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\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, 2 May 2006}}}
\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\\
Final Exam\\
Tuesday, 2 May 2006, 4:30-6:30pm or 6:00-8:00pm \\
\end{center}

{\bf Instructions}: This in-class exam is 120 minutes. No
calculators, notes, tables or books. No answer check is expected.
Details count $75$\%. The answer counts $25$\%.

\EJECT



\begin{exercises}
\prob{1.}{(ch1: Separable Equation)} ~\newline
 % u:=(2*x^2+4*x)/(1+x); int(u,x);
 Given $\dd y^2y' = \frac{2x^2+4x}{1+x}\left(\frac{27}{8}-y^3\right)$. ~\newline
 (a) [20\%]  Find all equilibrium solutions. ~\newline
 (b) [80\%] Find all non-equilibrium solutions in implicit form. ~\newline
 To save time, {\bf do not solve} for $y$ explicitly.

\end{exercises}

\ifsolutions
\begin{solution}{1} ~\newline
 There is one equilibrium solution $y=3/2$. \newline
 Long division gives the quadrature step $\int
 \frac{2x^2+4x}{1+x}dx=x^2+2x-2\ln|1+x|$. The left side quadrature
 step is $\int u^2du/(27/8-u^3)= (-1/3)\ln|27/8-u^3|$.\newline
 The implicit solution is
 $$
  -\frac13\ln\left| \frac{27}{8}-y(x)^3\right| =
  c+x^2+2x-2\ln|1+x|.
 $$
\end{solution}
\fi

\EJECT

\begin{exercises}
\prob{2.}{(ch1: Linear Equation)} ~\newline
% de:=2*diff(v(t),t)=-32+(2/(2*t+1))*v(t); ic:=v(0)=-16; dsolve({de,ic},v(t));
 (a) [60\%] Solve $\dd 2v'(t)=-32 + \frac{2}{2t+1}v(t)$, $v(0)=-16$. Show
 all integrating factor steps.\newline
 % de:=2*sqrt(x+2)*diff(y(x),x)=y(x)+Pi; ic:=y(0)=-Pi: dsolve({de,ic},y(x));
 (b) [40\%]  Solve $\dd 2\sqrt{x+2}\,\frac{dy}{dx}=y+\pi$, $y(0)=-\pi$.
\newline


\end{exercises}

\ifsolutions
\begin{solution}{2} ~\newline
 (a) $v(t) = -32t-16$. \newline
 (b) $y(x)=-\pi$.
\end{solution}
\fi

\EJECT



\begin{exercises}
\prob{3.}{(ch3: Recipe)} ~\newline (a) Using the {\em recipe} for
higher order constant-coefficient differential equations, write
out the general solutions:


 $\begin{array}{ll} {\bf (a.1)} ~[25\%] & y^{vi} - 9y^{iv} = 0
 \end{array}$,

$\begin{array}{ll} {\bf (a.2)} ~[25\%] & \mbox{Char. eq. ~}
 r^2(r-4)(r^3-16r)^2(r^2+16)^3=0
 \end{array}$.

(b) [50\%]  Given $4x''(t) + 4x'(t) + x(t) = 0$, which represents a
damped spring-mass system with $m=4$, $c=4$, $k=1$, solve the
differential equation [40\%] and classify the answer as
over-damped, critically damped or under-damped [10\%].

\ifsolutions
\begin{solution}{3} ~\newline
Part (a)\newline
 {\bf a.1}: $r^6-9r^4=0$, roots $r=0,0,0,0,3,-3$. Then
 $y=(c_1+c_2x+c_3x^2+c_4x^3)e^{0x}+c_5e^{3x}+c_6e^{-3x}$. \newline
 {\bf a.2}: Write as $r^4(r-4)^3(r+4)^2(r^2+16)^3=0$.
 Then $y=u_1e^{0x}+u_2e^{4x}+u_3e^{-4x} + u_4\cos 4x + u_5\sin 3x$. The
 polynomials are $u_1=c_1+c_2x+c_3x^2+c_4x^3$, $u_2=c_5+c_6x+c_7x^2$,
 $u_3=c_8+c_9x$, $u_4=c_{10}+c_{11}x+c_{12}x^2$, $u_5=c_{13}+c_{14}x+c_{15}x^2$.

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{4.}{(ch3: Undetermined Coefficients)} ~\newline Determine
for $ y^{vi} + 4y^{iv} = x + x^3 + e^{-x}+x\cos 2x$ the corrected
trial solution for $y_p$ according to the method of undetermined
coefficients. To save time, {\bf do not evaluate} the undetermined
coefficients!

\ifsolutions
\begin{solution}{4} ~\newline
The homogeneous solution is $y_h=c_1+c_2x+c_3x^2+c_4x^3+c_5\cos
2x+c_6\sin 2x$, because the characteristic polynomial has roots
$0$, $0$, $0$, $0$, $2i$, $-2i$.

{\bf \fbox{1}~} An initial trial solution $y$ is constructed from
the 9 atoms $1$, $x$, $x^2$, $x^3$, $e^{-x}$, $\cos 2x$, $\sin
2x$, $x\cos 2x$, $x\sin 2x$, giving
$$\begin{array}{lcl}
 y &=& y_1+y_2+y_3,\\
 y_1 &=& (d_1+d_2x+d_3x^2+d_4x^3)e^{0x},\\
 y_2 &=& d_5e^{-x},\\
 y_3 &=& (d_6+d_7x)\cos 2x+(d_8+d_9x)\sin 2x.
\end{array}
$$
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
$y_1$, $y_2$, $y_3$ to give the \BF{corrected trial solution}
$$\begin{array}{lcl}
 y &=& y_1+y_2+y_3,\\
 y_1 &=& x^4(d_1+d_2x+d_3x^2+d_4x^3)e^{0x},\\
 y_2 &=& d_5e^{-x},\\
 y_3 &=& x(d_6+d_7x)\cos 2x+x(d_8+d_9x)\sin 2x.
\end{array}
$$
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
$y_2$ 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{5.}{(ch5: Eigenanalysis Method)} ~\newline
% The eigenanalysis method says that the $4\times 4$ system
%$\.x' = A\.x$ has general solution $\.x(t)=c_1\.v_1e^{\lambda_1
%t}+c_2\.v_2e^{\lambda_2 t}+c_3\.v_3e^{\lambda_3
%t}+c_4\.v_4e^{\lambda_4 t}$. In the solution formula,
%$(\lambda_i,\.v_i)$, $i=1,2,3,4$, is an eigenpair of $A$.
Given
$$A=\left[\begin{array}{rrrr}
 5 & 1 &  1  & 0\\
 1 & 5 &  1  & 0 \\
 0 & 0 &  7  & 0 \\
 1 & 0 &  5  & 1
 \end{array}\right],$$
 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}{5} ~\newline
%A:=matrix([[ 5 , 1 ,  1  , 0],
% [ 1 , 5 ,  1  , 0],
% [ 0 , 0 ,  7  , 0],
% [ 1 , 0 ,  5  , 1]]);
% eigenvals(A);
%#                              1, 4, 6, 7
% eigenvects(A);
% #  [1, 1, {[0, 0, 0, 1]}], [7, 1, {[1, 1, 1, 1]}],
% #        [4, 1,{[3, -3, 0, 1]}], [6, 1, {[5, 5, 0, 1]}]
Answer (a): The eigenpairs are
$$
 1,  \colvector{0\\ 0\\ 0\\ 1};\quad 7, \colvector{1\\ 1\\ 1\\ 1};\quad
  4, \colvector{3\\ -3\\ 0\\ 1};\quad 6, \colvector{5\\ 5\\ 0 \\ 1}.
$$
Answer (b): The eigenanalysis method implies
$$
\.x(t)=c_1e^t\colvector{0\\ 0\\ 0\\ 1}+ c_2e^{ 7t}\colvector{1\\ 1\\
1\\ 1}+c_3e^{4t}\colvector{3\\ -3\\ 0\\ 1}+c_4e^{6t}\colvector{5\\
5\\ 0 \\ 1}.
$$
\end{solution}
\fi


\end{exercises}
\EJECT

\begin{exercises}
\prob{6.}{(ch5; Putzer Spectral Formula)} ~\newline
 Let $4\times 4$ real matrix $A$ have eigenvalues $-2$, $0$, $2$,
 $3$.
% dsolve({diff(y(x),x)=-2*y(x),y(0)=1},y(x));
% #                          y(x) = exp(-2 x)
% dsolve({diff(y(x),x)=0*y(x)+exp(-2*x),y(0)=0},y(x));
% #                     y(x) = - 1/2 exp(-2 x) + 1/2
% dsolve({diff(y(x),x)=2*y(x)-1/2*exp(-2*x)+1/2,y(0)=0},y(x));
% #              y(x) = 1/8 exp(-2 x) - 1/4 + 1/8 exp(2 x)
% dsolve({diff(y(x),x)=3*y(x)+1/8*exp(-2*x)-1/4+1/8*exp(2*x),y(0)=0},y(x));
% #    y(x) = - 1/40 exp(-2 x) + 1/12 - 1/8 exp(2 x) + 1/15 exp(3 x)

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 $ r_4(t) = -\frac{1}{40} e^{-2t}+\frac{1}{12}e^{0t} -\frac18
e^{2t} + \frac{1}{15} e^{3t}$.

\ifsolutions
\begin{solution}{6} ~\newline
Let
$$B=\left[\begin{array}{rrrr}
 -2 & 0 & 0 & 0 \\
 1 & 0 & 0 & 0 \\
 0 & 1 &  2 & 0 \\
 0 & 0 &  1 & 3
 \end{array}\right].$$
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,
$$ r_1=e^{-2t},\quad r_2=-\frac12e^{-2t}+\frac12, \quad r_3=\frac18e^{-2t}-\frac14+\frac18e^{2t},$$
$$ r_4=-\frac{1}{40}e^{-2t} + \frac{1}{12}e^{0t} - \frac{1}{8}e^{2t} + \frac{1}{15}e^{3t}.$$
Define
$$P_1=I,\quad P_2=A+2I,\quad P_3=(A+2I)A, \quad P_4=(A+2I)A(A-2I).$$
Then $ \.u=\left(r_1P1+r_2P_2+r_3P_3+r_4P_4\right)\.u_0$ implies
$$ \.u=\left(r_1I+r_2(A-+2I)+r_3(A+2I)A+r_4(A+2I)A(A-2I)\right)\.u_0$$
\end{solution}
\fi
\end{exercises}
\EJECT

\begin{exercises}
\prob{7.}{(ch6: Stability and Phase Diagram)} ~\newline
 (a) [40\%] Draw a figure to illustrate an {\em asymptotically stable equilibrium} for $\.u'=\.f(\.u)$,
 a 2-dimensional system. Draw a second figure to illustrate a {\em
 stable equilibrium} that is not asymptotically stable.
\newline
 (b) [60\%] Define saddle, spiral, center and node for a linear $2\times
 2$ system $\.u'=A\.u$, in terms of the eigenvalues of $A$.
 \newline



\ifsolutions
\begin{solution}{7} ~\newline

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


 \EJECT
\begin{exercises}
\prob{8.}{(ch7: Laplace Theory)} ~\newline
 (a) [20\%] Display the 4 entries of the basic Laplace Table.\newline
 (b) [40\%] State 6 Laplace rules. \newline
 (c) [40\%] State Lerch's Theorem [cancellation law]. Please include
 hypotheses on the integrands of the Laplace integrals in the
 cancellation law.

\ifsolutions
\begin{solution}{8} ~\newline

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

\begin{exercises}
\prob{9.}{(ch7: Laplace Method)} ~\newline
  (a) [50\%] Solve for $f(t)$:
  $$\L(f)=\dd\frac{d}{ds}\left(\frac{s}{(s+1)^2+4}\right) +
  \frac{s+2}{s^2+2s+10} +
  \left.\frac{d}{ds}\left( \frac{s}{(s(s+1)(s-1)} \right)\right|_{s\to
  s+1}.$$
 (b ) [25\%] Solve $y''+4y'+4y=t$, $y(0)=y'(0)=0$ by Laplace's Method.
  \newline
 (c) [25\%] Solve the system $x'=x-y$, $y'=x+y+2$, $x(0)=0$, $y(0)=0$
 by Laplace's Method for $x(t)$ {\bf only}. Don't solve for
 $y(t)$!


\ifsolutions
\begin{solution}{9} ~\newline


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

 \EJECT
\begin{exercises}
\prob{10.}{(ch9: Fourier Series)} ~\newline
 (a) [30\%] Find the Fourier sine and cosine coefficients for the $1$-periodic
 function $f(t)$ equal to $t-1$ on $0\le t \le 1$.
 \newline
 (b) [20\%] Define a function $f$ piecewise continuous on $a<t<b$. \newline
 (c) [20\%] State Fourier's convergence theorem. \newline
 (d) [30\%] Find the (standard) Fourier coefficients for the $2\pi$-periodic
 function $f(t)= \sin^2(t) - \cos(2t) + \sin(2t)\cos(3t)$.

\ifsolutions
\begin{solution}{10} ~\newline


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



 \EJECTEND

\end{document}
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