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{\bf Name:}  \vrule height .4pt depth 0 pt width 2.0
in  \hskip .5 in {\bf Professor: }
  \vrule height .4pt depth 0 pt width 1.0
in     \hskip 5.0 in {\bf Section: }  \vrule height  .4pt depth 0 pt width 1.0
in  \hskip .5 in
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\bigskip

 \centerline{\bf {Mathematics 325}}
\centerline{\bf Fall Semester 2002}
\centerline{\bf Final Exam.}
\centerline{\bf {December 20, 2002}}
\bigskip
\bigskip
\bigskip
\hrule
\bigskip
\noindent
1. This Examination contains 25 problems worth a total of  150  points.
  The test will be exactly
120 minutes in length.
\bigskip
\noindent
2. For each multiple choice question, please mark an $X$ on the correct
answer on the answer sheet.
{\bf Do not circle it.}
\bigskip
\noindent
3.  Calculators, books and notes are
not allowed.
\bigskip
\noindent
4. A table of Laplace transforms and a  table giving type and stability
properties in terms of eigenvalues
are included at the end of the booklet.
\bigskip
\noindent
5. Hand in the entire test.
\bigskip
\hrule
\bigskip
\bigskip



{\bf Sign the pledge:}  \lq \lq {On my honor,
I have neither given nor received unauthorized aid on this Exam}":
\bigskip
\hskip .7 in\vrule height .10pt depth 0 pt width 2 in
\bigskip
\bigskip
\centerline{\bf HAVE A NICE VACATION}



\newpage
\noindent
1. The Wronskian of $\{1,\cos t,\sin t\}$ is
\medskip
(a). $\cos t$
\medskip
(b). $\sin t$
\medskip
(c). 1
\medskip
(d). $-1$
\medskip
(e).  $\sin^2 t-\cos^2 t$

\bigskip
\noindent
2. Determine a suitable form for  a particular solution of the equation
$$y''''-3y'' -4y=  \cos 2t.$$
\medskip
(a). $A \sin 2t$
\medskip
(b). $A\cos 2t+B\sin 2t$
\medskip
(c). $ B \cos 2t$
\medskip
(d). $(At+B)\cos 2 t$
\medskip
(e). $ At\cos 2t+Bt\sin 2t $

\bigskip
\noindent
3. Find  the general solution  of the equation
$$y'''+y''-y'-y = 0.$$
\medskip
(a). $C_1e^{ t}+C_2e^{-t}+C_3te^{-t} $
\medskip
(b). $C_1e^{ t}+C_2e^{-t}+C_3 e^{-t} $
\medskip
(c). $C_1e^{ t}+C_2te^{ t}+C_3 e^{-t} $
\medskip
(d).  $C_1e^{ -t}+C_2te^{-t}+C_3t^2e^{-t} $
\medskip
(e). $C_1e^{ t}+C_2e^{-t} $

\newpage
\noindent
4. Let $X_1(t)=3+t,\ X_2(t)=-t+2t^2,\ X_3(t)=2-t^2, X_4(t)=3-t+4t^2$.
Which of the following is {\it false}?
\medskip
(a). $X_1,X_2,X_4$ are linearly independent
\medskip
(b). $X_1, X_2,X_3$ are linearly independent
\medskip
(c). $X_1,X_2 $ are linearly independent
\medskip
(d). $X_1, X_3$ are linearly independent
\medskip
(e). $X_2,X_3,X_4$ are linearly independent


\bigskip
\noindent
5.  Let
$$ f(t)=\int_0^t(t-\tau)^3e^{2\tau}d\tau.
 $$
Find the Laplace transform of $f$.
\medskip
(a).    $ \frac 6{s^4(s+2)}$
\medskip
(b). $ \frac 6{s^4(s-2)}$
\medskip
(c). $s^3e^{-2s}$
\medskip
(d).   $ \frac 3{s^4(s+2)}$
\medskip
(a). $ \frac 6  { (s+2)^4}$

\bigskip
 \noindent
6. The  Laplace transform of $$f(t)=t^2\delta(t-3) $$ is
\medskip
(a). $\frac 2{s^3}$
\medskip
(b).  $9e^{-3s}$
\medskip
(c). $e^{-3s}$
\medskip
(d)  $\frac 2{(s-3)^3}$
\medskip
(e).  $3e^{-9s}$.

\newpage
\noindent
7. Solve the initial value problem $$y''=u_1(t),\  y(0)=1, y'(0)=0.$$
\medskip
(a). $y=1+ 2u_1(t) t^2$
\medskip
(b). $y=1+ u_1(t) (t-1)^2$
\medskip
(c).  $y=1+\frac 12u_1(t) t^2$
\medskip
(d).  $y=1+\frac 12u_1(t) (t-1)^2$
\medskip
(e).  $y= \frac 12(1+u_1(t)t^2)$

\bigskip
\noindent
8. Solve the initial value problem
$$y''-4y=8\delta (t-1),\quad y(0)=y'(0)=0 .$$

\medskip
(a). $y= 2 (e^{2t}+e^{-2t})$
\medskip
(b). $y= 2 u_1(t)(e^{2t}+e^{-2t})$
\medskip
(c). $y= 2 u_1(t)(e^{2t}-e^{-2t})$
\medskip
(d). $y= 2 u_1(t)(e^{2(t-1)}+e^{-2(t-1)})$
\medskip
(e).  $y= 2 u_1(t)(e^{2(t-1)}-e^{-2(t-1)})$

\bigskip
\noindent
9. Consider the initial value problem  $$y'=5y,\  y(0)=1.$$  Using
step size
$h=\frac 1{10}$ and two steps of the (forward) Euler method, what is the
error at $t=\frac 15$
compared to   the actual solution to the equation?
\medskip
(a). $|1-\frac 94|$
\medskip
(b). $|e-\frac 94|$
\medskip
(c). $|1-\frac 32|$
\medskip
(d). $|e-\frac 32|$
\medskip
(e).  $|1-e|$

\newpage
\noindent
10. Find the general solution of

  $$\frac {d\Bbb X}{dt}=\left(\matrix   1&5\\
3&-1\endmatrix \right)\Bbb X.$$
 (a).  $$C_1 e^{4t}\left(\matrix   5\\
3\endmatrix \right)+C_2e^{-4t}\left(\matrix   1\\
-1\endmatrix \right).$$
(b).  $$C_1 e^{4t}\left(\matrix   5\\
3\endmatrix \right)+C_2e^{ 4t}\left(\matrix   1\\
-1\endmatrix \right).$$
 (c).  $$C_1 e^{4t}\left(\matrix   5\\
3\endmatrix \right)+C_2e^{-4t}\left(\matrix   0\\
-1\endmatrix \right).$$
(d).  $$C_1 e^{4t}\left(\matrix  3\\
5\endmatrix \right)+C_2e^{-4t}\left(\matrix   1\\
 1\endmatrix \right).$$
(e).  $$ e^{ 4t}\left(\matrix   1\\
-1\endmatrix \right).$$
\bigskip
\noindent
11. 	The trajectory corresponding to the solution of
$$ \aligned &\frac {dx}{dt}=4y,\quad  x(0)=1,\\&\frac {dy}{dt}=-x,
\quad  y(0)=0 \endaligned$$

is
\medskip
(a). $ x^2 +4y^2 = 1$, traveling clockwise.
\medskip
	(b). $4x^2    + y^2    = 1$, traveling  clockwise.
\medskip
	(c). $ x^2   + 4y^2 = 1$, traveling counterclockwise.
\medskip
(d).  $4x^2 +   y^2  = 1	$, traveling counterclockwise.
\medskip
			(e).  $x^2   +4 y^2 =4$, traveling counterclockwise.

\newpage
\noindent
12. The matrix     $ \left(\matrix   1&2\\
2&1\endmatrix \right) $
has eigenvalues 3 and $-1$ and eigenvectors  $\left(\matrix   1\\
 1\endmatrix \right)$ and $\left(\matrix   1\\
 -1\endmatrix \right)$ respectively. Find the general solution to
 $\dsize\frac {d\Bbb X}{dt}=\left(\matrix   1&2\\
2&1\endmatrix \right)\Bbb X+\left(\matrix   0\\
 {e^t}\endmatrix \right).$

\medskip
\noindent
 (a).$$C_1 e^{ t}\left(\matrix   1\\
0\endmatrix \right)+C_2e^{- t}\left(\matrix   0\\
 1\endmatrix \right)+ e^{ 3t}\left(\matrix   1\\
1\endmatrix \right).$$
(b).$$C_1 e^{3 t}\left(\matrix   1\\
1\endmatrix \right)+C_2e^{- t}\left(\matrix   1\\
 -1\endmatrix \right)- \frac 12 e^{  t}\left(\matrix   0\\
1\endmatrix \right).$$
(c).$$C_1 e^{3 t}\left(\matrix   1\\
1\endmatrix \right)+C_2e^{- t}\left(\matrix   1\\
- 1\endmatrix \right)+ \frac 12 e^{  t}\left(\matrix   0\\
1\endmatrix \right).$$
(d).$$C_1 e^{3 t}\left(\matrix   1\\
1\endmatrix \right)+C_2e^{- t}\left(\matrix   1\\
 -1\endmatrix \right)+ \frac 12 e^{  t}\left(\matrix   1\\
0\endmatrix \right).$$
(e).$$C_1 e^{ 3t}\left(\matrix   1\\
 1\endmatrix \right)+C_2e^{- t}\left(\matrix   1\\
 -1\endmatrix \right)- \frac 12 e^{  t}\left(\matrix   1\\
0\endmatrix \right).$$



\bigskip
\noindent
13. The system of nonlinear differential equations:

$$\aligned \frac{dx}{dt}&= x-xy\\
           \frac{dy}{dt}&=-y+xy\endaligned$$
 has an isolated critical point at $(1,1)$. The linearization of this system
around the critical point $(1,1)$, using the coordinates $u=x-1$,
$v=y-1$, is

\medskip
(a).  $$\frac{d}{dt} \left(\matrix u\\v \endmatrix \right)=
      \left(\matrix 1 &0\\ 0&-1\endmatrix \right)\left(\matrix u\\v \endmatrix
\right)$$

(b). $$\frac{d}{dt} \left(\matrix u\\v \endmatrix \right)=
      \left(\matrix 0&1\\ -1&0\endmatrix \right)\left(\matrix u\\v
\endmatrix\right)$$

(c). $$\frac{d}{dt} \left(\matrix u\\v \endmatrix \right)=
      \left(\matrix 1&-1\\ 1&1\endmatrix \right)\left(\matrix u\\v \endmatrix
\right)$$

(d).  $$\frac{d}{dt} \left(\matrix u\\v \endmatrix \right)=
      \left(\matrix 0&-1\\ 1&0\endmatrix \right)\left(\matrix u\\v \endmatrix
\right)$$

(e).  $$\frac{d}{dt} \left(\matrix u\\v \endmatrix \right)=
      \left(\matrix 0&1\\  1&0\endmatrix \right)\left(\matrix u\\v \endmatrix
\right)$$

\bigskip
\noindent
14. Consider the nonlinear system of differential equations
$$\aligned \frac{dx}{dt}&=y+x^2\\
           \frac{dy}{dt}&= -\sin x- y.\endaligned$$
Classify the type and stability of the critical point at $(0,0)$.

\medskip
(a). $(0,0)$ is a spiral point and is unstable.
\medskip
(b). $(0,0)$ is an improper node and  unstable.
\medskip
(c). $(0,0)$ is a saddle point and is unstable.
\medskip
(d). $(0,0)$ is a center and is stable.
\medskip
(e). $(0,0)$ is a spiral point and is asymptotically stable.

\bigskip
\noindent
15. Find the eigenvalues of the matrix $\left(\matrix   0&2&16\\-2&0&9\\ 0&0&11
 \endmatrix \right)$

\medskip

(a). $2i$, -$2i$, 11
\medskip
(b). 2, 2,  11
\medskip
(c). $2i$, -$2i$, 16
\medskip
(d). 2, -2, 11
\medskip
(e). $2i$, -$2i$, $3i$







\newpage
\noindent
16.  Find all critical points for the following system of nonlinear
differential equations:
$$\aligned \frac{dx}{dt}&= x- x^2-xy\\
           \frac{dy}{dt}&=2y- y^2-3xy\endaligned$$
\medskip
(a). $(0,0)$
\medskip
(b). (0,0), (1,0)
\medskip
(c). $(0,0), (1,0), (0,2) $
\medskip
(d).  $(0,0), (1,0), (0,1), (0,2)$
\medskip
(e). $(0,0), (1,0), (0,2), (\frac 12,\frac 12)$





\bigskip
\noindent
17. Use the Liapunov  function $V(x,y)=x^2+y^2$ to decide the stability of the
critical point (0,0) of the
system
$$\aligned \frac{dx}{dt}&=x^3-y^3\\
           \frac{dy}{dt}&=xy^2+4y^3.\endaligned$$


\medskip
(a). $(0,0)$   is unstable.
\medskip
(b).  $(0,0)$ is   asymptotically stable.
\medskip
(c). $(0,0)$ is stable but not asymptotically stable.
\medskip
(d). $(0,0)$ is a center and is stable.
\medskip
(e). One cannot conclude the stability from the Liapunov function.



 \bigskip
\noindent
18. Which of the following functions is  not periodic?
\medskip
(a). $\sin \pi x$
\medskip
(b). $(\cos x)^2$
\medskip
(c). $e^x-e^{-x}$
\medskip
(d). $\dsize e^{\cos x}$
\medskip
(e). 100

\newpage
\noindent
19. Let
$  f(x)=x^3+1$ on $ -1<x\le 1$ and  $f(x+2 )=f(x)$. What value does the Fourier
series of $f$ converges to at $x=1$?

\medskip
(a). $2 $
\medskip
(b). $0$
\medskip
(c). $1 $

\medskip
(d). $-1 $
\medskip
(e). The Fourier series diverges at $x=1$.
\bigskip
\noindent
20.
What is the solution of the heat equation
$$
\left\{
\aligned  &4u_{xx}=u_t, \quad \quad 0< x<2,\ t>0\\
& u(0,t)=u(2,t)=0,\quad 0<t \\
& u(x,0)=\sin \pi x,\quad 0<x<2 \endaligned \right.
$$

(a). $$u= e^{\frac{-\pi^2 t  }2} \sin \pi x  .$$

(b). $$u= e^{ {-4\pi^2 t  } } \sin \pi x  .$$

(c). $$u= e^{\frac{-\pi^2 t  }4} \sin \pi x  .$$

(d). $$u= e^{\frac{-\pi^2 t  }4} \sin \frac{\pi x}2  .$$

(e).   $$u= e^{ {4\pi^2 t  } } \sin \pi x  .$$

\newpage
\noindent
21. Consider the two-point boundary value problem $$y''+9y=0, \ y(0)=1,\
y(\pi)=\alpha.$$
For which value(s) of $\alpha$ does this problem have infinitely many
solutions?
\medskip
(a). 0
\medskip
(b). $-1$
\medskip
(c). $1$
\medskip
(d). $1$ or $-1$
\medskip
(e). No $\alpha$ will satisfy the conditions.
\bigskip
\noindent
22. Let $f(x)=x ,\  0\le x< 1$. Consider the even extension of $f$  and
$f(x+2)=f(x)$.  The
Fourier coefficient  $a_1$ in
 $$f(x)= \frac  {a_0}2+\sum_{n=1}^\infty a_n\cos
  {n\pi x} +\sum_{n=1}^\infty b_n\sin   {n\pi x} $$ is
\medskip
(a). 0
\medskip
(b). $-\frac 2\pi$
\medskip
(c). $\frac 2\pi$
\medskip
(d).  $-\frac 4{\pi^2}$
\medskip
(e). $ \frac 4 {\pi^2}$
\bigskip
\noindent
23. Consider the equation
$$u_{xx}-u_{tt}=0.$$
Let $u=X(x)T(t)$ be a solution.  Then $X$ and $T$ satisfy    which of the
following pairs of ordinary differential equations:
\medskip
(a). $X' +\lambda X=0,\  T' +\lambda T=0$
 \medskip
(b). $X''+\lambda X=0,\  T''-\lambda T=0$
\medskip
(c). $X''-\lambda X=0,\  T' +\lambda T=0$
\medskip
(d). $X''+\lambda X=0,\  T''+\lambda T=0$
\medskip
(e). $X''+\lambda X=0,\  T''+\lambda^2 T=0$

\newpage
\noindent
24. Let
 $$ f(x)=
\left\{
\aligned  &3, \quad \quad 0< x<1 \\
& 0,\quad\ \ \  1<x<2.\endaligned
\right.
$$
Consider the odd extension of
$f(x)$ with $f(x+4)=f(x)$.
    The Fourier coefficient  $a_5$ in
 $$f(x)= \frac  {a_0}2+\sum_{n=1}^\infty a_n\cos
\frac {n\pi x}2+\sum_{n=1}^\infty b_n\sin \frac {n\pi x}2$$ is

(a). $\dsize\frac {1 }{5\pi}\ \ \ $
 (b).  $\dsize\frac {2}{5\pi}\ \ \ $
 (c). $\dsize\frac {4}{5 }\ \ \ $
 (d). $0\ \ \ $
 (e).$\dsize\frac {-2}{5\pi}$


\bigskip
\noindent
25. The sine series of $f(x)=x$, $0\le x\le 2$ is
$$ \sum_{n=1}^\infty\frac 4{n\pi}(-1)^{n+1}\sin\frac{n\pi x}2.$$
What is the solution of the wave equation
$$
\left\{
\aligned  &u_{xx}=u_{tt}, \quad \quad 0< x<2,\ t>0\\
& u(0,t)=u(2,t)=0,\quad t>0 \\
& u(x,0)=x, \  u_t(x,0)=0,\quad 0< x< 2 \endaligned \right.
$$

(a). $$u=\frac 4\pi \sum_{n=1}^\infty\frac {(-1)^{n+1}}{n
}\sin\frac{n\pi x}2\sin\frac{n\pi t}2.$$

(b). $$u=\frac 4\pi \sum_{n=1}^\infty\frac {(-1)^{n+1}}{n
}\sin\frac{n\pi x}2\cos\frac{n\pi t}2 .$$

(c). $$u=\frac 4\pi \sum_{n=1}^\infty\frac {(-1)^{n+1}}{n }
\sin\frac{n\pi x}2\sin {n\pi  t} .$$

(d). $$u=\frac 4\pi \sum_{n=1}^\infty\frac {(-1)^{n+1}}{n }e^{ -\pi^2n^2t
} \sin\frac{n\pi x}2.$$

(e). $$u=\frac 4\pi \sum_{n=1}^\infty\frac {(-1)^{n+1}}{n
}e^{\frac{-\pi^2n^2t  }4} \sin
  2n\pi x .$$



 \end{document}
 
