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\begin{center}
MA 262 \\
FINAL EXAM INSTRUCTIONS \\
December 15, 1997
\end{center}
\bigskip
\leftline{NAME\ \ {\hbox to 2truein{\hrulefill}}\hfil
INSTRUCTOR\ \ {\hbox to 2truein{\hrulefill}}}
\bigskip
\bigskip

\begin{enumerate}[{\bfseries 1.}]


\item You must use a \underbar{\#2 pencil} on the mark--sense sheet
(answer sheet).

\bigskip
\item If the cover of your question booklet is WHITE, write 01
in the TEST/QUIZ NUMBER boxes and blacken in the appropriate spaces
below.  If the cover is BLUE, write 02
in the TEST/QUIZ NUMBER boxes and darken the spaces below.

\bigskip
\item On the mark-sense sheet, fill in the \underbar{instructor's} name
and the \underbar{course number}.

\bigskip
\item Fill in your \underbar{NAME} and \underbar{STUDENT IDENTIFICATION
NUMBER} and blacken in the appropriate spaces.

\bigskip
\item Fill in the \underbar{SECTION NUMBER} boxes with the division and
section number of your class.
For example, for division 02, section 03, fill in 0203 and blacken the
corresponding circles, including the circles for the zeros.
(If you do not know your division and section number ask your instructor.)

\bigskip
\item Sign the mark--sense sheet.

\bigskip
\item Fill in your name and your instructor's name on the question
sheets above.

\bigskip
\item There are 25 questions, each worth 8 points.
Blacken in your choice of the correct answer in the spaces provided for
questions 1--25.
Do all your work on the question sheets.
\underbar{Turn in both the mark--sense sheets and the question sheets
when you are finished}.

\bigskip
\item \underbar{Show your work} on the question sheets.
Although no partial credit will be given, any disputes about grades or
grading will be settled by examining your written work on the question
sheets.

\bigskip
\item NO CALCULATORS, BOOKS, OR PAPERS ARE ALLOWED.
Use the back of the test pages for scrap paper.
\end{enumerate}

\vfill\eject

\begin{enumerate}[{\bfseries 1.}]

\item
The general solution to $xy'+y=e^{5x}$ is
  \begin{enumerate}[A.]
\item $y=\frac15 e^{5x} + c$
\item \correct{$y=\frac{1}{5x}e^{5x}+cx^{-1}$}
\item $y=\frac{1}{6}e^{5x}+ce^{-x}$
\item $y=ce^{5x}$
\item $y=\frac{5}{x}e^{5x}+c$
  \end{enumerate}

\vfill

\item
Solutions to $2xy+(x^2+1)\frac{dy}{dx}=0$ satisfy
  \begin{enumerate}[A.]
\item \correct{$x^2y+y=c$}
\item $x^2y^2+x=c$
\item $y=e^{\tan^{-1} (x)}+c$
\item $x^2y+1=c$
\item $\ln y=-2x \tan^{-1}(x)+c$
  \end{enumerate}

\vfill

\item
The substitution $v=y/x$ transforms the equation
$\frac{dy}{dx}=\sin(y/x)$ into
\begin{enumerate} [A.]
\item $v'= \sin(v)$
\item $v'= x \sin(v)$
\item $v'+ v = \sin(v)$
\item \correct{$x v'+ v = \sin(v)$}
\item $v'+ x v = \sin(v)$
\end{enumerate}

\vfill\eject

\item
The general solution to $y''=\frac{2}{x}y'+4x^2$ is
\begin{enumerate} [A.]
\item $y=\frac29 x^6+ c_1x^{3}+c_2$
\item $y=2x^2+c_1x+c_2$
\item $y=x^{4}+c_1x +c_2$
\item $y=\frac13 x^4+c_1x+c_2$
\item \correct{$y=x^4+c_1x^3+c_2$}
\end{enumerate}

\vfill

\item
A fish tank contains 20 gallons of a salt solution with a
concentration of 5 grams of salt per gallon.  A salt solution with a
concentration of 10 grams/gallon is added to the tank at a rate of
2 gallons per minute.  At the same time, water is drained from the
tank at a rate of 2 gallons per minute. How many grams of salt are in
the tank after 10 minutes?
\begin{enumerate} [A.]
\item $200e^{-1}+100$
\item \correct{$200-100e^{-1}$}
\item $200e-100$
\item $100$
\item $200$
\end{enumerate}

\vfill

\item
Which of the following sets forms a basis for ${\mathbb R}^3$?
\newline
(i) \ \ $\{(2,-1,0), (1,-2,0),(1,-3,1)\}$
\newline
(ii) \ \ $\{({\frac12},1,1), (0,3,0),(0,0,2),(1,1,0)\}$
\newline
(iii) \ \ $\{(1,1,1), (0,{\frac12}, {\frac12}), (2,0,0)\}$
\newline
(iv) \ \ $\{({\frac12},1,1), (0,1,0), (1,2,0)\}$
\begin{enumerate} [A.]
\item
(i) only
\item
(i), (ii) and (iv)
\item
(i), (iii) and (iv)
\item
\correct{(i) and (iv)}
\item
None of the above.
\end{enumerate}

\vfill\eject

\item
Let $T:{\mathbb R}^4\rightarrow {\mathbb R}^3$ be defined by $T({\mathbf x})=
A{\mathbf x}$ where $A=\left[\matrix
1  & 1 & -1  &-3 \\
0  & 1 & 1  &-4 \\
2  & 2 & -2 &-6 
\endmatrix\right]$. Then the dimension of Ker\,$(T)$ is
\begin{enumerate} [A.]
\item
$0$
\item
$1$
\item
\correct{$2$}
\item
$3$
\item
$4$
\end{enumerate}

\vfill

\item
Let $\cal S$ denote the subspace of ${\mathbb R}^3$ equal to the set
of all solutions to $A\vec x=0$ where
$$A=\left[\matrix
0  & 1 & 1\\
0  & 1 & 1\\
-1 & 0 & 2
\endmatrix\right].$$
Which of the following vectors forms a basis for $\cal S$?
\begin{enumerate} [A.]
\item
$(-2,1,1)$
\item
$(2,1,0)$
\item
\correct{$(2, -1,1)$}
\item
$(-1,0,1)$
\item
None of the above.
\end{enumerate}

\vfill

\item
If $T:{\mathbb R}^3\rightarrow {\mathbb R}^3$ is a linear transformation
satisfying
$$T(1,0,0)=(0,0,-1), \quad T(0,1,0)=(1,1,0), \quad T(0,0,1)=(1,1,2),$$
then $T(1,2,3)=$
\begin{enumerate} [A.]
\item
\correct{$(5,5,5)$}
\item
$(2,1,0)$
\item
$(2, -1,1)$
\item
$(0,0,0)$
\item
$(1,2,3)$
\end{enumerate}

\vfill\eject

\item
For which value(s) of $k$ are the vectors
$( 2,-k,0)$, $(1,2,2)$, and $(0,1,-k)$ linearly {\it dependent}?
  \begin{enumerate}[A.]
\item
No values of $k$.
\item
\correct{$k = - 2$}
\item
$k \ne - 2$
\item
$k = 0$
\item
All values of $k$.
\end{enumerate}

\vfill

\item
For which value(s) of $k$ are the vectors
$( 2,-k,0)$ and $(1,2,2)$ linearly {\it dependent}?
  \begin{enumerate}[A.]
\item
\correct{No values of $k$.}
\item
$k = - 2$
\item
$k \ne - 2$
\item
$k = 0$
\item
All values of $k$.
\end{enumerate}

\vfill

\item
For which value(s) of $k$ are the vectors
$( 2,-k,0)$, $(1,2,2)$, $(0,1,-k)$, and $(0,1,k)$ linearly {\it dependent}?
  \begin{enumerate}[A.]
\item
No values of $k$.
\item
$k = - 2$
\item
$k \ne - 2$
\item
$k = 0$
\item
\correct{All values of $k$.}
\end{enumerate}

\vfill\eject

\item
If $A=\left[\matrix
1  & 2 & 3   \\
0  & 2 & 4 \\
0  & 0 & 3
\endmatrix
\right]$, and $B=A^{-1}$, then the entry $b_{23}$ of $B$ is
\begin{enumerate} [A.]
\item
$0$
\item
$-4$
\item
$4$
\item
$1/6$
\item
\correct{$-2/3$}
\end{enumerate}

\vfill

\item
Determine all values of $k$ so that the following system has {\bf no
solution}.
\begin{eqnarray*}
x_1+x_2+3 x_3 &=& 1\\
2x_1+4x_2+5 x_3 &=& 1\\
x_1-x_2+k^2x_3 &=& -k
\end{eqnarray*}
  \begin{enumerate}[A.]
\item
$k\ne\pm2$
\item
$k=\pm 2$
\item
\correct{$k=2$}
\item
$k=-2$
\item
$-2<k<2$
\end{enumerate}

\vfill

\item If one solution of \( y'' + y' - 2 y = f(x) \) is
      \( y(x) = \ln x \), then the general solution is
  \begin{enumerate}[A.]
    \item \( c_1 \ln x \)
    \item \( c_1 \ln x + c_2 e^x + c_3 e^{-2x} \)
    \item \correct{\( c_1 e^x + c_2 e^{-2x} + \ln x \)}
    \item \( c_1 e^x + c_2 e^{-2x} \)
    \item \( c_1 e^{-x} + c_2 e^{2x} + \ln x \)
  \end{enumerate}

\vfill\eject
  
\item If \( y(x) \) is the solution of \( y'' - y' - 2y = 0 \)
satisfying \( y(0) = 1 \) and \( y'(0) = -1 \), then \( y(1) = {} \)
  \begin{enumerate}[A.]
    \item \( e^{-1} + 2e^{2} \)
    \item \correct{\( e^{-1} \)}
    \item \( e^{2} \)
    \item \( e^{2} - e^{-1} \)
    \item \( 2e^{2} \)
  \end{enumerate}

\vfill
  
\item The general solution of the equation \( y'' + 2y' + 5y = 0 \) is
  \begin{enumerate}[A.]
    \item \correct{\( c_1 e^{-x} \cos 2x + c_2 e^{-x} \sin 2x \)}
    \item \( c_1 e^{(-1+i)x} + c_2 e^{(-1-i)x} \)
    \item \( c_1 e^{-x} + c_2 x e^{-x} \)
    \item \( c_1 e^{2x} \cos x + c_2 \, i \, e^{2x} \sin x \)
    \item \( c_1 e^x + c_2 e^{-3x} \)
  \end{enumerate}

\vfill
 
\item The general solution of \( D (D^2 + 1) (D + 1)^2 y = 0 \) is
  \begin{enumerate}[A.]
    \item \( c_1 x + c_2 \cos x + c_3 \sin x + e^{-x} (c_4 + c_5 x) \)
    \item \( c_1 + c_2 e^x + e^{-x} (c_3 + c_4 x + c_5 x^2) \)
    \item \( c_1 \cos x + c_2 \sin x + e^{-x} (c_3 + c_4 x) \)
    \item \( c_1 \cos x + c_2 \sin x + c_3 e^{-x} + c_4 \)
    \item \correct{\( c_1 \cos x + c_2 \sin x + e^{-x} (c_3 + c_4 x) + c_5 \)}
  \end{enumerate}

\vfill\eject

\item In the method of undetermined coefficients, the appropriate trial
      solution for \newline \( (D^2 + 2D + 2)(D + 1) y = \sin x + x e^{-x} \) is
  \begin{enumerate}[A.]
    \item \( A \sin x + B x e^{-x} \)
    \item \( A \sin x + B \cos x + C x e^{-x} \)
    \item \( A \sin x + B \cos x + (C + D x)e^{-x} \)
    \item \correct{\( A \sin x + B \cos x + x(C + D x)e^{-x} \)}
    \item \( A \sin x + B x^2 e^{-x} \)
  \end{enumerate}

\vskip1in
  
\item A particular solution for \( y'' + 2y' + y = x^{-1} e^{-x} \)
is
  \begin{enumerate}[A.]
    \item \( (\ln x - x^2) e^{-x} \)
    \item \( A x^{-1} e^{-x} \)
    \item \( A x e^{-x} \)
    \item \( \ln x \,e^{-x} \)
    \item \correct{\( x\ln x \,e^{-x} \)}
  \end{enumerate}

\vfill\eject
      
\item For which values of the parameter \( \alpha \) will the equation
      \( y'' + \alpha y' + y = 0 \) have solutions whose graphs are
      similar to the following graph?
      \begin{center}
      \includegraphics{pic.eps}
      \end{center}
  \begin{enumerate}[A.]
    \item \( \alpha < 2 \)
    \item \( \alpha = 0 \)
    \item \correct{\( 0 < \alpha < 2 \)}
    \item \( \alpha > 2 \)
    \item all \( \alpha \)
  \end{enumerate}

\vfill

\item
All the solutions of a $2\times2$ homogeneous linear system of
differential equations ${\vec x}\,'=A{\vec x}$ tend to zero as
${t\to\infty}$ if the eigenvalues of $A$ are equal to:
  \begin{enumerate}[A.]
\item
$1$ and $-1$
\item
$i$ and $-i$
\item
$1+i$ and $1-i$
\item
\correct{$-1+i$ and $-1-i$}
\item
None of the above.
  \end{enumerate}

\vfill\eject

\item
The general solution of the linear system of differential equations
\begin{eqnarray*}
x_1'&=&x_1+2x_2\\
x_2'&=&4x_1+3x_2
\end{eqnarray*}
is equal to
  \begin{enumerate}[A.]
\item
\correct{
$c_1\bmatrix e^{-t}\\ -e^{-t}\endbmatrix +
c_2\bmatrix e^{5t}\\ 2e^{5t}\endbmatrix$
}
\item
$c_1\bmatrix e^{-t}\\ e^{-t}\endbmatrix +
c_2\bmatrix e^{5t}\\ -2e^{5t}\endbmatrix$
\item
$c_1\bmatrix e^{-t}\\ -e^{-t}\endbmatrix +
c_2\bmatrix e^{5t}\\ -2e^{5t}\endbmatrix$
\item
$c_1\bmatrix e^{-t}\\ e^{-t}\endbmatrix +
c_2\bmatrix e^{5t}\\ 2e^{5t}\endbmatrix$
\item
None of the above.
  \end{enumerate}

\vfill

\item
The function $x_2(t)$ determined by the initial value problem
\begin{eqnarray*}
x_1'&=&x_2\\
x_2'&=&-x_1
\end{eqnarray*}
with initial conditions $x_1(0)=1$ and $x_2(0)=1$ is given by
  \begin{enumerate}[A.]
\item
\correct{$x_2=-\sin t + \cos t$}
\item
$x_2=\sin t + \cos t$
\item
$x_2=\frac12(e^{t}+e^{-t})$
\item
$x_2= \cos t$
\item
$x_2= ie^{it}-ie^{-it}$
  \end{enumerate}

\vfill\eject

\item
The $2\times2$ matrix
$A=\left[\matrix -3 & -1 \\ 2 & -1\endmatrix\right]$ has complex
eigenvalues $r=-2\pm i$.  An eigenvector corresponding to $r=-2+i$
is $\left(\matrix 1 \\ -1-i \endmatrix\right)$.
The system
$$\vec x\,'=A\vec x+\left(\matrix 3\\-4\endmatrix\right)e^{-2t}$$
has one solution given by
$\vec x(t)=\left(\matrix 1 \\ 2 \endmatrix\right)e^{-2t}$.
What is the general solution to the system?
  \begin{enumerate}[A.]
\item
$c_1
\left(\matrix 1 \\  -1-i \endmatrix\right)e^{(-2+i)t}
+c_2
\left(\matrix 1 \\ 2 \endmatrix\right)e^{2t}$
\item\correct{
$c_1
\left(\matrix \cos t \\ \sin t-\cos t \endmatrix\right)e^{-2t}
+
c_2
\left(\matrix \sin t \\ -\sin t-\cos t \endmatrix\right)e^{-2t}
+
\left(\matrix 1 \\ 2 \endmatrix\right)e^{-2t}$
}
\item
$c_1
\left(\matrix \cos t \\ -\cos t \endmatrix\right)e^{-2t}
+
c_2
\left(\matrix 0 \\ -\sin t \endmatrix\right)e^{-2t}
+
\left(\matrix 1 \\ 2 \endmatrix\right)e^{-2t}$
\item
$c_1
\left(\matrix 1 \\  -1-i \endmatrix\right)e^{(-2+i)t}
+c_2
\left(\matrix -1 \\  1+i \endmatrix\right)e^{(-2-i)t}
+
\left(\matrix 1 \\ 2 \endmatrix\right)e^{-2t}$
\item
$c_1
\left(\matrix \cos t \\ -\cos t \endmatrix\right)e^{-2t}
+
c_2
\left(\matrix 0 \\ -\sin t \endmatrix\right)e^{-2t}
+
c_3\left(\matrix 1 \\ 2 \endmatrix\right)e^{-2t}$
  \end{enumerate}

\vfill\eject
  
\end{enumerate}


\end{document}