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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 S2009}}}
\def\EJECTEND{\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 S2009}

\begin{center}\Large\bf
Differential Equations 2280\\
Final Exam\\
Wednesday, 6 May 2009, 8:00-10:00am \\
\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 $25$\%.


\begin{exercises}
\prob{1.}{(ch3)} ~\newline
% de:=2*diff(v(t),t)=-8+(2/(2*t+1))*v(t); ic:=v(0)=-4; dsolve({de,ic},v(t));
 (1a) [20\%] Solve $\dd 2v'(t)=-8 + \frac{2}{2t+1}v(t)$, $v(0)=-4$. Show
 all integrating factor steps.

 (1b) [10\%] Solve for the general solution: $y'' + 4y' + 6y=0$.

 (1c) [10\%] Solve for the general solution of the homogeneous constant-coefficient differential equation whose characteristic equation is $r(r^2+r)^2(r^2+9)^2=0$.

 (1d) [20\%] Find a linear homogeneous constant coefficient differential equation of lowest order which has a particular solution $y=x + \sin \sqrt2 x + e^{-x}\cos 3x$.

 (1e) [15\%] A particular solution of the equation $
mx'' + cx' + kx = F_0\cos(2t)$ happens to be $x(t)=11\cos 2t+ e^{-t}\sin \sqrt{11}t  -\sqrt{11} \sin 2t $.
Assume $m, c, k$ all positive. Find the unique periodic steady-state solution $x_{\mbox{ss}}$.

(1f) [25\%] Determine for $ y''' + y'' =
100 + 2e^{-x}+\sin x$ the shortest trial solution for
$y_p$ according to the method of undetermined coefficients. Do not
evaluate the undetermined coefficients!

\Answer{~\newline
 (1a) $v(t) = -4 -8t$
 \newline
 (1b) $r^2+4r+6=0$, $y=c_1y_1+c_2y_2$, $y_1=e^{-2x}\cos \sqrt2 x$, $y_2=e^{-2x}\sin \sqrt2 x$.
\newline
 (1c) Write as $r^3(r+1)^2(r^2+9)^2=0$.
 Then $y$ is a linear combination of the atoms
 $1$, $x$, $x^2$, $e^{-x}$, $xe^{-x}$,
 $\cos 3x$, $x\cos 3x$,
 $\sin 3x$, $x\sin 3x$.
\newline
 (1d) It has to the terms left over after striking out the transient terms, those terms with limit zero at infinity. Then $x_{\mbox{ss}}(t)=11\cos 2t  -\sqrt{11} sin 2t $.
 \newline
 (1e) The homogeneous solution is a linear combination of the atoms $1$, $x$, $e^{-x}$
because the characteristic polynomial has roots $0$, $0$, $-1$.
\newline
{\bf \fbox{1}~} An initial trial solution $y$ is constructed for
atoms $1$, $x$, $x^2$, $\cos x$, $\sin x$ giving
$$\begin{array}{lcl}
 y &=& y_1+y_2+y_3,\\
 y_1 &=& d_1+d_2x+d_3x^2,\\
 y_2 &=& d_4\cos x,\\
 y_3 &=& d_5\sin x.
\end{array}
$$
Linear combinations of the listed independent atoms are supposed
to reproduce, by specialization of constants, all derivatives of the
right side of the differential equation.
\newline
{\bf \fbox{2}~} The correction rule is applied individually to each of
$y_1$, $y_2$, $y_3$.\newline
The result is the {\bf shortest trial
solution}
$$\begin{array}{lcl}
 y &=& y_1+y_2+y_3,\\
 y_1 &=& d_3x^2+d_6x^3+d_7x^4,\\
 y_2 &=& d_4\cos x,\\
 y_3 &=& d_5\sin x.
\end{array}
$$
Some facts:
\begin{itemize}
\item The number of terms in each of $y_1$ to $y_3$ is unchanged.
\item If an atom of the homogeneous equation appears in a group, then it is removed. The crossed-out term is replaced by adding another term on {\bf the end} of that group.
\item Suppose a group has base atom $A$. The number $s$ of crossed-out terms for this group is
    exactly the number of atoms of the homogeneous equation having base atom $A$.
 \newline
 This number $s$ is the corresponding root multiplicity for base atom $A$ in the characteristic equation. The value $s$ appears in the Edwards-Penney table containing the mystery factor $x^s$.
\end{itemize}
}
\end{exercises}
\EJECT

\begin{exercises}
\prob{2.}{(ch5)} ~\newline
  (2a) [25\%] A certain $2\times 2$ matrix $A$ has eigenpairs
 $
 \left(4, \colvector{-1\\ 1}\right)$, $\left(6,  \colvector{1\\ 1}\right)$.
The general solution of $\.x'(t)=A\.x(t)$ is $\.x(t)=e^{At}\.x(0)$. Eigenanalysis gives a different formula for $\.x(t)$. Combine these facts to find $e^{At}$.

 (2b) [25\%] Let $A=\left(\begin{array}{rr}
 4 & 1  \\
 1 & 4
 \end{array}\right)$. The Cayley-Hamilton method says that the solution of $\.x'=A\.x$, $\.x(0)=\colvector{1\\ 2}$ is given by $\.x(t)=e^{3t}\.c_1 + e^{5t}\.c_2$. Find the constant vectors $\.c_1$, $\.c_2$ which produce the solution with $\.x(0)=\colvector{1\\ -1}$.

 (2c) [20\%] Let $A=\left(\begin{array}{rrrr}
 3 & 1  \\
 1 & 3  \\
 \end{array}\right)$. Find the general solution of $\.x'=A\.x$ using eigenanalysis.

 (2d) [20\%] Let $A=\left(\begin{array}{rrrr}
 3 & 1  \\
 0 & 3  \\
 \end{array}\right)$. Find the general solution of $\.x'=A\.x$ using the exponential matrix.
 The Laplace resolvent formula implies $\L(e^{At})=(sI-A)^{-1}$. Putzer's formula also applies.
 Use any method available to find $e^{At}$.

 (2e) [10\%]  Let $e^{At}=\twobytwo{e^t & 0}{te^t & e^t}$. Find $A$.
 
\Answer{~\newline
% A:=matrix([[5,1,1],[1,5,1],[0,0,7]]);eigenvects(A);
 (2a) to be completed.
\newline
(2b) to be completed.
\newline
(2c) to be completed.
\newline
 (2d) to be completed.
 \newline
 (2e) to be completed.
}
\end{exercises}
\EJECT


\begin{exercises}
\prob{3.}{(ch6)} ~\newline
 (3a) [20\%]  Compute all equilibrium points of $x'=14x-x^2/2-xy$,
 $y'=16y-y^2/2-xy$.

 (3b) [40\%]  For the nonlinear competition system $x'=14x-x^2/2-xy$,
 $y'=16y-y^2/2-xy$, compute the linearization at $(12,8)$, which is a linear equation $\.u'=A\.u$.  Classify the unique equilibrium $(0,0)$ of $\.u'=A\.u$ as a node,
 spiral, center, saddle. What classification can be deduced for
 the nonlinear system?

 (3c) [20\%] Write the nonlinear pendulum equation $x''+\sin x = 0$ as a nonlinear dynamical system. Then find all equilibrium points of the dynamical system.

  (3d) [20\%] The dynamical system $x'=y+1$, $y'=x+y$ can be solved by various methods. Solve it and classify its unique equilibrium point $x=1$, $y=-1$ as a node, saddle, center or spiral.

\Answer{~\newline
(3a) The equilibria are constant solutions, which are found from the equations
$$
 \begin{array}{lcl}
  0 &=& (14-x/2-y)x \\
  0 &=& (16-y/2 -x)y
 \end{array}
$$
Considering when a zero factor can occur leads to the four equilibria
$(0,0)$, $(0,32)$, $(28,0)$, $(12,8)$. The last equilibrium comes from solving the system of equations
$$
 \begin{array}{lcl}
  x/2+y &=& 14\\
  x+y/2 &=& 16
 \end{array}
$$
(3b) {\bf Linearization}\newline
The Jacobian matrix $J$ is the augmented matrix of partial derivatives $\partial_x\vec{\.F}$,  $\partial_y\vec{\.F}$ (column vectors) computed from
$$\vec{\.f}(x,y) = \colvector{14x-x^2/2-yx \\ 16y-y^2/2 -xy}.$$
Then
$$
 J(x,y) = \twobytwo{14-x-y &   -x  }
                   {  -y   & 16-y-x}.
$$
The four matrices below are $J(x,y)$ when $(x,y)$ is replaced by an equilibrium point. Included in the table are the roots of the characteristic equation for each matrix and its classification based on the roots. No book was consulted for the classifications. The idea in each is to examine the limits at $t=\pm\infty$, then eliminate classifications. No matrix has complex
eigenvalues, and that eliminates the center and spiral. The first three are stable at either $t=infty$ or $t=infty$, which eliminates the saddle and leaves the node as the only possible classification.
$$
 \begin{array}{lclclcl}
  A_1 &=& J(0,0) &=& \twobytwo{14 & 0 }{0 &16 }     & r=14,16 & \mbox{node} \\
  A_2 &=& J(0,32) &=& \twobytwo{-18 &0 }{-32 &-16 } & r=-18,-16 & \mbox{node}\\
  A_3 &=& J(28,0) &=& \twobytwo{-14 & -28}{ 0& -12} & r=-14,-12 & \mbox{node} \\
  A_4 &=& J(12,8) &=& \twobytwo{-6 & -12}{-8 &-4 }  & r=-5+\sqrt {97},-5-\sqrt {97} & \mbox{saddle}
 \end{array}
$$
 (3c) to be completed.
 \newline
  (3d) to be completed.
}
\end{exercises}

\EJECT
\begin{exercises}
\prob{4.}{(ch7)} ~\newline
  (4a) [20\%] Explain Laplace's Method, as applied to the differential
 equation $x'(t) + 2x(t) = e^t$, $x(0)=1$.

  (4b) [15\%] Solve $\L(f(t))=\dd\frac{10}{(s^2+4)(s^2+9)}$ for $f(t)$.

  (4c) [15\%] Solve for $f(t)$ in the equation
  $\L(f(t))=\dd\frac{1}{s^2(s+3)}$.

  (4d) [10\%] Find $\L(f)$ given $f(t)=(-t)e^{2t}$.

  (4e) [20\%] Solve $x'''+x''=0$, $x(0)=1$, $x'(0)=0$, $x''(0)=0$ by
 Laplace's Method.

  (4f) [20\%] Solve the system $x'=x+y$, $y'=x-y+2$, $x(0)=0$, $y(0)=0$
 by Laplace's Method.


\Answer{~\newline
(a) Laplace's method explained.\newline
The first step transforms the equation using the parts formula and initial data to get
$$(s+2)\L(x) = 1 + \L(e^t).$$
The forward Laplace table applies to write, after a division, the isolated formula for $\L(x)$:
$$\L(x) = \frac{1+1/(s-1)}{s+2}=\frac{s}{(s-1)(s+2)}.$$
Partial fraction methods imply
$$\L(x) = \frac{a}{s-1}+\frac{b}{s+2}=\L(ae^t+be^{-2t})$$
and then $x(t)=ae^t+be^{-2t}$ by Lerch's theorem. The constants are $a=1/3$, $b=2/3$.
\newline
(b) $\L(f)=\frac{100}{(u+1)(u+4)}=\frac{100/3}{u+1}+\frac{-100/3}{u+4}$ where $u=s^2$.
Then $\L(f)=\frac{100}{3}(\frac{1}{s^2+1}-\frac{1}{s^2+4})=\frac{100}{3}\L(\sin t - \frac12\sin 2t)$ implies $f(t)=\frac{100}{3}(\sin t- \frac12\sin 2t)$.
\newline
(c) $\L(f)=\frac{a}{s}+\frac{b}{s^2}+\frac{c}{s-3}=\L(a+bt+ce^{3t})$ implies $f(t)=a+bt+ce^{3t}$.
The constants, by Heaviside coverup, are $a=-1/9$, $b=-1/3$, $c=1/9$.
\newline
(d) $\L(f)=\frac{d}{ds}\L(e^{2t}\sin 3t)$ by the $s$-differentiation theorem.
The first shifting theorem implies $\L(e^{2t}\sin 3t)=\left.\L(\sin 3t)\right|_{s\to(s-2)}$. Finally, the forward table
implies $\L(f)=\frac{d}{ds}\left(\frac{1}{(s-2)^2+9}\right)=\frac{-2(s-2)}{((s-2)^2+9)^2}$.
}
\end{exercises}



 \EJECTEND

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