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\def\newproblem{\advance\problemcount by 1
		\leavevmode\llap{\the\problemcount.~}}

\def\endproblem{\vskip\interproblemspace\vfil\goodbreak}

\def\answers#1#2#3#4#5{\smallskip\theanswers{#1}{#2}{#3}{#4}{#5}\par
    \vskip\interproblemspace\vfil\goodbreak}

\def\TF{\smallskip
    {\it (T)~True\qquad (F)~False}\vskip\interproblemspace\vfil\goodbreak}

\def\theanswers#1#2#3#4#5{\leavevmode
  {\it (a)~}#1\hfil{\it (b)~}#2\hfil{\it (c)~}#3\hfil{\it (d)~}#4\hfil{\it (e)~}#5}

\def\endanswerline{\endgraf\hfilneg} % to force next line in \theanswers

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\def\answersheet#1#2#3#4#5{{
%#1 = start, #2 = stop, #3 = group size, #4 = no. of columns, #5 text for each answer
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    \ifnum#2<\columnend \global\columnend=#2\fi 
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      \leavevmode\llap{\number\answercount.~}\enspace
      #5\endgraf
				  \ifnum\groupcount<#3 \advance\groupcount by 1\else\groupcount=1\bigskip\fi
				  \repeat}\hfill\groupcount=1
  \repeat}}

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{\bf Math 226: Calculus IV}\hfill Name:\hbox to 2.5truein{\hrulefill}\medskip
{\bf Final Exam}\quad{\sl May 11, 1990}\hfill
Instructor:\hbox to 2.5truein{\hrulefill}\medskip
\leavevmode\hfill Section:\hbox to 2.5truein{\hrulefill}

\bigskip
Record your answers to the multiple choice problems by placing an $\times$
through one letter for each problem on this answer sheet. There are 25
questions worth 6 points each.
\bigskip\bigskip

{
  \baselineskip=20pt
  \hrule
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  \answersheet{1}{25}{1}{2}{\theanswers{}{}{}{}{}}
  \hrule
}

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\leavevmode
\hfill Score \hbox to 1truein{\hrulefill}

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%    Questions
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\pageno=1
\footline={\hfil{\rm \folio}\hfil}
\bigskip\bigskip


%Matrix Operations

\newproblem
Let 
$$A = \pmatrix{1&2&0\cr0&5&1\cr},\quad B = \pmatrix{3&-1\cr1&2\cr1&0\cr}$$
Compute the matrix product $AB$.
\answers
{$\pmatrix{3&-2\cr0&10\cr}$}
{$\pmatrix{5&3\cr6&10\cr}$}
{$\pmatrix{5&4&0\cr6&10&0\cr}$\endanswerline}
{$\pmatrix{5&-1\cr6&10\cr0&0\cr}$}
{{\sl Not defined}}

\newproblem
Find the general solution to the following system of linear equations.
$$\matrix{2x_1&-&2x_2&+&6x_3&=&-4\cr x_1&+&x_2&+&2x_3&=&0\cr -3x_1&+&3x_2&-&9x_3&=&1\cr}$$ \answers
{\vtop{\hbox{$x_1 = 0$}\hbox{$x_2 = 0$}\hbox{$x_3 = 0$}}}
{\vtop{\hbox{$x_1 = 0$}\hbox{$x_2 = 0$}\hbox{$x_3 = t$}}}
{\vtop{\hbox{$x_1 = 2 + 2t$}\hbox{$x_2 = 4 + 5t$}\hbox{$x_3 = t$}}}
{\vtop{\hbox{$x_1 = -2-2t$}\hbox{$x_2 = t$}\hbox{$x_3 = t$}}}
{\vtop{\hbox{\sl System is}\hbox{\sl inconsistent}}}

\newproblem
Calculate the determinant of the matrix
$$\pmatrix{1&2&2&-1\cr2&-3&1&5\cr1&2&0&-2\cr0&0&0&1\cr}$$
\answers
{$14$}
{$-120$}
{$480$}
{$-24$}
{$0$}


\newproblem
The vectors resulting from the Gram-Schmidt process applied to $X_1 = (1,1,0,1)$
and $X_2 = (0,1,1,1)$ are
\answers
{$Q_1 = (1,1,0,1),\quad Q_2 = (-1,1,1,0)$\endanswerline}
{$Q_1 = {1\over\sqrt{3}}(1,1,0,1),\quad Q_2 = {1\over\sqrt{3}}(-1,1,1,0)$\endanswerline}
{$Q_1 = (1,0,0,0),\quad Q_2 = (0,1,0,0)$\endanswerline}
{$Q_1={1\over\sqrt{3}}(1,1,0,1),\quad Q_2={3\over\sqrt{7}}(-{2\over3},{1\over3},-1,{1\over3})$\endanswerline}
{$Q_1 ={1\over\sqrt{3}}(1,1,0,1),\quad Q_2 = {1\over\sqrt{15}}(-2,1,3,1)$}


 %First Order Equations

\newproblem
Find an integrating factor for the differential equation
$$y' - {2\over x} y = x^3$$
\answers
{$x$}
{$x^{2}$}
{$x^{-2}$}
{$x^{-1}$}
{$2\ln(x)$}

\newproblem
Which function below is an integrating factor of
$${1\over x}e^{xy}\left(1 + y - {1\over x}\right) dx + e^{xy} dy = 0$$
\answers
{$\mu = x^2$}
{$\mu = e^x$}
{$\mu = x e^y + y e^x$}
{$\mu =e^{y^2}$}
{$\mu = e^{x+y}$}

\newproblem
Solve the initial value problem
$$ x y' + y = \sin(x),\qquad y(\pi/2) = 2$$
\answers
{$ y = x^{-1} \cos(x) + 2$}
{$ y = x^{-1} \cos(x) + \pi x^{-1} $}
{$y = -x^{-1} \cos(x) + \pi x^{-1}$}
{$y = x^{-1}  \sin(x) + {(2\pi - 2)\over \pi}$}
{$y = \pi x^{-1} \sin(x)$}

\newproblem
Find the solution to the initial value problem
$$y' + 2 x y = x,\qquad y(0) = 0$$
\answers
{$y = {1\over2}\left(1 - e^{-x^2}\right)$}
{$y = e^{x^2} - 1$}
{$y = 2x + x e^{-x^2}$\endanswerline}
{$y = x^2 + x\left(e^{x^2} - 1\right)$}
{{\sl None of the above}}

\newproblem
The critical points of the differential equation
$${dN\over dt} = N(N^2 - 5N + 6)$$
are $N_0 = 0$, $2$, and $3$. Their stability properties are as follows:
\answers
{$N_0 = 0$ unstable, $N_0 = 2$ unstable, $N_0 = 3$ stable\endanswerline}
{$N_0 = 0$ unstable, $N_0 = 2$ stable, $N_0 = 3$ unstable\endanswerline}
{$N_0 = 0$ stable, $N_0 = 2$ stable, $N_0 = 3$ stable\endanswerline}
{$N_0 = 0$ stable, $N_0 = 2$ unstable, $N_0 = 3$ unstable\endanswerline}
{$N_0 = 0$ unstable, $N_0 = 2$ unstable, $N_0 = 3$ unstable\endanswerline}


\newproblem
Determine the largest region in the plane where the existence of a unique solution through
any point is guaranteed for the differential equation
$$y' = {\ln|xy|\over x^2 - y^2}$$
\answers
{$\{ (x,y) \, | \, x \ne \pm y \}$}
{$\{ (x,y) \, | \,  x > 0, \space y > 0, \space x \ne y \}$}
{$\{ (x,y) \, | \,  x > y, \space y > 0 \}$}
{$\{ (x,y) \, | \, x \ne \pm y, \space x \ne 0, \space y \ne 0 \}$}
{{\sl None of the above}}

\newproblem
The solution of
$$(1 + y \cos(xy) )dx + x \cos(xy) dy = 0$$
is given implicitly by
\answers
{$y + \cos(xy) = c$}
{$2xy + \tan(xy) = c$}
{$3x + y\sin(xy) = c$\endanswerline}
{$\sin(x)\cos(y) = c$}
{$x + \sin(xy) = c$}

\newproblem
Solve the differential equation
$$ y' = {y^4 + x^4\over x y^3}$$
by making a substitution $v = {y\over x}$.
\answers
{$y = c \ln(x) + {1\over x^4}$}
{$y^4 = x^4 + c \ln(x)$\endanswerline}
{$y^3 = c x^3$}
{$\ln(y) = x^4 + c$\endanswerline}
{$y^4 = x^4(\ln(x^4) + c)$}

\newproblem
Suppose that a certain sum of money is deposited in a bank that pays
interest at an annual rate of 8\% compounded continuously. Find the time
$T$ required for this original sum to triple.
\answers
{8.66 years}
{10.14 years}
{11.45 years}
{12.64 years}
{13.73 years}

% Second Order Equations

\newproblem
Each of the following is a set of solutions to $y'' - y = 0$. Find the one which is
{\bf not} a fundamental set of solutions.
\answers
{$\{ e^x, e^{-x}\}$}
{$\{ e^x + e^{-x}, e^x - e^{-x}\}$}
{$\{ e^{1-x}, e^{1+x}\}$\endanswerline}
{$\{ e^{x-1}, e^{x+1}\}$}
{$\{ e^x, 2e^{-x}\}$}


\newproblem
Compute the Wronskian of the functions $y_1 = e^{2x}$ and $y_2 = x e^{2x}$
\answers
{$e^{2x}(1+2x)$}
{$e^{4x}(1+2x)$}
{$e^{4x}$}
{$2xe^{2x}$}
{$e^{2x}$}

\newproblem
Solve the differential equation with initial conditions
$$y'' + 2y' + 5 y = 0,\qquad y(0) = 1,\quad y'(0) = 1$$
\answers
{$y = e^{-x}(\cos(2x) + \sin(2x))$}
{$y = e^{-x}(\cos(x) + 2\sin(x))$\endanswerline}
{$y = e^{-x}(\cos(\sqrt{5}x) + {2\over\sqrt{5}}\sin(\sqrt{5}x))$}
{$y = e^{-2x}(\cos(2x) -{1\over2} \sin(2x))$\endanswerline}
{{\sl None of the above}}

\newproblem
Given that $y_1 = x^{-1}$ is a solution of
$$x^2 y'' + 3 x y' + y = 0$$
find another independent solution of the form $y_2 = v y_1$.
\answers
{$y_2 = x^{-1}(\cos(|\ln(x)|) + \sin(|\ln(x)|))$}
{$y_2 = x$}
{$y_2 = x\ln(x)$\endanswerline}
{$y_2 = x^{-1}\ln(x)$}
{$y_2 = x^{-1}\sum_{n=0}^\infty{(n+1) x^n}$}

\newproblem
The form of the particular solution of 
$$y'' - 2y' + y = x + 1 + xe^x$$
given by the method of undetermined coefficients is
\answers
{$y_p = Ax + xe^x$}
{$y_p = Ae^x + Bxe^x$}
{$y_p = Ax^2e^x$\endanswerline}
{$y_p = Ax + B + x(Cx + D)e^x$}
{$y_p =Ax + B + x^2(Cx + D)e^x$}

\newproblem
The particular solution of
$$y'' + y = \tan(x)$$
given by the method of variation of parameters is
\answers
{$y_p = \ln|\tan(x)|$}
{$y_p = (\tan(x) + 1)\cos(x)$}
{$y_p = -\cos(x)\ln|\tan(x) + \sec(x)|$}
{$y_p = \tan^2(x) + 1$}
{{\sl None of the above}}

% Power Series Solutions

\newproblem
Compute the radius of convergence of the series expansion
$${1\over3 + x^2} = {1\over3} - {1\over3^2}x^2 + {1\over3^3}x^4 - {1\over3^4}x^6 +
\ldots$$
\answers
{$1$}
{$\sqrt{3}$}
{$3$}
{${1\over3}$}
{$\infty$}

\newproblem
Determine the power series expansion around $x_0 = 0$ for $f(x) = {1\over
(1+x)^2}$.
\answers
{$1 -2x + 3x^2 - 4x^3 + \ldots$}
{$1 + x^2 + x^4 + x^6 + \ldots$\endanswerline}
{$1 + {1\over2}x + {1\over3}x^2 + {1\over4}x^3 + \ldots$}
{$1 - {1\over2}x^2 + {1\over4}x^4 - {1\over6}x^6 + \ldots$\endanswerline}
{$1 - x^2 + x^4 - x^6 + \ldots$}


\newproblem
Find  a power series solution around $x_0 = 0$ for the
differential equation
$$ y'' - x y' - 2 y = 0$$
which satisfies $y(0) = 0$, $y'(0) = 1$.
\answers
{$y = x + {1\over3\cdot2}x^3 + {1\over5\cdot4}x^5 + {1\over7\cdot6}x^7 +\ldots$\endanswerline}
{$y = x - {1\over2}x^3 + {1\over4\cdot2}x^5 - {1\over6\cdot4\cdot2}x^7+\ldots$\endanswerline}
{$y = x + {1\over3}x^3 + {1\over5\cdot3}x^5 +{1\over7\cdot5\cdot3}x^7 +\ldots$\endanswerline}
{$y = x - {1\over2}x^3 + {1\over4\cdot2}x^5 - {1\over6\cdot4\cdot2}x^7 +\ldots$\endanswerline}
{$y = x + {1\over2}x^3 + {1\over4\cdot2}x^5 + {1\over6\cdot4\cdot2}x^7 +\ldots$}


% Regular Singular points 

\newproblem
Find a fundamental set of solutions for the following Euler equation
$$ x^2 y'' - x y' - 3 y = 0$$
\answers
{$\left\{x^3, x^3\ln(x)\right\}$}
{$\left\{x^3, x^{-1}\right\}$\endanswerline}
{$\left\{x^3, x^{-1}\ln(x)\right\}$}
{$\left\{x, x^{-3}\right\}$\endanswerline}
{$\left\{\sum_{n=0}^\infty{1\over n!}x^{n+1}, \sum_{n=0}^\infty{(-1)^n\over n!}x^{n-1}\right\}$}

\newproblem
The following differential equation has a regular singular point at $x_0 = 0$.
$$x^2 y'' + 4xy' + (2 + x^2)y = 0$$
Which of the following functions is a solution to this equation.
\answers
{$y = x^{-1}(1 - {1\over3!}x^2 + {1\over5!}x^4 - \ldots)$\endanswerline}
{$y = (x - {1\over3!}x^3 + {1\over5!}x^5 - \ldots)$\endanswerline}
{$y = x^{-1}(1 + {1\over3!}x^3 + {1\over5!}x^5 + \ldots)$\endanswerline}
{$y = x^{-3}(1 - {1\over2!}x^2 + {1\over4!}x^4 - \ldots)$\endanswerline}
{$y = x^{-2}(1 + {1\over2!}x^2 + {1\over4!}x^4 + \ldots)$}

\newproblem
The differential equation
$$x^2 y'' + \sin(x) y' - 4y = 0$$
is guaranteed to have solution(s) of the form
\answers
{$ y = |x|^{-2}\left(1 +\sum_{n=1}^\infty a_n x^n\right)$\endanswerline}
{$ y = |x|^{-1}\left(1 +\sum_{n=1}^\infty a_n x^n\right)$\endanswerline}
{$ y = \left(1 +\sum_{n=1}^\infty a_n x^n\right)$\endanswerline}
{$ y = |x|^2\left(1 +\sum_{n=1}^\infty a_n x^n\right)$\endanswerline}
{$ y = |x|^r$ for some $r$}



\bye