% 226S98Test2 Version 1, Tue Mar 17 09:29:54 1998
\documentclass[11pt]{article}
\usepackage{graphicx}
\usepackage{amssymb}
\usepackage{epstopdf}
\DeclareGraphicsRule{.tif}{png}{.png}{`convert #1 `dirname #1`/`basename #1 .tif`.png}

\textwidth = 6.5 in
\textheight = 9 in
\oddsidemargin = 0.0 in
\evensidemargin = 0.0 in
\topmargin = 0.0 in
\headheight = 0.0 in
\headsep = 0.0 in
\parskip = 0.2in
\parindent = 0.0in

\newtheorem{theorem}{Theorem}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{definition}{Definition}


\begin{document}






% Problems

\vbox{
%1

\noindent\llap{1.\ }Let $y(t)$ be the solution of \quad $2y''+y'=0$ \quad
for which \quad $y(0)=y'(0)=1$.\quad  Find
$\lim\limits_{t\rightarrow\infty}y(t).$

\beginanswers
{(a)~3\answerspace}
{(b)~${3\over 2}$\answerspace}
{(c)~1\answerspace}
{(d)~4\answerspace}
{(e)~$\infty$\answerspace}
\endanswers\problemskip}\vfil\goodbreak

\vbox{
%2

\noindent\llap{2.\ }The existence and uniqueness theorem for second order
linear differential equations guarantees a unique
solution to the equation
$$(t-1)^2y''+(\ln t)y'+3y=2$$
\noindent
satisfying

\beginanswers
{(a)~$y(2)=1, y'(2)=0$\answerspace}
{(b)~$y(0)=1, y'(0)=2$\answerspace}
{(c)~$y(1)=2, y'(1)=3$\answerspace}
{(d)~$y(3)=2, y(5)=4$\answerspace}
{(e)~$y(0)=0, y'(0)=0$\answerspace}
\endanswers\problemskip}\vfil\goodbreak

\vbox{
%3

\noindent\llap{3.\ }Each of the following is a set of solutions of the
equation \quad $y''-9y=0$.\quad Find the one that is
${\bf NOT}$ a fundamental set of solutions.

\beginanswers
{(a)~$\{e^{3t-1},e^{3t+1}\}$\answerspace}
{(b)~$\{e^{3t},e^{-3t}\}$\answerspace}
{(c)~$\{e^{3t}+e^{-3t},e^{3t}-e^{-3t}\}$\answerspace}
{(d)~$\{e^{1-3t},e^{1+3t}\}$\answerspace}
{(e)~$\{3e^{3t},-e^{-3t}\}$\answerspace}
\endanswers\problemskip}\vfil\goodbreak

\vbox{
%4

\noindent\llap{4.\ }Let $W(t)$ be the Wronskian of a fundamental set of
solutions of the equation
$$(1+t^2)y''+ty'-(3t+2)y=0.$$
\noindent
If $W(0)=4$, find $W(\sqrt3)$.

\beginanswers
{(a)~2\answerspace}
{(b)~1\answerspace}
{(c)~8\answerspace}
{(d)~0.707\answerspace}
{(e)~$-4$\answerspace}
\endanswers\problemskip}\vfil\goodbreak

\vbox{
%5

\noindent\llap{5.\ }Let $y(t)$ be the solution of the initial value problem
$$y''-2y'+17y=0,\quad y(0)=1=y'(0).$$
\noindent
Find $y({\pi\over 16})$.

\beginanswers
{(a)~0.861\answerspace}
{(b)~1.721\answerspace}
{(c)~-2.139\answerspace}
{(d)~1.217\answerspace}
{(e)~0\answerspace}
\endanswers\problemskip}\vfil\goodbreak

\vbox{
%6

\noindent\llap{6.\ }Find the general solution of the equation \quad
$4y''+12y'+9y=3e^{{3\over2}t}$.

\beginanswers
{(a)~$e^{-{3\over2}t}(c_1+c_2t)+{1\over12}e^{{3\over2}t}$\answerspace}
{(b)~$e^{-{3\over2}t}(c_1+c_2t)+{t\over6}e^{{3\over2}t}$\answerspace}
{(c)~$e^{-{3\over2}t}(c_1+c_2t+{t^2\over9})$\answerspace}
{(d)~$c_1e^{-{3\over2}t}+c_2t+{1\over 18}e^{{3\over2}t}$\answerspace}
{(e)~$e^{{2\over3}t}(c_1+c_2t)-{1\over 12}e^{{3\over2}t}$\answerspace}
\endanswers\problemskip}\vfil\goodbreak

\vbox{
%7

\noindent\llap{7.\ }The function \quad $y_1(t)=e^t$\quad is a solution of
the differential equation
$$ty''-y'-(t-1)y=0,\quad t>0.$$
\noindent
The function \quad $y_1(t)v(t)$\quad will be a second linearly independent
solution if $v(t)$ satisfies
the differential equation

\beginanswers
{(a)~$v''+(2-{1\over t})v'=0$.\answerspace}
{(b)~$tv''+(1-2t)v'=0$.\answerspace}
{(c)~$v''+{e^{2t}\over t}v'=0$.\answerspace}
{(d)~$(te^{-2t}v')'=0$.\answerspace}
{(e)~$v''+(1+{1\over t})v'=0$.\answerspace}
\endanswers\problemskip}\vfil\goodbreak

\vbox{
%8

\noindent\llap{8.\ }Using the method of undetermined coefficients, find the
correct form for a particular solution $Y(t)$ of
the equation
$$y''+4y=e^{2t}\cos 2t-\sin 2t.$$

\beginanswers
{(a)~$e^{2t}(A\cos 2t+B\sin 2t)+t(C\cos 2t+D\sin 2t)$\answerspace}
{(b)~$e^{2t}(A\cos 2t+B\sin 2t)+C\cos 2t+D\sin 2t$\answerspace}
{(c)~$te^{2t}(A\cos 2t+B\sin 2t)$\answerspace}
{(d)~$(At+B)e^{2t}\cos 2t+(Ct+D)\sin 2t$\answerspace}
{(e)~$te^{2t}(A\cos 2t+B\sin 2t)+t(C\cos 2t+D\sin 2t)$\answerspace}
\endanswers\problemskip}\vfil\goodbreak

\vbox{
%9

\noindent\llap{9.\ }Suppose that $y_1(x)=x$ and $y_2(x)=x^2$ are solutions
of the differential equation
$$y''+p(x)y'+q(x)y=0,$$
\noindent
where $p(x)$ and $q(x)$ are continuous for $x>0$.  Find a particular
solution to the nonhomogeneous
equation
$$y''+p(x)y'+q(x)y=10x^4.$$

\beginanswers
{(a)~${x^6\over 2}$\answerspace}
{(b)~${x^5\over 10}$\answerspace}
{(c)~${x^6\over 5}$\answerspace}
{(d)~${x^5\over 12}$\answerspace}
{(e)~${x^7\over 20}$\answerspace}
\endanswers\problemskip}\vfil\goodbreak

\vbox{
%10

\noindent\llap{10.\ }An object which weighs 32 pounds is attached to a
spring with spring constant 1 pound/foot and is immersed
in a viscous medium with damping constant 2 pound-seconds/foot.  At time
$t=0$ the mass is pulled down
$1\over4$ foot from the equilibrium position and given an initial velocity
of one foot/second in the
downward direction.  Solve the spring equation and choose the best
description of what happens.  (g=32.)

\beginanswers
{(a)~The mass will move away from and then slowly return toward the
equilibrium position without ever passing
through it.\answerspace}
{(b)~The mass will pass through the equilibrium position once, and then
slowly return toward it.\answerspace}
{(c)~The mass will oscillate about the equilibrium position, moving the
same distance above and below it over
and over.\answerspace}
{(d)~The mass will pass through the equilibrium position exactly
twice.\answerspace}
{(e)~The mass will oscillate about the equilibrium position, but with
constantly decreasing distance above and
below it.\answerspace}
\endanswers\problemskip}\vfil\goodbreak

\vbox{
%11

\noindent\llap{11.\ }The expression \quad $-2\cos 5t+7\sin 5t$\quad can be
rewritten in the form \quad $R\cos(5t-\delta)$,\quad
where an approximate value for $\delta$ in radians is

\beginanswers
{(a)~1.849\answerspace}
{(b)~1.292\answerspace}
{(c)~4.990\answerspace}
{(d)~2.863\answerspace}
{(e)~-1.292\answerspace}
\endanswers\problemskip}\vfil\goodbreak

\vbox{
%12

\noindent\llap{12.\ }For the equation \quad $(1-x^2)y''-6xy'-4y=0$,\quad
find the first three nonzero terms of the
series solution in powers of $x$ for which \quad $y(0)=1$\quad and \quad
$y'(0)=0$.

\beginanswers
{(a)~$1+2x^2+3x^4$\answerspace}
{(b)~$2+{5\over 3}x^3+{7\over3}x^5$\answerspace}
{(c)~$x^2+3x^4+4x^6$\answerspace}
{(d)~$1+x+2x^2$\answerspace}
{(e)~$1+x^2+{3\over 2}x^4$\answerspace}
\endanswers\problemskip}\vfil\goodbreak

\vbox{
%13

\noindent\llap{13.\ }Find the recurrence relation for the series solution
in powers of \quad ${\bf x-1}$\quad of the equation
$$(2x-x^2)y''-y=0.$$

\beginanswers
{(a)~$a_{n+2}={n^2-n+1\over n^2+3n+2}a_n$\answerspace}
{(b)~$a_{n+2}={n-1\over n+2}a_n$\answerspace}
{(c)~$a_{n+2}={(n-1)a_{n-1}+a_n\over (n+2)(n+1)}$\answerspace}
{(d)~$a_{n+2}={n+1\over n+2}a_n$\answerspace}
{(e)~$a_{n+2}={(n-1)a_{n-1}+na_n\over (n+2)(n+1)}$\answerspace}
\endanswers\problemskip}\vfil\goodbreak

\vbox{
%14

\noindent\llap{14.\ }The equation \quad
$(x-2)(x^2+4x+8)y''+(x^2+2)y'+(x-2)y=0$\hfil\break
has a power series solution in powers of $x+2$.  The general theory
guarantees that its radius of
convergence is at least

\beginanswers
{(a)~$2$.\answerspace}
{(b)~$2\sqrt 2$.\answerspace}
{(c)~$2\sqrt 5$.\answerspace}
{(d)~$4$.\answerspace}
{(e)~$3\sqrt 2$.\answerspace}
\endanswers\problemskip}\vfil\goodbreak

\vbox{
%15

\noindent\llap{15.\ }The equation \quad $(x^2-1)^2y''+(x^2-1)y'+y=0$ \quad has

\beginanswers
{(a)~two regular singular points.\answerspace}
{(b)~two irregular singular points.\answerspace}
{(c)~one regular singular point and one irregular singular point.\answerspace}
{(d)~just one singular point, which is regular.\answerspace}
{(e)~just one singular point, which is irregular.\answerspace}
\endanswers\problemskip}\vfil\goodbreak

\vbox{
%16

\noindent\llap{16.\ }Find the general solution of the Euler equation
$$x^2y''+7xy'+13y=0.$$

\beginanswers
{(a)~$x^{-3}(c_1\cos (2\ln|x|)+c_2 \sin(2\ln |x|))$\answerspace}
{(b)~$x^{3}(c_1\cos (\ln|x|)+c_2 \sin(\ln |x|))$\answerspace}
{(c)~$x^{-3}(c_1\cos (\ln2|x|)+c_2 \sin(\ln 2|x|))$\answerspace}
{(d)~$|x|^{-{7\over2}}(c_1\cos ({{\sqrt 3}\over 2}\ln|x|)+c_2 \sin({{\sqrt
3}\over 2}\ln |x|))$\answerspace}
{(e)~$|x|^{-{7\over2}}(c_1\cos (\ln{{\sqrt 3}\over 2}|x|)+c_2 \sin(\ln
{{\sqrt 3}\over 2}|x|))$\answerspace}
\endanswers\problemskip}\vfil\goodbreak


% Appendix

{}

% Blank answer sheet

\vfil\eject\output={}
\vbox{{\noindent
{\bf Math 226: Calculus IV}\hfill
Sign your Name:\hbox to 2.5truein{\hrulefill}\medskip\noindent
%Put in the correct exam number and date%%%%%%%%
{\bf Exam II}\quad{\sl March 20, 1998}\hfill
Print your Name:\hbox to 2.5truein{\hrulefill}\medskip\hfill
Section:\hbox to 2.5truein{\hrulefill}\medskip\hfill
TA:\hbox to 2.5truein{\hrulefill}

\bigskip\noindent
Do not turn this page until you are told to begin.
\bigskip\noindent
{\bf Do not detach the answer sheet from the test.}
\bigskip\noindent
You are required to hand in both the answer sheet and the problems.
\bigskip\noindent
Record your answers to the multiple choice problems by placing an $\times$
through one letter for each problem on this answer sheet. There are 16 multiple
choice questions worth 6 points each. You start with 10 points, and the
highest
possible score is 106.  Fill in the answers as you go along.  You will not be
allowed to fill in the answers after the time is up.
\bigskip\noindent
You may use a calculator, but only the standard functions found on very
inexpensive
scientific calculators.  In particular you may not use graphing,
integration, formula, matrix or
program capabilities.
\bigskip\noindent



\def\answerskip{\vskip0.25in}


\divide\hsize by3
\leavevmode\hfil\vtop{
\answerskip\noindent\llap{1.}\enskip\ansbox{15pt}{a}\enskip\ansbox{15pt}{b}\ensk
ip\ansbox{15pt}{c}\enskip\ansbox{15pt}{d}\enskip\ansbox{15pt}{e}\par
\answerskip\noindent\llap{2.}\enskip\ansbox{15pt}{a}\enskip\ansbox{15pt}{b}\ensk
ip\ansbox{15pt}{c}\enskip\ansbox{15pt}{d}\enskip\ansbox{15pt}{e}\par
\answerskip\noindent\llap{3.}\enskip\ansbox{15pt}{a}\enskip\ansbox{15pt}{b}\ensk
ip\ansbox{15pt}{c}\enskip\ansbox{15pt}{d}\enskip\ansbox{15pt}{e}\par
\answerskip\noindent\llap{4.}\enskip\ansbox{15pt}{a}\enskip\ansbox{15pt}{b}\ensk
ip\ansbox{15pt}{c}\enskip\ansbox{15pt}{d}\enskip\ansbox{15pt}{e}\par
\answerskip\noindent\llap{5.}\enskip\ansbox{15pt}{a}\enskip\ansbox{15pt}{b}\ensk
ip\ansbox{15pt}{c}\enskip\ansbox{15pt}{d}\enskip\ansbox{15pt}{e}\par
\answerskip\noindent\llap{6.}\enskip\ansbox{15pt}{a}\enskip\ansbox{15pt}{b}\ensk
ip\ansbox{15pt}{c}\enskip\ansbox{15pt}{d}\enskip\ansbox{15pt}{e}\par
\answerskip\noindent\llap{7.}\enskip\ansbox{15pt}{a}\enskip\ansbox{15pt}{b}\ensk
ip\ansbox{15pt}{c}\enskip\ansbox{15pt}{d}\enskip\ansbox{15pt}{e}\par
\answerskip\noindent\llap{8.}\enskip\ansbox{15pt}{a}\enskip\ansbox{15pt}{b}\ensk
ip\ansbox{15pt}{c}\enskip\ansbox{15pt}{d}\enskip\ansbox{15pt}{e}\par
}\hfil\vtop{
\answerskip\noindent\llap{9.}\enskip\ansbox{15pt}{a}\enskip\ansbox{15pt}{b}\ensk
ip\ansbox{15pt}{c}\enskip\ansbox{15pt}{d}\enskip\ansbox{15pt}{e}\par
\answerskip\noindent\llap{10.}\enskip\ansbox{15pt}{a}\enskip\ansbox{15pt}{b}\ens
kip\ansbox{15pt}{c}\enskip\ansbox{15pt}{d}\enskip\ansbox{15pt}{e}\par
\answerskip\noindent\llap{11.}\enskip\ansbox{15pt}{a}\enskip\ansbox{15pt}{b}\ens
kip\ansbox{15pt}{c}\enskip\ansbox{15pt}{d}\enskip\ansbox{15pt}{e}\par
\answerskip\noindent\llap{12.}\enskip\ansbox{15pt}{a}\enskip\ansbox{15pt}{b}\ens
kip\ansbox{15pt}{c}\enskip\ansbox{15pt}{d}\enskip\ansbox{15pt}{e}\par
\answerskip\noindent\llap{13.}\enskip\ansbox{15pt}{a}\enskip\ansbox{15pt}{b}\enskip\ansbox{15pt}{c}\enskip\ansbox{15pt}{d}\en
skip\ansbox{15pt}{e}\par
\answerskip\noindent\llap{14.}\enskip\ansbox{15pt}{a}\enskip\ansbox{15pt}{b}\ens
kip\ansbox{15pt}{c}\enskip\ansbox{15pt}{d}\enskip\ansbox{15pt}{e}\par
\answerskip\noindent\llap{15.}\enskip\ansbox{15pt}{a}\enskip\ansbox{15pt}{b}\ens
kip\ansbox{15pt}{c}\enskip\ansbox{15pt}{d}\enskip\ansbox{15pt}{e}\par
\answerskip\noindent\llap{16.}\enskip\ansbox{15pt}{a}\enskip\ansbox{15pt}{b}\ens
kip\ansbox{15pt}{c}\enskip\ansbox{15pt}{d}\enskip\ansbox{15pt}{e}\par
}}}


% Marked answer sheet

\vfil\eject
\vbox{\vbox to0pt{\vskip-22.5pt\hbox to \hsize{\hfil\ansbox{70pt}{\ Answer
Key 1\ }\hfil}\vss}
\prevdepth=-1000pt
{\noindent
{\bf Math 226: Calculus IV}\hfill
Sign your Name:\hbox to 2.5truein{\hrulefill}\medskip\noindent
%Put in the correct exam number and date%%%%%%%%
{\bf Exam II}\quad{\sl March 20, 1998}\hfill
Print your Name:\hbox to 2.5truein{\hrulefill}\medskip\hfill
Section:\hbox to 2.5truein{\hrulefill}\medskip\hfill
TA:\hbox to 2.5truein{\hrulefill}

\bigskip\noindent
Do not turn this page until you are told to begin.
\bigskip\noindent
{\bf Do not detach the answer sheet from the test.}
\bigskip\noindent
You are required to hand in both the answer sheet and the problems.
\bigskip\noindent
Record your answers to the multiple choice problems by placing an $\times$
through one letter for each problem on this answer sheet. There are 16 multiple
choice questions worth 6 points each. You start with 10 points, and the
highest
possible score is 106.  Fill in the answers as you go along.  You will not be
allowed to fill in the answers after the time is up.
\bigskip\noindent
You may use a calculator, but only the standard functions found on very
inexpensive
scientific calculators.  In particular you may not use graphing,
integration, formula, matrix or
program capabilities.
\bigskip\noindent



\def\answerskip{\vskip0.25in}


\divide\hsize by3
\leavevmode\hfil\vtop{
\answerskip\noindent\llap{1.}\enspace\ansbox{15pt}{$\bullet$}\enskip\ansbox{15pt
}{b}\enskip\ansbox{15pt}{c}\enskip\ansbox{15pt}{d}\enskip\ansbox{15pt}{e}\par\a
nswerskip\noindent\llap{2.}\enspace\ansbox{15pt}{$\bullet$}\enskip\ansbox{15pt}{
b}\enskip\ansbox{15pt}{c}\enskip\ansbox{15pt}{d}\enskip\ansbox{15pt}{e}\par
\answerskip\noindent\llap{3.}\enspace\ansbox{15pt}{$\bullet$}\enskip\ansbox{15pt
}{b}\enskip\ansbox{15pt}{c}\enskip\ansbox{15pt}{d}\enskip\ansbox{15pt}{e}\par\a
nswerskip\noindent\llap{4.}\enspace\ansbox{15pt}{$\bullet$}\enskip\ansbox{15pt}{
b}\enskip\ansbox{15pt}{c}\enskip\ansbox{15pt}{d}\enskip\ansbox{15pt}{e}\par
\answerskip\noindent\llap{5.}\enspace\ansbox{15pt}{$\bullet$}\enskip\ansbox{15pt
}{b}\enskip\ansbox{15pt}{c}\enskip\ansbox{15pt}{d}\enskip\ansbox{15pt}{e}\par\a
nswerskip\noindent\llap{6.}\enspace\ansbox{15pt}{$\bullet$}\enskip\ansbox{15pt}{
b}\enskip\ansbox{15pt}{c}\enskip\ansbox{15pt}{d}\enskip\ansbox{15pt}{e}\par
\answerskip\noindent\llap{7.}\enspace\ansbox{15pt}{$\bullet$}\enskip\ansbox{15pt
}{b}\enskip\ansbox{15pt}{c}\enskip\ansbox{15pt}{d}\enskip\ansbox{15pt}{e}\par\a
nswerskip\noindent\llap{8.}\enspace\ansbox{15pt}{$\bullet$}\enskip\ansbox{15pt}{
b}\enskip\ansbox{15pt}{c}\enskip\ansbox{15pt}{d}\enskip\ansbox{15pt}{e}\par
}\hfil\vtop{
\answerskip\noindent\llap{9.}\enspace\ansbox{15pt}{$\bullet$}\enskip\ansbox{15pt
}{b}\enskip\ansbox{15pt}{c}\enskip\ansbox{15pt}{d}\enskip\ansbox{15pt}{e}\par\a
nswerskip\noindent\llap{10.}\enspace\ansbox{15pt}{$\bullet$}\enskip\ansbox{15pt}
{b}\enskip\ansbox{15pt}{c}\enskip\ansbox{15pt}{d}\enskip\ansbox{15pt}{e}\par
\answerskip\noindent\llap{11.}\enspace\ansbox{15pt}{$\bullet$}\enskip\ansbox{15p
t}{b}\enskip\ansbox{15pt}{c}\enskip\ansbox{15pt}{d}\enskip\ansbox{15pt}{e}\par\
answerskip\noindent\llap{12.}\enspace\ansbox{15pt}{$\bullet$}\enskip\ansbox{15pt
}{b}\enskip\ansbox{15pt}{c}\enskip\ansbox{15pt}{d}\enskip\ansbox{15pt}{e}\par\a
nswerskip\noindent\llap{13.}\enspace\ansbox{15pt}{$\bullet$}\enskip\ansbox{15pt}
{b}\enskip\ansbox{15pt}{c}\enskip\ansbox{15pt}{d}\enskip\ansbox{15pt}{e}\par
\answerskip\noindent\llap{14.}\enspace\ansbox{15pt}{$\bullet$}\enskip\ansbox{15pt}{b}\enskip\ansbox{15pt}{c}\enskip\
ansbox{15pt}{d}\enskip\ansbox{15pt}{e}\par
\answerskip\noindent\llap{15.}\enspace\ansbox{15pt}{$\bullet$}\enskip\ansbox{15p
t}{b}\enskip\ansbox{15pt}{c}\enskip\ansbox{15pt}{d}\enskip\ansbox{15pt}{e}\par\
answerskip\noindent\llap{16.}\enspace\ansbox{15pt}{$\bullet$}\enskip\ansbox{15pt
}{b}\enskip\ansbox{15pt}{c}\enskip\ansbox{15pt}{d}\enskip\ansbox{15pt}{e}\par}}
}

 \end{document}


