\documentclass[12pt]{article}
\usepackage{linalg}
\setlength{\oddsidemargin}{0in} \setlength{\evensidemargin}{0in}
\setlength{\textwidth}{6.4in} \setlength{\textheight}{9in}
\setlength{\parskip}{5pt} \setlength{\topmargin}{-1in}
\setlength{\parindent}{0pt}
\newenvironment{solution}[1]{\protect\par{\bf Notes on Problem #1}.~\small\rm}{\protect\par\normalsize\rm}
%\long\def\answer#1{\begin{quote}\sf {\large\sf Answer:}\par #1\end{quote}}
\long\def\answer#1{}
%\def\SeparatedPages#1{#1}
\def\SeparatedPages#1{}
\def\prob#1#2{\item[{\normalsize\bf #1}] {\bf #2}}
\def\L{{\cal L}}
\def\dd{\displaystyle}
\newif\ifsolutions
\solutionsfalse

\begin{document}
\def\EJECT{\SeparatedPages{\vfill Use this page to start your solution.
Attach extra pages as needed, then staple.\eject
\thispagestyle{empty} {\large\bf Name}.
\underline{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~} \hfill{2280
Final Exam Sample}}}
\def\EJECTEND{\SeparatedPages{\vfill Use this page to start your solution.
Attach extra pages as needed, then staple.\eject}}

\begin{center}\Large\bf
Applied Differential Equations 2280\\
Sample Final Exam\\
Tuesday, 2 May 2006, 4:30-8:00pm \\
\end{center}

{\bf Instructions}: This in-class exam is 50 minutes. No
calculators, notes, tables or books. No answer check is expected.
Details count $75$\%. The answer counts $25$\%.

\EJECT

\begin{exercises}
\prob{1.}{(Quadrature Equation)} ~\newline
        Solve for the general solution $y(x)$ in the equation
 $\dd y'=
 2\cot x
 + \frac{1250 x^{3}}{1+25x^{2}}
 + x\ln(1+x^2)$.
 %  f:=1250*x^(3)/(1+25*x^(2));

\iftrue {\small [The required integration talent includes
 basic formulae, integration by parts, substitution and college algebra.]
}\fi

\end{exercises}
\EJECT


\begin{exercises}
\prob{2.}{(Separable Equation Test)} ~\newline
 The problem $y'=f(x,y)$ is said to be separable provided
$f(x,y)=F(x)G(y)$ for some functions $F$ and $G$.

 (a) [75\%]  Check (\fbox{X}) the problems that can be put into separable form,
 but don't supply any details.
\begin{center}\large\begin{tabular}{|lp{0.45\linewidth}|lp{0.3\linewidth}|}
\hline
 \fbox{$\phantom{X}$} & $ y'=-y(2xy+1)+(2x+3)y^2$ & \fbox{$\phantom{X}$} &   $yy'=xy^2+5x^2y$ \\[2ex]  \hline
  \fbox{$\phantom{X}$} &  $y'=e^{x+y}+e^y$ & \fbox{$\phantom{X}$} &  $3y'+5y=10y^2$ \\[2ex] \hline
  \end{tabular}
\end{center}

 (b) [25\%] State a test which can verify that an equation is not separable.
 Use the test to verify that $y'=x+\sqrt{|xy|}$ is not separable.

\end{exercises}
\EJECT

\begin{exercises}
\prob{3.}{(Solve a Separable Equation)} ~\newline
 % u:=(2*x^2+4*x)/(1+x); int(u,x);
 Given $\dd y^2y' = \frac{2x^2+3x}{1+x^2}\left(\frac{125}{64}-y^3\right)$. ~\newline
 (a) Find all equilibrium solutions. ~\newline
 (b) Find the non-equilibrium solution in implicit form. ~\newline
 To save time, {\bf do not solve} for $y$ explicitly.

\end{exercises}
\EJECT

\begin{exercises}
\prob{4.}{(Linear Equations)} ~\newline
% de:=2*diff(v(t),t)=-32+(2/(3*t+1))*v(t); ic:=v(0)=-8; dsolve({de,ic},v(t));
 (a) [60\%] Solve $\dd 2v'(t)=-32 + \frac{2}{3t+1}v(t)$, $v(0)=-8$. Show
 all integrating factor steps.\newline
 (b) [30\%]  Solve $\dd 2\sqrt{x+2}\,\frac{dy}{dx}=y$. The answer contains symbol $c$.
\newline
 (c) [10\%] The problem $\dd 2\sqrt{x+2}\,y'=y-5$ can be solved using
 the answer $y_h$ from part (b) plus superposition $y=y_h+y_p$. Find $y_p$. Hint:
 If you cannot write the answer in a few seconds, then return here after
 finishing all problems on the exam.

\end{exercises}
\EJECT

\begin{exercises}
\prob{5.}{(Stability)} ~\newline
 (a) [50\%] Draw a phase line diagram for the differential equation
 $$dx/dt = 1000\left(2-\sqrt[5]{x}\right)^3(2+3x)(9x^2-4)^{8}.$$
 Expected in the diagram are equilibrium
 points and signs of $x'$ (or flow direction markers $<$ and
 $>$).\newline
 (b) [40\%] Draw a phase diagram using the phase line diagram of (a). Add these labels
 as appropriate: funnel, spout, node, source, sink, stable, unstable. Show at
 least 8 threaded curves. A direction field is not expected or required.
 \newline
 (c) [10\%] Outline how to solve for non-equilibrium solutions, without doing any
  integrations or long details.
\end{exercises}
\EJECT



\begin{exercises}
\prob{6.}{(ch3)} ~\newline (a) Using the {\em recipe} for higher
order constant-coefficient differential equations, write out the
general solutions:

 $ \begin{array}{ll} {\bf (a.1)} ~[25\%] & y'' + 4y' + 4y=0 \end{array}$,

 $\begin{array}{ll} {\bf (a.2)} ~[25\%] & y^{vi} + 4y^{iv} = 0
 \end{array}$,

$\begin{array}{ll} {\bf (a.3)} ~[25\%] & \mbox{Char. eq. ~}
 r(r-3)(r^3-9r)^2(r^2+4)^3=0
 \end{array}$.

(b)  Given $6x''(t) + 7x'(t) + 2x(t) = 0$, which represents a
damped spring-mass system with $m=6$, $c=7$, $k=2$, solve the
differential equation [15\%] and classify the answer as
over-damped, critically damped or under-damped [5\%]. Illustrate
in a physical model drawing the meaning of constants $m$, $c$, $k$
[5\%].

\ifsolutions
\begin{solution}{6} ~\newline
Part (a)\newline
 {\bf 1}: $r^2+r+1=0$, $y=c_1y_1+c_2y_2$, $y_1=e^{-x/2}\cos(\sqrt3
x/2)$, $y_2=e^{-x/2}\sin(\sqrt3 x/2)$.\newline
 {\bf 2}: $r^{iv}+4r^2=0$, roots $r=0,0,2i,-2i$. Then
 $y=(c_1+c_2x)e^{0x}+c_3\cos 2x + c_4\sin 2x$. \newline
 {\bf 3}: Write as $(r-a)^2(r+a)^2(r^2+16)^3=0$ where $a=\sqrt3$.
 Then $y=u_1e^{ax}+u_2e^{-ax}+u_3\cos 4x + u_5\sin 3x$. The
 polynomials are $u_1=c_1+c_2x$, $u_2=c_3+c_4x$,
 $u_3=c_5+c_6x+c_7x^2$, $u_4=c_8+c_9x+c_{10}x^2$.

Part (b)\newline Use $4r^2+4r+1=0$ and the quadratic formula to
obtain roots $r=-1/2,-1/2$. Case 2 of the recipe gives
$y=(c_1+c_2t)e^{-t/2}$. This is critically damped. The
illustration shows a spring, dampener and mass with labels $k$,
$c$, $m$, $x$ and the equilibrium position of the mass.
\end{solution}
\fi

\end{exercises}
\EJECT

\begin{exercises}
\prob{7.}{(ch3)} ~\newline Determine for $ y^{vi} + y^{iv} =
x+2x^2+ x^3 + e^{-x}+x\sin x$ the corrected trial solution for
$y_p$ according to the method of undetermined coefficients. Do not
evaluate the undetermined coefficients!

\ifsolutions
\begin{solution}{7} ~\newline
The homogeneous solution is $y_h=c_1+c_2x+c_3e^{3x}+c_4e^{-3x}$,
because the characteristic polynomial has roots $0$, $0$, $3$,
$-3$.

{\bf \fbox{1}~} An initial trial solution $y$ is constructed for
atoms $1$, $x$, $e^{3x}$, $e^{-3x}$, $\cos x$, $\sin x$ giving
$$\begin{array}{lcl}
 y &=& y_1+y_2+y_3+y_4,\\
 y_1 &=& (d_1+d_2x)e^{3x},\\
 y_2 &=& d_3+d_4x+d_5x^2+d_6x^3,\\
 y_3 &=& d_7e^{-3x},\\
 y_4 &=& d_8\cos x + d_9\sin x.
\end{array}
$$
Linear combinations of the listed independent atoms are supposed
to reproduce, by assignment of constants, all derivatives of the
right side of the differential equation.

{\bf \fbox{2}~} The fixup rule is applied individually to each of
$y_1$, $y_2$, $y_3$, $y_4$ to give the \BF{corrected trial
solution}
$$\begin{array}{lcl}
y &=& y_1+y_2+y_3,\\
 y_1 &=& x(d_1+d_2x)e^{3x},\\
 y_2 &=& x^2(d_3+d_4x+d_5x^2+d_6x^3),\\
 y_3 &=& x(d_7e^{-3x}),\\
 y_4 &=& d_8\cos x + d_9\sin x.
\end{array}
$$
The powers of $x$ multiplied in each case are designed to
eliminate terms in the initial trial solution which duplicate
atoms appearing in the homogeneous solution $y_h$. The factor is
exactly $x^s$ of the Edwards-Penney table, where $s$ is the
multiplicity of the characteristic equation root $r$ that produced
the related atom in the homogeneous solution $y_h$. By design,
unrelated atoms are unaffected by the fixup rule, and that is why
$y_4$ was unaltered.

{\bf \fbox{3}~} Undetermined coefficient step skipped, according
to the problem statement.

{\bf \fbox{4}~} Undetermined coefficient step skipped, according
to the problem statement.


\end{solution}
\fi

\end{exercises}
\EJECT

\begin{exercises}
\prob{8.}{(ch3)} ~\newline (a) [50\%] Find by undetermined
coefficients the steady-state periodic solution for the equation $
x'' + 4x' + 6x = 10\cos(2t)$.
\newline
(b) [50\%] Find by variation of parameters a particular solution
$y_p$ for the equation $y''+3y'+2y=xe^{2x}$.

\ifsolutions
\begin{solution}{8} ~\newline
Solve $x''+2x'+6x=0$ by the recipe to get $x_h=c_1x_1+c_2x_2$,
$x_1=e^{-t}\cos\sqrt5 t$, $x_2=e^{-t}\sin\sqrt5 t$.  Compute the
Wronskian $W=x_1x_2'-x_1'x_2=\sqrt5 e^{-2t}$. Then for
$f(t)=5\cos(3t)$,
$$
 x_p=x_1\int x_2\frac{-f}{W}dt+x_2\int x_1\frac{f}{W}dt.
 $$
 The integrations are horribly difficult, so the
 method of choice is undetermined coefficients.

The trial solution is $x=d_1\cos 3t + d_2\sin 3t$. Substitute the
trial solution to obtain the answers $d_1=-1/3$, $d_2=2/3$. The
unique periodic solution $x_{\mbox{ss}}$ is extracted from the
general solution $x=x_h+x_p$ by crossing out all negative
exponential terms (terms which limit to zero at infinity). If
$x_p=d_1\cos 3t + d_2\sin 3t = (1/3)(-\cos 3t + 2\sin 3t)$, then
$$
x_{\mbox{ss}}=\frac{-1}{3}\cos 3t + \frac{2}{3}\sin 3t.
$$
\end{solution}
\fi

\end{exercises}
\EJECT

\begin{exercises}
\prob{9.}{(ch5)} ~\newline
 The eigenanalysis method says that the system
$\.x' = A\.x$ has general solution $\.x(t)=c_1\.v_1e^{\lambda_1
t}+c_2\.v_2e^{\lambda_2 t}+c_3\.v_3e^{\lambda_3 t}$. In the
solution formula, $(\lambda_i,\.v_i)$, $i=1,2,3$, is an eigenpair
of $A$.  Given
$$A=\left[\begin{array}{rrr}
 5 & 1 &  1  \\
 1 & 5 &  1  \\
 0 & 0 &  7
 \end{array}\right],$$
 then \newline
 (a) [75\%] Display eigenanalysis details for $A$. \newline
 (b) [25\%] Display the solution $\.x(t)$ of $\.x'(t)=A\.x(t)$.


\ifsolutions
\begin{solution}{9} ~\newline
Answer (1): The eigenpairs are
$$
 5,  \colvector{1\\ 1\\ 0};\quad 4, \colvector{-1\\ -1\\ 1};\quad
  3, \colvector{1\\ -1\\ 0}.
$$
Answer (2): The eigenanalysis method implies
$$
\.x(t)=c_1e^{5t}  \colvector{1\\ 1\\ 0}+ c_2e^{ 4t} \colvector{-1\\ -1\\
1}+ c_3e^{3t} \colvector{1\\ -1\\ 0}.
$$
\end{solution}
\fi


\end{exercises}
\EJECT

\begin{exercises}
\prob{10.}{(ch5)} ~\newline (a) [40\%]  Find the eigenvalues of
the matrix $A=\left[\begin{array}{rrrr}
 4 & 1 & -1 & 0 \\
 1 & 4 & -2 & 1 \\
 0 & 0 &  2 & 0 \\
 0 & 0 &  2 & 4
 \end{array}\right]$.
%with(linalg):
%B:=matrix([[4,1,-1,0],[1,4,-2,1],[0,0,2,0],[0,0,2,4]]);
%eigenvals(B);
%dsolve({diff(y(x),x)=2*y(x),y(0)=1},y(x));
%dsolve({diff(y(x),x)=3*y(x)+exp(2*x),y(0)=0},y(x));
%dsolve({diff(y(x),x)=4*y(x)+(-exp(-x)+1)*exp(3*x),y(0)=0},y(x));
%dsolve({diff(y(x),x)=5*y(x)+(1/2*exp(-2*x)-exp(-x)+1/2)*exp(4*x),y(0)=0},y(x));

(b) [60\%] Display the general solution of $\.u'=A\.u$ according
to Putzer's spectral formula. Don't expand matrix products, in
order to save time. However, do compute the coefficient functions
$r_1$, $r_2$, $r_3$, $r_4$.
 The correct answer for $r_4$, using $\lambda$ in increasing magnitude,
is $ y(x) = -\frac16 e^{2x}+\frac12 e^{3x}-\frac12 e^{4x}+\frac16
e^{5x}$.

\ifsolutions
\begin{solution}{10} ~\newline
(a) Subtract $\lambda$ from the diagonal elements of $A$ and
expand the determinant $\det(A-\lambda I)$ to obtain the
characteristic polynomial
$(1-\lambda)(1-\lambda)(4-\lambda)(1-\lambda)=0$. The eigenvalues
are the roots: $\lambda=1,1,1,4$. Used here was the {\em cofactor
rule} for determinants. Sarrus' rule does not apply for $4\times
4$ determinants (an error) and the triangular rule likewise does
not directly apply (another error).

% exp(t), t*exp(t), 1/2*t^2*exp(t),
% 1/27*exp(4*t)-1/27*exp(t)-1/9*t*exp(t)-1/6*t^2*exp(t)])

(b) Let
$$B=\left[\begin{array}{rrrr}
 1 & 0 & 0 & 0 \\
 1 & 1 & 0 & 0 \\
 0 & 1 &  1 & 0 \\
 0 & 0 &  1 & 4
 \end{array}\right].$$
Define functions $r_1$, $r_2$, $r_3$, $r_4$ to be the components
of the vector solution $\.r(t)$ to the initial value problem
$$\.r'=B\.r,\quad \.r(0)=\colvector{1\\ 0\\ 0\\ 0}.$$
Solving,
$$ r_1=e^t,\quad r_2=te^t, \quad r_3=t^2e^t/2,$$
$$ r_4=\frac{1}{27}e^{4t} - \frac{1}{27}e^t - \frac{1}{9}te^t - \frac{1}{6}t^2e^t.$$
Define
$$P_1=I,\quad P_2=A-I,\quad P_3=(A-I)^2, \quad P_4=(A-I)^3.$$
Then $ \.u=\left(r_1P1+r_2P_2+r_3P_3+r_4P_4\right)\.u_0$ implies
$$ \.u=\left(r_1I+r_2(A-I)+r_3(A-I)^2+r_4(A-I)^3\right)\.u_0$$
\end{solution}
\fi
\end{exercises}
\EJECT
\begin{exercises}
\prob{11.}{(ch6)} ~\newline
 (a) Define {\em asymptotically stable equilibrium} for $\.u'=\.f(\.u)$, a 2-dimensional system.
\newline
 (b) Give examples of 2-dimensional systems of type saddle, spiral, center and node.
 \newline
 (c) Give a $2$-dimensional predator-prey example $\.u'=\.f(\.u)$ and
 explain the meaning of the variables in the model.


\ifsolutions
\begin{solution}{11} ~\newline

\end{solution}
\fi
\end{exercises}


 \EJECT
\begin{exercises}
\prob{12.}{(ch6)} ~\newline
  Find the equilibrium points of $x'=14x-x^2/2-xy$,
 $y'=16y-y^2/2-xy$ and classify the linearizations as node,
 spiral, center, saddle. What classifications can be deduced for
 the nonlinear system?

\ifsolutions
\begin{solution}{12} ~\newline

\end{solution}
\fi
\end{exercises}

 \EJECT
\begin{exercises}
\prob{13.}{(ch7)} ~\newline
 (a) Define the direct Laplace Transform. \newline
 (b) Define Heaviside's unit step function. \newline
 (c) Derive a Laplace integral formula for Heaviside's unit step
 function. \newline
 (d) Explain Laplace's Method, as applied to the differential
 equation $x'(t) + 2x(t) = e^t$, $x(0)=1$.

\ifsolutions
\begin{solution}{13} ~\newline

\end{solution}
\fi
\end{exercises}
 \EJECT
\begin{exercises}
\prob{14.}{(ch7)} ~\newline
  (a) Solve $\L(f(t))=\dd\frac{100}{s^2+1)(s^2+4)}$ for $f(t)$.
  \newline
  (b) Solve for $f(t)$ in the equation
  $\L(f(t))=\dd\frac{1}{s^2(s-3)}$.
  \newline
  (c) Find $\L(f)$ given $f(t)=(-t)e^{2t}\sin(3t)$.
  \newline
  (d) Find $\L(f)$ where $f(t)$ is the periodic function of period
  $2$ equal to $t/2$ on $0\le t \le 2$ (sawtooth wave).


\ifsolutions
\begin{solution}{14} ~\newline

\end{solution}
\fi
\end{exercises}

 \EJECT
\begin{exercises}
\prob{15.}{(ch7)} ~\newline
 (a) Solve $y''+4y'+4y=t^2$, $y(0)=y'(0)=0$ by Laplace's Method.
  \newline
 (b) Solve $x'''+x''-6x'=0$, $x(0)=x'(0)=0$, $x''(0)=1$ by
 Laplace's Method.
  \newline
 (c) Solve the system $x'=x+y$, $y'=x-y+e^t$, $x(0)=0$, $y(0)=0$
 by Laplace's Method.


\ifsolutions
\begin{solution}{15} ~\newline


\end{solution}
\fi
\end{exercises}

 \EJECT
\begin{exercises}
\prob{16.}{(ch9)} ~\newline
 (a) Find the Fourier sine and cosine coefficients for the $2$-periodic
 function $f(t)$ equal to $t/2$ on $0\le t \le 2$.
 \newline
 (b) State Fourier's convergence theorem.
 \newline
 (c) State the results for term-by-term integration and differentiation
 of Fourier series.
 
\ifsolutions
\begin{solution}{16} ~\newline


\end{solution}
\fi
\end{exercises}

 \EJECT
\begin{exercises}
\prob{17.}{(ch10)} ~\newline
 (a) Find a steady-state periodic solution by Fourier's method for
 $x''+x=F(t)$, where $F(t)$ is $2$-periodic and equal to $10$ on
 $0<t<1$, equal to $-10$ on $1 < t < 2$.
 \newline
 (b) Display Fourier's Model for the solution to the heat problem
 $u_t=u_{xx}$, $u(0,0)=u(1,0)=0$, $u(x,0)=f(x)$ on $0\le x \le 1$, $t\ge
 0$.
 \newline 
 (c) Solve $u_t=u_{xx}$, $u(0,0)=u(\pi,0)=0$, $u(x,0)=80\sin^3 x$ on
 $0\le x \le \pi$, $t\ge 0$.

\ifsolutions
\begin{solution}{17} ~\newline


\end{solution}
\fi
\end{exercises}


 \EJECT
\begin{exercises}
\prob{18.}{(ch10)} ~\newline
 (a) D'Alembert's solution to the wave equation can be displayed as the 
superposition of two waves, one moved left and one moving right. Explain 
this with an example and a snapshot sequence of $4$ frames.
 \newline
 (b) Solve by Fourier's Method the plucked string equation $y_{tt}=a^2 
y_{xx}$ on $0<x<1$, $t\ge 0$, $y(0,t)=y(1,t)=0$, $y(x,0)=f(x)$, 
$y_t(x,0)=0$ with $f(x)=2(1-x)$.
 (c) Solve by Fourier's Method the Dirichlet steady-state heat problem 
$u_{xx}+u_{yy}=0$, $u(0,y)=u(1,y)=u(x,2)=0$ on the rectangle $0\le x \le 
1$, $0\le y \le 2$, with initial data $u(x,0)=f(x)$.

\ifsolutions
\begin{solution}{18} ~\newline


\end{solution}
\fi
\end{exercises}

 \EJECTEND

\end{document}
\endinput
