subsection*{Chapters 1 and 2: First Order Differential Equations}

\begin{quote}\small\sf
{\bf Definitions}. An {\bf equilibrium solution} is a constant solution, found by replacing
all derivatives by zero, then solve for $y$. If $y$ found by this method is not constant, then the method fails. For $y'+py=q$, the {\bf homogeneous equation} is $y'+py=0$. An equation $y'=f(x,y)$ is {\bf separable} provided functions $F,G$ exist such that $f(x,y)=F(x)G(y)$.
\end{quote}

\vspace*{-4pc}
\begin{exercises}
\prob{}{} ~\newline

 {\bf (a)} [20\%] Apply a test to the equation $y'=x+y+y^2$, showing it fails to be separable.

\myspace{-1pc}\hrulefill
\myspace{16pc}

 {\bf (b)} [30\%] The problem $\dd x\frac{dy}{dx}=xy+3x+2y+6$ is both linear and separable. It can be solved by
 superposition $y=y_h+y_p$, where $y_h$ is the homogeneous solution and $y_p$ is an equilibrium solution. Show details for the answers $y_h=c\,x^2\,e^x$ and $y_p=-3$.
% de:=(x)^(2)*diff(y(x),x)=x*y(x)+y(x)+2*x+2; dsolve(de,y(x));



\myspace{-1pc}\hrulefill
\mynewpage


 {\bf (c)} [20\%]  Solve the linear homogeneous equation $\dd x^2\,\frac{dy}{dx} + 2y=xy$.
% de:=(x)^(2)*diff(y(x),x)+2*y(x)=x*y(x); dsolve(de,y(x));

\hrulefill
\myspace{16pc}

% de:=2*diff(v(t),t)=5*exp(-t)+(1/(2))*v(t); ic:=v(0)=0; 
% dsolve({de,ic},v(t));
 {\bf (d)} [30\%] Solve $\dd 2\,\frac{d}{dt}v(t)=5\,e^{-t}+ \frac{1}{2}\,v(t)$, $v(0)= 0$ by the linear integrating factor method. Show
 all steps.

\myspace{-1pc}\hrulefill

\end{exercises}
\EJECT
\vspace*{-3ex}


\subsection*{Chapter 3: Linear Equations of Higher Order}
\vspace*{-3ex}

\begin{exercises}
\prob{}{} ~\newline
{\bf (a)} ~[20\%] Solve for the general solution: $ y'' + 6y' + 73y=730$
% de:=diff(y(x),x,x)+6*diff(y(x),x)+73*y(x)=730; dsolve(de,y(x));

\myspace{-1pc}\hrulefill
\myspace{20pc}

\myspace{-1pc}\hrulefill

{\bf (b)} ~[30\%]  Given $5x''(t) + 2 x'(t) + 10x(t) = F_0\cos(\omega t)$, which represents a
damped forced spring-mass system with $m=5$, $c=2$, $k=10$, answer questions {\bf (c1), (c2)}.
  \begin{quote}\sf
    {\large\bf (c1)} Compute the frequency $\omega$ for practical mechanical resonance.
   % q:=subs(m=5,c=2,k=10,(k/m-c^2/m^2/2)); 48/75, omega=4/5 * sqrt(3)

\myspace{8pc}

   {\large\bf (c2)} Classify the homogeneous problem as over-damped (non-oscillatory), critically-damped or under-damped (oscillatory).
   % subs(m=5,c=2,k=10,(c^2-4*k*m)); -196

  \end{quote}

\mynewpage

{\bf (c)} ~[20\%] Define $y(x)=x\cos(3x)+3x^3e^x$. Construct the characteristic equation of a linear $n$th order homogeneous differential equation
of least order $n$ which has $y(x)$ as a particular solution.

\myspace{-1pc}\hrulefill\myspace{10pc}

{\bf (d)} [30\%] An $n$th order non-homogeneous differential equation is specified by its characteristic equation
$(r+1)^3(r^2+100)=0$ and the forcing term
$f(x)=x^2 + x^2\,e^{-x} + x\,e^{x} + \sin (10x)$. Find the shortest trial solution for
$y_p$ according to the method of undetermined coefficients. {\bf Do not
evaluate} undetermined coefficients.

\myspace{-1pc}\hrulefill
\end{exercises}
\EJECT


\vspace*{-3ex}
\subsection*{Chapters 4 and 5: Systems of Differential Equations}
\vspace*{-3ex}

\begin{exercises}
\prob{}{} ~\newline
 {\bf (a)} [20\%] Let
$A=\left(\begin{array}{rrr}
 0 & 2 &  1  \\
 2 & 0 &  1  \\
 0 & 0 & 5
 \end{array}\right)$, $\.v_1=\colvector{-1\\ 1 \\ 0}, \.v_2=\colvector{1\\ 1 \\ 0}, \.v_3=\colvector{1\\ 1 \\ 3}$.

 The eigenpairs of $A$ are
 $(-2,\.v_1), (2,\.v_2), (5,\.v_3)$.

 \begin{quote}\sf
   {\bf (a1)} Apply the Eigenanalysis Method to solve
 $\dd\frac{d}{dt}\.x(t)=A\.x(t)$.

\myspace{8pc}

  {\bf (a2)} Show details for computing eigenpair $(2,\.v_2)$.

   \begin{small} {\bf Expected}: Show linear algebra details for computing $\.v_2$ for eigenvalue $\lambda=2$. This involves row reduction plus display of the scalar solution and the vector solution. \end{small}
 \end{quote}
% with(LinearAlgebra):
% A:=Matrix([[0,2,1],[2,0,1],[0,0,5]]);Eigenvectors(A);

\mynewpage

{\bf (b)} [20\%] Find the scalar general solution of the $2\times 2$ system
$$
\left\{
\begin{array}{lcl}
x' &=& 7x+2y,\\
y' &=& 2x + 7y
\end{array}
\right. \quad\mbox{or}\quad\quad
\frac{d}{dt}\colvectorC{x(t)\\ y(t)}=\left(\begin{array}{rr}
 7 & 2  \\
 2 & 7
 \end{array}\right)\colvectorC{x(t)\\ y(t)}$$
  by the Cayley-Hamilton-Ziebur Method, using the textbook's Chapter 4 shortcut.
% des:=diff(x(t),t)=7*x(t)+2*y(t),diff(y(t),t)=2*x(t)+7*y(t); dsolve([des],[x(t),y(t)]);
% x(t) = -_C1*exp(5*t)+_C2*exp(9*t), y(t) = _C1*exp(5*t)+_C2*exp(9*t)

\myspace{-1pc}\hrulefill
\mynewpage

{\bf (c)} [30\%] Assume a $3\times 3$ system $\dd\frac{d}{dt}\.{u}=A\.{u}$ has a vector general solution
$$
\.u(t)=\colvector{u_1\\ u_2 \\ u_3} = 
\colvector{-c_1e^{5t} + c_2 e^{8t} \\ c_1e^{5t} + c_2e^{8t}\\ c_3e^t}.$$

\myspace{-1pc}\hrulefill
\begin{quote}\sf
 {\large\bf (c1)} Compute a $3\times 3$ fundamental matrix $\Phi(t)$.

\myspace{10pc}

 {\large\bf (c2)} Write a formula for the exponential matrix $e^{At}$ as an explicit matrix product. Do not multiply or simplify the product.

\myspace{8pc}

 {\large\bf (c3)} Let $\.c=\colvector{c_1\\c_2\\c_3}$. Display an explicit matrix-vector product for the solution $\.u(t)$ of the initial value problem  $\dd\frac{d}{dt}\.{u}=A\.{u}$, $\.u(0)=\.c$. Do not multiply or simplify the product.
\end{quote}

\mynewpage

{\bf (d)} [30\%] Consider the $3\times 3$ linear homogeneous system
$$\left\{\begin{array}{lcr}
x' &=&6x-2y \\
y' &=&-2x+6y, \\
z' &=& y+ z
\end{array}\right. \quad\mbox{~or~}\quad \frac{d}{dt}\.u(t) = \threebythree{6&-2&0}{-2&6&0}{0&1&1}\,\.u(t).$$
 Solve the system by the most efficient method.
 
 \myspace{-1pc}\hrulefill
% f1:=(x,y,z)->6*x-2*y;f2:=(x,y,z)->-2*x+6*y;f3:=(x,y,z)->y+z;
% des:=diff(x(t),t)=f1(x(t),y(t),z(t)),diff(y(t),t)=f2(x(t),y(t),z(t)),
% diff(z(t),t)=f3(x(t),y(t),z(t));
% dsolve([des],[x(t),y(t),z(t)]);

\end{exercises}
 \EJECT

\vspace*{-3ex}
 \subsection*{Chapter 6: Dynamical Systems}
\vspace*{-3ex}

\begin{exercises}
\prob{}{}~\newline

\vspace*{-1.5pc}
Consider the nonlinear dynamical system
\begin{equation}\label{nonlinear-system}
\left\{\begin{array}{lcl}
x' &=& 16x-4x^2-xy, \\
y'&=& 7y-y^2-xy
\end{array}\right.
 \end{equation}
%F:=(x,y)->16*x-4*x*x-1*x*y;G:=(x,y)->7*y-1*y*y-1*x*y; solve([F(x,y)=0,G(x,y)=0],[x,y]);
%[[x = 0, y = 0], [x = 0, y = 7], [x = 4, y = 0], [x = 3, y = 4]]
%J:=unapply(
%Matrix([[diff(F(x,y),x),diff(F(x,y),y)],
%        [diff(G(x,y),x),diff(G(x,y),y)]]),(x,y));
%J(x,y);
%J(0,0),J(0,7),J(4,0),J(3,4);evalf(Eigenvalues(J(3,4)));

\vspace*{-2ex}
 {\bf (a)} [20\%]  Find the four equilibrium points for nonlinear system (\ref{nonlinear-system}). One of these is $x=3,y=4$.
% [x = 0, y = 0], [x = 0, y = 7], [x = 4, y = 0], [x = 3, y = 4]

\myspace{-1pc}\hrulefill
\myspace{20pc}

{\bf (b)} [20\%]  Compute the Jacobian matrix $J(x,y)$ for nonlinear system (\ref{nonlinear-system}). Then
evaluate $J(x,y)$ at equilibrium point $x=3, y=4$.

\myspace{-1pc}\hrulefill
\mynewpage

 {\bf (c)} [30\%] Consider nonlinear system (\ref{nonlinear-system}).
 Classify the linearization at equilibrium point $x=3, y=4$ as a node,
 spiral, center, saddle. Do not sub-classify a node.

\myspace{-1pc}\hrulefill\myspace{24pc}

 {\bf (d)} [30\%] Consider nonlinear system (\ref{nonlinear-system}).
 Determine the possible classification of node, spiral, center or saddle and corresponding stability for equilibrium $x=3, y=4$
 according to the {\bf Pasting Theorem}, which is Theorem 2 in section 6.2 (Stability of Almost Linear Systems). State precisely the {\bf two exceptions} of the pasting theorem, then explain how the theorem applies to nonlinear system (\ref{nonlinear-system}) at $x=3, y=4$.

\myspace{-1pc}\hrulefill
\Answer{
}
\iffalse
Some maple code for checking the answers:
\begin{verbatim}
F:=unapply([16*x-2*x^2 -x*y,14*y-2*y^2-y*x ],(x,y));
Fx:=unapply(map(u->diff(u,x),F(x,y)),(x,y));
Fy:=unapply(map(u->diff(u,y),F(x,y)),(x,y));
Fx(0,0);Fy(0,0);Fx(0,7);Fy(0,7);Fx(8,0);Fy(8,0);Fx(6,4);Fy(6,4);
J:=unapply(Matrix([Fx(x,y),Fy(x,y)])^+,(x,y));
\end{verbatim}
\fi



\end{exercises}
\EJECT

\vspace*{-3ex}
\subsection*{Chapter 7: Laplace Theory}
\vspace*{-1ex}

Symbol $\delta(t)$ is the Dirac impulse. Symbol $u(t)$ is the unit step. Assumed below is experience with the following. Rules have precise hypotheses, omitted here for brevity.
\begin{small}
\begin{quote}\sf
Convolution Theorem. $\L(g_1)\L(g_2)=\L\left(\int_0^t g_1(t-x)g_2(x)dx\right)$

Periodic Function Theorem. $f(t+p)=f(t)$ implies
 $\L(f(t))=\dd\frac{\int_0^p f(t)dt}{1-e^{-ps}}$

Second Shifting Theorem Forward. $\L(g(t)u(t-a))=e^{-as}\L\left(\left. g(t)\right|_{t->t+a}\right)$

Second Shifting Theorem Backward. $e^{-as}\L(f(t))=\L(f(t-a)u(t-a))$

Dirac Impulse Identity.  $\int_{0}^\infty W(x)d u(t-a) = W(a)$. Formally $\delta(t) = du(t)$. Then $\L(\delta(t-a))=e^{-as}$ for $a\ge 0$.

Resolvent Identity. $\.u'=A\.u+\.F(t)$ has identity $(sI-A)\L(\.u)=\.u(0)+\L(\.F)$.

Exponential Order. This means $|f(x)|\le M\,e^{\alpha x}$ for some $M>0$ and real number $\alpha$.
\end{quote}
\end{small}

\begin{exercises}
\prob{}{}

  {\bf  (a)} [20\%] Let $f(t)$ be continuous and of exponential order. Define $F(s)=\L(f(t))$. Prove the {\bf Final Value Theorem}: $\lim_{s\to\infty} F(s)=0$ (succinctly $F(\infty)=0$).

\myspace{-1pc}\hrulefill
\mynewpage


  {\bf (b)} [20\%] Illustrate the convolution theorem by solving for $f(t)$ in the equation $\L(f(t)) = \dd\frac{1}{s}\,\frac{1}{s+1}$. Check the answer with partial fractions.

\hrulefill
\myspace{20pc}

  {\bf (c)}  [20\%] Solve for $f(t)$ using the second shifting theorem: $\L(f(t))=\dd e^{-2s}\frac{1}{s+1}$.

\hrulefill
\mynewpage

{\bf (d)}  [20\%] Symbol $\delta(t)$ is the Dirac impulse. Derive an expression for $\L(x(t))$ for the impulse problem \\ $$x''(t)+100x(t) = 5\delta(t-\pi),\quad x(0)=0,\quad x'(0)=1.$$
 To save time, do not solve for $x(t)$. 

\myspace{-1pc}\hrulefill
\mynewpage

 {\bf (e)} [20\%] {\bf Laplace Theory} applied to the forced linear dynamical system
$$\left\{\begin{array}{l}
 x'=\phantom{-}4x-2y+2t,\\
 y' = -2x+4y, \\
 x(0)=0, y(0)=0,
\end{array}\right.\quad\mbox{or}\quad
\left\{\begin{array}{l}
\.u' = \twobytwo{4&-2}{-2&4}\.u+\colvector{2t \\ 0},\\
\.u(0)=\colvector{0\\ 0}.
\end{array}\right.
$$
%Laplace theory subject to initial states $x(0)=0$, $y(0)=0$
produces the formulas
$$
\L(x(t)) = \frac{2s-8}{s^2(s-2)(s-6)},
\quad \L(y(t)) = \frac{-4}{s^2(s-2)(s-6)}.
$$
% A:=<4,-2|-2,4>^+;II:=<1,0|0,1>; 
% simplify((1/(s*II-A)) . <2/s^2,0>);factor(Determinant(s*II-A));

Display the {\bf Resolvent Method} solution steps that produce these formulas. To save time, {\bf do not solve for $x(t)$ or $y(t)$}.

\myspace{-1pc}\hrulefill
%\smallskip

%{\bf Expected}: \begin{small}\sf Define the coefficient matrix $A$ and forcing term $\.F$ in $\.u' = A\.u + \.F(t)$. Use the resolvent equation $(sI-A)\L(\.u)=\.u(0)+\L(\.F)$ and matrix inversion. \end{small}




\end{exercises}
\EJECT

\subsection*{Chapter 9: Fourier Series and Partial Differential Equations}

\begin{exercises}
\prob{}{}

\begin{quote}\sf
In part (a), let $f_0(x)=2$ on the interval $1<x<2$, $f_0(x)=0$ for  all other values of $x$ on $-2\le x \le 2$. Let $f(x)$ be the
periodic extension of $f_0$ to the whole real line, of period $4$. The Fourier series of $f(x)$ on $|x| \le L$ is

\centerline{$a_0 + \sum_{n=1}^\infty a_n\cos(n\pi x/L) + b_n\sin(n\pi x/L).$}

Formulas exist for $a_n$, $b_n$ expressed in terms of $f$, using inner product spaces.
\end{quote}

{\bf (a)} [20\%] Compute the Fourier coefficients $a_5$ and $b_5$ of $f(x)$ on $[-2,2]$.
Warning: $f$ is neither even nor odd.

\myspace{-1pc}\hrulefill
\mynewpage

\begin{quote}\sf
In part (b), let $g_0(x)=1$ on the interval $-2 < x < 0$, $g_0(x)=2$ on the interval $0<x<2$, $g_0(x)=0$ for  all other values of $x$ on $-2\le x \le 2$. Let $g(x)$ be the
periodic extension of $g_0$ to the whole real line, of period $4$.
\end{quote}

{\bf (b)} [10\%] Find all values of $x$ in $-3 < x < 5$ for which the Fourier series of $g$ will exhibit Gibb's over-shoot.

\myspace{-1pc}\hrulefill
\myspace{14pc}

{\bf (c)} [10\%] Assume $h(x)$ is a piecewise continuous function on $(-\infty,\infty)$. Let $H(x)$ be the Fourier series of $h(x)$ on $-L \le x \le L$. Does $H(0)=h(0)$ hold no matter the choice of $h$? Cite a theorem or invent a counterexample.

\myspace{-1pc}\hrulefill
\mynewpage

 {\bf (d)} [30\%] {\bf Heat Conduction in a Rod}.\\
   Let $L=2$ (rod length), $k=1$ (conduction constant). Solve the rod problem on
 $0\le x \le L$, $t\ge 0$:
 $$\left\{\begin{array}{lcl}
 \dd u_t &=& \dd k\,u_{xx},\\
 u(0,t) &=& 0,\\
 u(L,t) &=& 0,\\
 u(x,0)&=&\dd\sum_{n=1}^\infty \frac{1}{n+1}\sin(2n\pi x)
 \end{array}\right.
 $$

\hrulefill
\mynewpage

\begin{quote}\small\sf
{\bf DEFINITION}. The {\bf normal modes} for the string equation $u_{tt}=c^2 u_{xx}$ on $0<x<L$, $t>0$ are
given by the functions
$$
\sin\left( \frac{n\pi x}{L} \right)
\cos\left( \frac{n\pi ct}{L} \right),\quad
\sin\left( \frac{n\pi x}{L} \right)
\sin\left( \frac{n\pi ct}{L} \right)
.$$
Each normal mode is a solution of the string equation. A superposition of normal modes is also a solution of the string equation.
\end{quote}

{\bf (e)} [30\%] {\bf Vibration of a Finite String}.\\
 Let $L=4$ (string length), $c=4$ (wave speed). Solve the finite string vibration problem on $0 \le x \le L$, $t>0$:
$$
\left\{ \begin{array}{lcl}
 \dd u_{tt}(x,t) &=& \dd c^2\, u_{xx}(x,t), \\
  u(0,t) &=& 0, \\
  u(4,t) &=& 0, \\
 \dd u(x,0) &=& \sin(5\pi x) + 5\sin(7\pi x), \\
\dd  u_t(x,0) &=& \sin(7\pi x) + 12\sin(15\pi x).
 \end{array}\right.
$$

\myspace{-1pc}\hrulefill
\end{exercises}


\EJECTEND

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