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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 Sample}}}
\def\EJECTEND{\SeparatedPages{\vfill Use this page to start your solution.
Attach extra pages as needed, then staple.\eject}}

\begin{center}\Large\bf
Differential Equations 2280\\
Sample Final Exam\\
Wednesday, 6 May 2015, 12:45pm-3:15pm \\
\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.}{(Quadrature Equation)} ~\newline
        Solve for the general solution $y(x)$ in the equation
 $\dd y'=
 2\cot x
 + \frac{1250 x^{3}}{1+25x^{2}}
 + x\ln(1+x^2)$.
 %  f:=1250*x^(3)/(1+25*x^(2));

\iftrue {\small [The required integration talent includes
 basic formulae, integration by parts, substitution and college algebra.]
}\fi

\Answer{~\newline
   % int(2*cot(x)+(1250*x^3)/(1+25*x^2)+x*ln(1+x^2),x); latex(%);
$$y=2\,\ln  \left( \sin \left( x \right)  \right) +{\frac {49}{2}}\,{x}^{2
}-\ln  \left( 1+25\,{x}^{2} \right) +1/2\, \left( 1+{x}^{2} \right)
\ln  \left( 1+{x}^{2} \right) -1/2  +C$$
}

\end{exercises}
\EJECT


\begin{exercises}
\prob{2.}{(Separable Equation Test)} ~\newline
 The problem $y'=f(x,y)$ is said to be separable provided
$f(x,y)=F(x)G(y)$ for some functions $F$ and $G$.

 (a) [75\%]  Check (\fbox{X}) the problems that can be put into separable form,
 but don't supply any details.
\begin{center}\large\begin{tabular}{|lp{0.45\linewidth}|lp{0.3\linewidth}|}
\hline
 \fbox{$\phantom{X}$} & $ y'=-y(2xy+1)+(2x+3)y^2$ & \fbox{$\phantom{X}$} &
$yy'=xy^2+5x^2y$ \\[2ex]  \hline
  \fbox{$\phantom{X}$} &  $y'=e^{x+y}+e^y$ & \fbox{$\phantom{X}$} &  $3y'+5y=10y^2$ \\[2ex] \hline
  \end{tabular}
\end{center}

 (b) [25\%] State a test which can verify that an equation is not separable.
 Apply the test to verify that $y'=x+\sqrt{|xy|}$ is not separable.

\Answer{~\newline
(a)   $yy'=xy^2+5x^2y$ is not separable, but the other three are separable.
\newline
(b) Test: $f_y/f$ not independent of $x$ implies not separable.
\newline
If the equation is separable for all allowed $x$ and $y$, then it is separable on a smaller domain.
Let $f=x+\sqrt{|xy|}$ and assume $x>0$ and $y>0$. Then $f=x+\sqrt{xy}=\sqrt{x}\sqrt{y}$.
We have $f_y=\frac12\sqrt{x}/\sqrt{y}$ and $f_y/f=\frac12\sqrt{x}/(x\sqrt{y}+y\sqrt{x})$. When $y=1$, then this fraction depends on $x$, so the DE
is not separable.
}
\end{exercises}
\EJECT

\begin{exercises}
\prob{3.}{(Solve a Separable Equation)} ~\newline
 % u:=(2*x^2+4*x)/(1+x); int(u,x);
 Given $\dd y^2y' = \frac{2x^2+3x}{1+x^2}\left(\frac{125}{64}-y^3\right)$. ~\newline
 (a) Find all equilibrium solutions. ~\newline
 (b) Find the non-equilibrium solution in implicit form. ~\newline
 To save time, {\bf do not solve} for $y$ explicitly.

\Answer{~\newline
 (a) $y=5/4$
\newline
 (b)
% int((2*x^2+3*x)/(1+x^2),x);
% int(y^2/(125/64-y^3),y);
$$
   -\frac13 \ln|125 - 64 y^3| =
   2 x + \frac32 \ln(1 + x^2 ) - 2 \arctan(x) + c
$$
}
\end{exercises}
\EJECT

\begin{exercises}
\prob{4.}{(Linear Equations)} ~\newline
% de:=2*diff(v(t),t)=-32+(2/(3*t+1))*v(t); ic:=v(0)=-8; dsolve({de,ic},v(t));
 (a) [60\%] Solve $\dd 2v'(t)=-32 + \frac{2}{3t+1}v(t)$, $v(0)=-8$. Show
 all integrating factor steps.\newline
 (b) [30\%]  Solve $\dd 2\sqrt{x+2}\,\frac{dy}{dx}=y$. The answer contains symbol $c$.
\newline
 (c) [10\%] The problem $\dd 2\sqrt{x+2}\,y'=y-5$ can be solved using
 the answer $y_h$ from part (b) plus superposition $y=y_h+y_p$. Find $y_p$. Hint:
 If you cannot write the answer in a few seconds, then return here after
 finishing all problems on the exam.

\Answer{~\newline
 (a) $v(t) = -24 t-8$
\newline
 (b)  %de:=2*sqrt(x+2)*diff(y(x),x)=y(x); dsolve(de,y(x));
  $ y(x) = C e^{\sqrt{x+2}}$
}
\end{exercises}
\EJECT

\begin{exercises}
\prob{5.}{(Stability)} ~\newline
 (a) [50\%] Draw a phase line diagram for the differential equation
 $$dx/dt = 1000\left(2-\sqrt[5]{x}\right)^3(2+3x)(9x^2-4)^{8}.$$
 Expected in the diagram are equilibrium
 points and signs of $x'$ (or flow direction markers $<$ and
 $>$).\newline
 (b) [40\%] Draw a phase portrait using the phase line diagram of (a). Add these labels
 as appropriate: funnel, spout, node, source, sink, stable, unstable. Show at
 least 8 threaded curves. A direction field is not expected or required.
 \newline
 (c) [10\%] Outline how to solve for non-equilibrium solutions, without doing any
  integrations or long details. This is a paragraph of text, with only a few equations.

\Answer{~\newline
  (a) and (b) See a handwritten exam solution for a similar problem on
  midterm 1.
\newline
(c) Put the DE into the form $y'/G(y) = F(x)$ and then apply the method of
quadrature.
}
\end{exercises}
\EJECT



\begin{exercises}
\prob{6.}{(ch3)} ~\newline (a)
Solve for the general solutions:

 $ \begin{array}{ll} {\bf (a.1)} ~[25\%] & y'' + 4y' + 4y=0 \end{array}$,

 $\begin{array}{ll} {\bf (a.2)} ~[25\%] & y^{vi} + 4y^{iv} = 0
 \end{array}$,

$\begin{array}{ll} {\bf (a.3)} ~[25\%] & \mbox{Char. eq. ~}
 r(r-3)(r^3-9r)^2(r^2+4)^3=0
 \end{array}$.

(b)  Given $6x''(t) + 7x'(t) + 2x(t) = 0$, which represents a
damped spring-mass system with $m=6$, $c=7$, $k=2$, solve the
differential equation [15\%] and classify the answer as
over-damped, critically damped or under-damped [5\%]. Illustrate
in a physical model drawing the meaning of constants $m$, $c$, $k$
[5\%].

\Answer{~\newline
 (a)\newline
 {\bf 1}: $r^2+4r+4=0$, $y=c_1y_1+c_2y_2$, $y_1=e^{-2x}$, $y_2=xe^{-2x}$.
\newline
 {\bf 2}: $r^{iv}+4r^2=0$, roots $r=0,0,2i,-2i$. Then
 $y=c_1e^{0x}+c_2xe^{0x}+c_3\cos 2x + c_4\sin 2x$. \newline
 {\bf 3}: Write as $r^3(r-3)^3(r+3)^2(r^2+4)^3=0$.
 Then $y$ is a linear combination of the atoms
 $1$, $x$, $x^2$, $e^{3x}$, $xe^{3x}$, $x^2e^{3x}$,
 $e^{-3x}$, $xe^{-3x}$, $\cos 2x$, $x\cos 2x$, $x^2\cos 2x$,
 $\sin 2x$, $x\sin 2x$, $x^2\sin 2x$.
\newline
Part (b)\newline Use $6r^2+7r+2=0$ and the quadratic formula to
obtain roots $r=-1/2,-2/3$. Then
$x(t)=c_1e^{-t/2}+c_2e^{-2t/3}$. This is over-damped. The
illustration shows a spring, dampener and mass with labels $k$,
$c$, $m$, $x$ and the equilibrium position of the mass.
}
\end{exercises}
\EJECT

\begin{exercises}
\prob{7.}{(ch3)} ~\newline Determine for $ y^{vi} + y^{iv} =
x+2x^2+ x^3 + e^{-x}+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
The homogeneous solution is a linear combination of the atoms $1$, $x$, $x^2$, $x^3$, $\cos x$, $\sin x$
because the characteristic polynomial has roots $0$, $0$, $0$, $0$, $i$,
$-i$.
\newline
{\bf \fbox{1}~} An initial trial solution $y$ is constructed by Rule I for
atoms $1$, $x$, $e^{3x}$, $e^{-3x}$, $\cos x$, $\sin x$ giving
$$\begin{array}{lcl}
 y &=& y_1+y_2+y_3+y_4,\\
 y_1 &=& d_1+d_2x+d_3x^2+d_4x^3,\\
 y_2 &=& d_5\cos x + d_6x\cos x,\\
 y_3 &=& d_7\sin x + d_8x\sin x, \\
 y_4 &=& d_9e^{-x}.
\end{array}
$$
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}~} Rule II is applied individually to each of
$y_1$, $y_2$, $y_3$, $y_4$.\newline
The result is the {\bf shortest trial
solution}
$$\begin{array}{lcl}
 y &=& y_1+y_2+y_3+y_4,\\
 y_1 &=& d_1x^4+d_2x^5+d_3x^6+d_4x^7,\\
 y_2 &=& d_5x\cos x + d_6x^2\cos x,\\
 y_3 &=& d_7x\sin x + d_8x^2\sin x, \\
 y_4 &=& d_9e^{-x}.
\end{array}
$$
}
\end{exercises}
\EJECT

\begin{exercises}
\prob{8.}{(ch3)} ~\newline (a) [50\%] Find by undetermined
coefficients the steady-state periodic solution for the equation $
x'' + 4x' + 6x = 10\cos(2t)$.
\newline
(b) [50\%] Find by variation of parameters a particular solution
$y_p$ for the equation $y''+3y'+2y=xe^{2x}$.

{\small\sf This instance has integration
difficulties, therefore it is for a practise exam only. See Midterm 3 for
an exam problem, required less integration effort.}

\Answer{~\newline
(a) Undetermined coefficients for $x'' + 4x' + 6x = 10\cos(2t)$.\newline
We solve $x''+4x'+6x=0$. The characteristic equation roots are $-2\pm \sqrt2 i$ and the atoms are
$x_1=e^{-2t}\cos\sqrt2 t$, $x_2=e^{-2t}\sin\sqrt2 t$.
\newline
The trial solution is computed from $f=10\cos 2t$. We find all atoms in the derivative list $f,f',f''$ \ldots, then take a linear combination to form the trial solution $x=d_1\cos 2t + d_2\sin 2t$.
No corrections are needed, because none of these terms are solutions of the homogeneous equation $x''+4x'+6x=0$.
\newline
Substitute the
trial solution to obtain the answers $d_1=5/17$, $d_2=20/17$. Cramer's rule was used to
solve for the unknowns $d_1$, $d_2$.
% de:=diff(x(t),t,t)+4*diff(x(t),t)+6*x(t)=10*cos(2*t);dsolve(de,x(t));
\newline
The unique periodic solution $x_{\mbox{ss}}$ is extracted from the
general solution $x=x_h+x_p$ by crossing out all negative
exponential terms (terms which limit to zero at infinity). Because $x_h(t)$ contains atoms $x_1$, $x_2$ (see above) containing a negative exponential factor, then the steady state solution is
$$
x_{\mbox{ss}}=\frac{5}{17}\cos 2t + \frac{20}{17}\sin 2t.
$$
(b) Variation of parameters for $y''+3y'+2y=xe^{2x}$.\newline
We solve $y''+3y'+2y=0$. The characteristic equation roots are $-2,-1$ and the atoms are
$y_1=e^{-2x}$, $y_2=e^{-x}$.
\newline
Compute the
Wronskian $W=y_1y_2'-y_1'y_2=e^{-x}$. Then for
$f(x)=xe^{2x}$,
$$
 y_p(x)=\left(\int y_2\frac{-f}{W}dx\right)y_1(x)+\left(\int y_1\frac{f}{W}dx\right)y_2(x).
 $$
 Substitution of $y_1$, $y_2$, $W$, $f$ gives
 $$
 y_p(x)=\left(\int e^{-x}\frac{-xe^{2x}}{e^{-3x}}dx\right)y_1(x)+\left(\int e^{-2x}\frac{xe^{2x}}{e^{-3x}}dx\right)y_2(x)=\frac{-7e^{2x}}{144} + \frac{xe^{2x}}{12}.
 $$
 % F1:=exp(-x)*(-x*exp(2*x))/exp(-3*x);F2:=exp(-2*x)*(x*exp(2*x))/exp(-3*x);
 % simplify(int(F1,x)*exp(-2*x)+int(F2,x)*exp(-x));
 % dsolve(diff(y(x),x,x)+3*diff(y(x),x)+2*y(x)=x*exp(2*x),y(x));simplify(expand(%));
}

\end{exercises}
\EJECT

\begin{exercises}
\prob{9.}{(ch5)} ~\newline
 The eigenanalysis method says that the system
$\.x' = A\.x$ has general solution $\.x(t)=c_1\.v_1e^{\lambda_1
t}+c_2\.v_2e^{\lambda_2 t}+c_3\.v_3e^{\lambda_3 t}$. In the
solution formula, $(\lambda_i,\.v_i)$, $i=1,2,3$, is an eigenpair
of $A$.  Given
$$A=\left[\begin{array}{rrr}
 5 & 1 &  1  \\
 1 & 5 &  1  \\
 0 & 0 &  7
 \end{array}\right],$$
 then \newline
 (a) [75\%] Display eigenanalysis details for $A$. \newline
 (b) [25\%] Display the solution $\.x(t)$ of $\.x'(t)=A\.x(t)$.


\Answer{~\newline
(1): The eigenpairs are
% A:=matrix([[5,1,1],[1,5,1],[0,0,7]]);eigenvects(A);
$$
 \left(4, \colvector{-1\\ 1\\ 0}\right),\quad
 \left(6,  \colvector{1\\ 1\\ 0}\right),\quad
  \left( 7, \colvector{1\\ 1\\ 1}\right).
$$
An expected detail is the cofactor expansion of $\det(A-\lambda I)$ and factoring to find eigenvalues $4,6,7$. Eigenvectors should be found by a sequence of swap, combo, mult operations on the augmented matrix, followed by taking the partial $\partial_{t_1}$ on invented symbol $t_1$ in the general solution to compute the eigenvector.
In short, the eigenvectors are {\sf Strang's Special Solutions}, and in general there can be many eigenvectors for a single eigenvalue.
\newline
(2): The eigenanalysis method for $\.x'=A\.x$ implies
$$
\.x(t)=c_1e^{4t}  \colvector{-1\\ 1\\ 0}+
       c_2e^{6t} \colvector{1\\ 1\\ 0}+
       c_3e^{7t} \colvector{1\\ 1\\ 1}.
$$
}


\end{exercises}
\EJECT

\begin{exercises}
\prob{10.}{(ch5)} ~\newline (a) [20\%]  Find the eigenvalues of
the matrix $A=\left[\begin{array}{rrrr}
 4 & 1 & -1  \\
 1 & 4 & -2  \\
 0 & 0 &  2  \\
 \end{array}\right]$.


(b) [40\%]
Putzer's formula
$$e^{At} = r_1(t)I + r_2(t)(A-\lambda_1 I)+r_3(t)(A-\lambda_1 I)(A-\lambda_2I)$$
uses the linear cascade
$$
\begin{array}{lcl}
 r_1' &=& 2r_1,\quad r_1(0)=1\\
 r_2' &=& 3r_2+r_1,\quad r_2(0)=0 \\
 r_3' &=& 5r_3+r2, \quad r_3(0)=0.
\end{array}
$$
The general solution of $\.u'=A\.u$ according
to Putzer's spectral formula is $\.u=e^{At}\.u(0)$, where $e^{At}$ is defined above.
Please compute all three coefficient functions
$r_1$, $r_2$, $r_3$. {\bf To save time}, don't write out $e^{At}$ or the general solution.

(c) [40\%] Display the general solution of $\.u'=A\.u$ according
to the Cayley-Hamilton-Ziebur Method. In particular, display the equations
that determine the three vectors in the general solution. {\bf To save time},
don't solve for the three vectors in the formula.

(d) [40\%] Display the general solution of $\.u'=A\.u$ according
to the Eigenanalysis Method. {\bf To save time}, find one eigenpair
explicitly, just to show how it is done, but don't solve for the last two eigenpairs.

(e) [40\%] Display the general solution of $\.u'=A\.u$ according
to Laplace's Method. {\bf To save time}, use symbols for partial fraction
constants and leave the symbols unevaluated.

\Answer{~\newline
(a) {\bf Eigenvalue Calculation}\newline
Subtract $\lambda$ from the diagonal elements of $A$ to obtain matrix $B=A-\lambda I$, then expand $\det(B)$ by cofactors to obtain the
characteristic polynomial. The roots are the eigenvalues
 $\lambda=2,3,5$.
%with(linalg):
%A:=matrix([[4,1,-1],[1,4,-2],[0,0,2]]);eigenvects(A);
% [2, 1, {vector([0, 1, 1])}], [3, 1, {vector([-1, 1, 0])}], [5, 1, {vector([1, 1, 0])}]
\newline
(b) {\bf Putzer Method for the Exponential Matrix}\newline
Let $A$ as given in the problem, which is $3\times 3$, and define for this case only the $3\times 3$ matrix
$$P=\left(\begin{array}{rrr}
 2 & 0 & 0  \\
 1 & 3 & 0 \\
 0 & 1 & 5
 \end{array}\right).$$
The diagonal elements of $P$ are the eigenvalues of $A$, and the sub-diagonal entries are ones.
Define functions $r_1$, $r_2$, $r_3$ to be the components
of the vector solution $\.r(t)$ of the initial value problem
$$\.r'=P\.r,\quad \.r(0)=\colvector{1\\ 0\\ 0}.$$
In expanded scalar form the equations are
$$
\begin{array}{lcl}
 r_1' &=& 2r_1,\quad r_1(0)=1\\
 r_2' &=& 3r_2+r_1,\quad r_2(0)=0 \\
 r_3' &=& 5r_3+r2, \quad r_3(0)=0.
\end{array}
$$
%dsolve({diff(y(x),x)=2*y(x),y(0)=1},y(x));
%dsolve({diff(y(x),x)=3*y(x)+exp(2*x),y(0)=0},y(x));
%dsolve({diff(y(x),x)=5*y(x)-exp(2*x)+exp(3*x),y(0)=0},y(x));
%dsolve({diff(y(x),x)=6*y(x)+1/3*exp(2*x)-1/2*exp(3*x)+1/6*exp(5*x),y(0)=0},y(x));
Solve by the linear integrating factor method:
$$ r_1=e^{2t},\quad r_2=e^{3t}-e^{2t}, \quad r_3=\frac13 e^{2t}- \frac12e^{3t}+\frac16 e^{4t}.$$
Then $\.u(t)=e^{At}\.u(0)$ implies
$$ \.u=\left(r_1I+r_2(A-2I)+r_3(A-2I)(A-3I)\right)\.u(0)$$
(c) {\bf Cayley-Hamilton-Ziebur Method}\newline
The eigenvalues $2,3,5$ from (a) are used to create the list of atoms $e^{2t},e^{3t},e^{5t}$. Then the Cayley-Hamilton-Ziebur method implies there are constant vectors $\vec{\.d}_1$, $\vec{\.d}_2$, $\vec{\.d}_3$
which depend on $\vec{\.u}(0)$ and $A$ such that
$$\vec{\.u}(t) = e^{2t}\vec{\.d}_1 + e^{3t}\vec{\.d}_2 + e^{5t}\vec{\.d}_3 .$$
The determining equations are formed from differentiation of this formula two times, then replace $\vec{\.u}'=A\vec{\.u}$, $\vec{\.u}''=A\vec{\.u}'=AA\vec{\.u}$. Finally, remove $t$ from the three equations by setting $t=0$, and define $\vec{\.u}_0=\vec{\.u}(0)$. Then the three equations are, with $\.u_0=\.u(0)$,
$$\begin{array}{rcrcrcrc}
 \vec{\.u}_0 &=& \vec{\.d}_1 &+& \vec{\.d}_2 &+& \vec{\.d}_3 \\
 A\vec{\.u}_0 &=& 2\vec{\.d}_1 &+& 3\vec{\.d}_2 &+&5\vec{\.d}_3 \\
 A^2\vec{\.u}_0 &=& 4\vec{\.d}_1 &+& 9\vec{\.d}_2 &+& 25\vec{\.d}_3
\end{array}
$$
This ends the solution to the problem. We continue, solving for the vectors $\vec{\.d}_j$, just to illustrate how it is done. The matrix of coefficients
$$C=\left(\begin{array}{rrr} 1&1&1 \\ 2&3&5 \\ 4&9&25 \end{array}\right)$$
and its transpose matrix $B=C^T$ give a formal relation
$$\AUG(\vec{\.u}_0,A\vec{\.u}_0,A^2\vec{\.u}_0)=\AUG(\vec{\.d}_1,\vec{\.d}_2,\vec{\.d}_3)B.$$
Multiplying this relation by $B^{-1}$ gives
$$\AUG(\vec{\.d}_1,\vec{\.d}_2,\vec{\.d}_3)=\AUG(\vec{\.u}_0,A\vec{\.u}_0,A^2\vec{\.u}_0)B^{-1}.$$
Then disassembling the formal matrix multiply implies
$$
\begin{array}{rcrcrcrc}
 \vec{\.d}_1 &=& 5\vec{\.u}_0 &-& \frac83A\vec{\.u}_0 &+& \frac13A^2\vec{\.u}_0 \\
 \vec{\.d}_2 &=& -5\vec{\.u}_0 &+& \frac{7}{2}A\vec{\.u}_0 &-&\frac12A^2\vec{\.u}_0 \\
 \vec{\.d}_3 &=& 5\vec{\.u}_0 &-& \frac56A\vec{\.u}_0 &+& \frac16A^2\vec{\.u}_0
\end{array}
$$
The matrix of coefficients is
$$
\left(\begin {array}{rrr} 5&-\frac83&\frac13\\\noalign{\medskip} -5&\frac72&-\frac12\\\noalign{\medskip} 1&-\frac56&\frac16\end{array} \right) = \left(B^{-1}\right)^T =
C^{-1}!
$$
This fact, that solving for $\vec{\.d}_1$, $\vec{\.d}_2$, $\vec{\.d}_3$ in the displayed equations
reduces to inverting the matrix of coefficients, can be used as a shortcut in the Cayley-Hamilton-Ziebur method.
%% C:=matrix([[1,1,1],[2,3,5],[4,9,25]]); inverse(C);
%% matrix([[5, -8/3, 1/3], [-5, 7/2, -1/2], [5, -5/6, 1/6]]);
(d) {\bf Eigenanalysis Method}\newline
For matrix
$$A=\left(\begin{array}{rrrr}
 4 & 1 & -1  \\
 1 & 4 & -2  \\
 0 & 0 &  2  \\
 \end{array}\right)$$
the eigenpairs are computed to be
%with(linalg):
%A:=matrix([[4,1,-1],[1,4,-2],[0,0,2]]);eigenvects(A);
% [2, 1, {vector([0, 1, 1])}], [3, 1, {vector([-1, 1, 0])}], [5, 1, {vector([1, 1, 0])}]
$$
  \left(2,\colvector{0\\ 1\\ 1} \right),\quad
  \left(3,\colvector{-1\\ 1\\ 0} \right),\quad
  \left(5,\colvector{1\\ 1\\ 0} \right).
$$
Then $\vec{\.u}'=A\vec{\.u}$ has general solution
$$
  \vec{\.u}(t)=c_1 e^{2t}\colvector{0\\ 1\\ 1} +
  c_2 e^{3t}\colvector{-1\\ 1\\ 0} +
  c_3 e^{5t}\colvector{1\\ 1\\ 0} .
$$
(e) {\bf Laplace's Method}\newline
The start is the Laplace resolvent formula for matrix differential equation $\vec{\.u}'=A\vec{\.u}$.
$$\left(s I - A\right)\L(\vec{\.u})=\vec{\.u}_0.$$
This formula expands to
$$
\threebythree{s-4 & -1 & 1}{-1 & s-4 & 2}{0 & 0 & s-2}\colvector{\L(x)\\ \L(y)\\ \L(z)}=
\colvector{a\\ b\\ c}
$$
where symbols $a$, $b$, $c$ are arbitrary constants for the initial data $\vec{\.u}_0$.
Let $W$ denote the coefficient matrix. Then the inverse of $W$ can be computed using the adjugate formula $W^{-1}=\adj(W)/\det(W)$. The answer for the inverse is
$$
 W^{-1} =\frac{1}{(s-5)(s-2)(s-3)} \left( \begin {array}{ccc} {s}^{2}-6\,s+8&s-2&-s+2\\s-2&{s}^{2}-6\,s+8&-2\,s+7\\0&0&
{s}^{2}-8\,s+15\end {array} \right)
$$
True, this formula can be derived and then followed by inverse Laplace methods to obtain an answer in variable $t$. However, we already know the outcome, because
this matrix is the Laplace of the exponential matrix $e^{At}$. The exponential matrix formula was already derived in (b) above. Expanding the matrix multiplies and collecting terms gives the final answer
$$W^{-1} = \L\left(e^{At}\right) = \frac12\L
\left( \begin {array}{ccc} {{\rm e}^{5\,t}}+{{\rm e}^{3\,t}}&{{\rm e}^{5\,t}}-{{\rm e}^{3\,t}}&-{{\rm e}^{5\,t}}+{{\rm e}^{3\,t}}
\\\noalign{\medskip}{{\rm e}^{5\,t}}-{{\rm e}^{3\,t}}&{{\rm e}^{5\,t}}
+{{\rm e}^{3\,t}}&-{{\rm e}^{5\,t}}+2\,{{\rm e}^{2\,t}}-{{\rm e}^{3\,t
}}\\\noalign{\medskip}0&0&2\,{{\rm e}^{2\,t}}\end {array} \right)
$$
Canceling the $\L$ with Lerch's Theorem implies the same answer as found in part (b), which is $$\vec{\.u}(t) = e^{At}\colvector{a\\ b\\ c} =
\frac12\left( \begin {array}{ccc} {{\rm e}^{5\,t}}+{{\rm e}^{3\,t}}&{{\rm e}^{5\,t}}-{{\rm e}^{3\,t}}&-{{\rm e}^{5\,t}}+{{\rm e}^{3\,t}}
\\\noalign{\medskip}{{\rm e}^{5\,t}}-{{\rm e}^{3\,t}}&{{\rm e}^{5\,t}}
+{{\rm e}^{3\,t}}&-{{\rm e}^{5\,t}}+2\,{{\rm e}^{2\,t}}-{{\rm e}^{3\,t
}}\\\noalign{\medskip}0&0&2\,{{\rm e}^{2\,t}}\end {array} \right)\colvector{a\\ b\\ c}.
$$
}% end Answer
\end{exercises}
\EJECT
\begin{exercises}
\prob{11.}{(ch5)} ~Do enough to make 100\%
\newline
(a) [50\%]  The eigenvalues are $4$, $6$ for the matrix
$A=\left[\begin{array}{rrrr}
 5 & 1   \\
 1 & 5
 \end{array}\right]$.
%with(linalg):
%C:=matrix([[5,1],[1,5]]); eigenvals(C);

Display the general solution of $\.u'=A\.u$. Show details from either
the eigenanalysis method or the Laplace method.


(b) [50\%] Using the same matrix $A$ from part (a), display the solution
of $\.u'=A\.u$ according to the Cayley-Hamilton Method. To save time,
write out the system to be solved for the two vectors, and then stop,
without solving for the vectors.

(c) [50\%] Using the same matrix $A$ from part (a), compute the
exponential matrix $e^{At}$ by any known method, for example, the
formula $e^{At}=\Phi(t)\Phi^{-1}(0)$ where $\Phi(t)$ is any fundamental matrix, or via Putzer's formula.


\Answer{~\newline
(a) Eigenanalysis method\newline
The eigenpairs of $A$ are
$$\left(4,\colvector{1\\ -1}\right),\quad \left(6,\colvector{1\\ 1}\right)$$
which implies the eigenanalysis general solution
$$\.u(t) = c_1 e^{4t}\colvector{1\\ -1} + c_2 e^{6t} \colvector{1\\ 1}.$$
\newline
(b)Cayley-Hamilton-Ziebur method\newline
 Then $\.u(t) = e^{4t}\vec{\.c}_1+e^{6t}\vec{\.c}_2$ for some constant vectors $\vec{\.c}_1$, $\vec{\.c}_2$ that depend on $\vec{\.u}(0)$ and $A$. Differentiate this equation once and use $\vec{\.u}'=A\vec{\.u}$, then set $t=0$. The resulting system is
$$
\begin{array}{lcl}
 \vec{\.u}_0 &=& e^{0}\vec{\.c}_1+e^{0}\vec{\.c}_2 \\
  A\vec{\.u}_0 &=& 4e^{0}\vec{\.c}_1+6e^{0}\vec{\.c}_2
  \end{array}
 $$
\newline
(c) Putzer Method\newline
The result is $e^{At}=e^{4t}I + \frac{e^{4t}-e^{6t}}{4-6}(A-4I)$.
Functions $r_1$, $r_2$ are computed from $r_1'=4r_1$, $r_1(0)=1$, $r_2'=6r_2+r_1$, $r_2(0)=0$.
$$ e^{At} =  \frac12
\left( \begin {array}{cc} {{\rm e}^{4\,t}}+{{\rm e}^{6\,t}}&{{\rm e}^{6\,t}}-{{\rm e}^{4\,t}}\\\noalign{\medskip}{{\rm e}^{6\,t}}-{{\rm e}^
{4\,t}}&{{\rm e}^{4\,t}}+{{\rm e}^{6\,t}}\end {array} \right).$$
}
\end{exercises}
\EJECT
\begin{exercises}
\prob{12.}{(ch5)} ~Do both
\newline
(a) [50\%] Display the solution of
$\.u'=\twobytwo{2&0}{1&2}\.u$, $\.u(0)=\colvector{0\\ 1}$, using any method that applies.


(b) [50\%] Display the variation of parameters formula for the system
below. Then integrate
to find $\.u_p(t)$ for $\.u'=A\.u$.
$$\.u'=\twobytwo{2&0}{1&2}\.u+\colvector{e^{2t}\\ 0}.$$

\Answer{~\newline
(a) {\bf Resolvent method}\newline
 The resolvent equation $(sI-A)\L(\vec{\.u})=\vec{\.u}(0)$ is the system
$$ \twobytwo{s-2 & 0}{-1 & s-2}\colvector{\L(x)\\ \L(y)} = \colvector{0\\ 1}.$$
The system is solved by Cramer's rule for unknowns $\L(x)$, $\L(y)$ to obtain
$$\L(x)=\frac{0}{(s-2)^2},\quad \L(y)=\frac{s-2}{(s-2)^2}.$$
The backward Laplace table implies
$$ x(t) = 0,\quad y(t)= e^{2t}.$$
{\bf Best method}. Look at the equations as scalar equations $x'=2x$, $x(0)=0$ and $y'=x+2y$, $y(0)=1$. Clearly $x(t)=0$ and then $y'=0+2y$, $y(0)=1$ implies $y(t)=e^{2t}$.
\newline
(b) Putzer Method\newline
Putzer's exponential formula gives
$$e^{At}=e^{2t}I+te^{2t}(A-2I)=\twobytwo{e^{2t} & 0}{te^{2t} & e^{2t}}.$$
Then
$\vec{\.u}_p(t) = e^{At}\int_0^t e^{-Au}\colvectorC{e^{2u}\\ 0}du=
 e^{At}\int_0^t\colvectorC{1\\ -u}du=\colvectorC{te^{2t}\\ t^2e^{2t}/2}$.
}
\end{exercises}
 \EJECT
\begin{exercises}
\prob{13.}{(ch6)} ~\newline
 (a) Define {\em asymptotically stable equilibrium} for $\.u'=\.f(\.u)$, a 2-dimensional system.
\newline
 (b) Give examples of 2-dimensional systems of type saddle, spiral, center and node.
 \newline
 (c) Give a $2$-dimensional predator-prey example $\.u'=\.f(\.u)$ and
 explain the meaning of the variables in the model.


\Answer{~\newline
(a) Definition\newline
An equilibrium point $\vec{\.u}=\vec{\.u}_0$ is {\bf asymptotically stable} at $t=\infty$ provided it is stable and in addition $\lim \vec{\.u}(t)=\vec{\.u}_0$ for all solutions $\vec{\.u}(t)$ with $\|\vec{\.u}(0)-\vec{\.u}_0\|$ sufficiently small.
\newline
An equilibrium point $\vec{\.u}=\vec{\.u}_0$ is {\bf stable} at $t=\infty$ provided for each $\epsilon>0$ there corresponds a number $\delta>0$, depending on $\epsilon$, such that $\|\vec{\.u}(0)-\vec{\.u}_0\|<\delta$ implies $\vec{\.u}(t)$ exists for $0\le t < \infty$ and for all such $t$-values $\|\vec{\.u}(t)-\vec{\.u}_0\|<\epsilon$.
\newline
(b) Examples\newline
The roots are considered for each type, to give the correct geometric picture. For instance, the example $x=e^t$, $y=e^{-t}$ traces out one branch of a saddle. Here's the answers:
$$
 \begin{array}{lclcl}
 A_1 &=& \twobytwo{1&0}{0&-1}  & \mbox{saddle} &  \mbox{roots = $1$, $-1$} \\
 A_2 &=& \twobytwo{1&0}{0&1}   & \mbox{node}   &  \mbox{roots = $1$, $1$} \\
 A_3 &=& \twobytwo{0&1}{-1&0}  & \mbox{center} &  \mbox{roots = $i$, $-i$} \\
 A_4 &=& \twobytwo{-1&1}{-1&1} & \mbox{spiral} &  \mbox{roots = $-1+i$, $-1-i$}
 \end{array}
$$
(c) Predator-Prey\newline
An example is
$$
 \begin{array}{lcl}
  x' &=& (1 - x - y)x, \\
  y' &=& (2 - y + x )y.
 \end{array}
$$
Then $x$ is the prey and $y$ is the predator. Without the interaction terms, each population would change according to a logistic equation, with carrying capacities of $1$ and $2$, respectively, for $x$ and $y$. There are four equilibria, three of which are extinction states and one that represents the ideal population sizes, about which the real populations oscillate.
}
\end{exercises}


 \EJECT
\begin{exercises}
\prob{14.}{(ch6)} ~\newline
  Find the equilibrium points of $x'=14x-x^2/2-xy$,
 $y'=16y-y^2/2-xy$ and classify each linearization at an equilibrium as a node,
 spiral, center, saddle. What classifications can be deduced for
 the nonlinear system, according to the Paste Theorem?

\Answer{~\newline
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
$$
 \begin{array}{lcl}
  x/2+y &=& 14\\
  x+y/2 &=& 16
 \end{array}
$$
{\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
$$
 J(x,y) = \left(\begin{array}{cc}14-x-y &   -x  \\ -y   & 16-y-x\end{array}\right).
$$
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.
$$
 \begin{array}{lclclcl}
  A_1 &=& J(0,0) &=& \twobytwo{14 & 0 }{0 &16 }     & r=14,16 & \mbox{node} \\
  A_2 &=& J(0,32) &=& \twobytwo{-18 &0 }{-32 &-16 } & r=-18,-16 & \mbox{node}\\
  A_3 &=& J(28,0) &=& \twobytwo{-14 & -28}{ 0& -12} & r=-14,-12 & \mbox{node} \\
  A_4 &=& J(12,8) &=& \twobytwo{-6 & -12}{-8 &-4 }  & r=-5+\sqrt {97},-5-\sqrt {97} & \mbox{saddle}
 \end{array}
$$
}
Some maple code for checking the answers:
\begin{verbatim}
F:=unapply([14*x-x^2/2-y*x , 16*y-y^2/2 -x*y],(x,y));
Fx:=unapply(map(u->diff(u,x),F(x,y)),(x,y));
Fy:=unapply(map(u->diff(u,y),F(x,y)),(x,y));
Fx(0,0);Fy(0,0);Fx(28,0);Fy(28,0);Fx(0,32);Fy(0,32);Fx(0,32);Fy(0,32);
\end{verbatim}
\end{exercises}

 \EJECT
\begin{exercises}
\prob{15.}{(ch6)} ~Do enough to make 100\%
\newline
 (a) [25\%] Which of the four types {\em center, spiral, node, saddle} can be unstable at $t=\infty$? Explain your answer.
\newline
 (b) [25\%] Give an example of a linear 2-dimensional system $\.u'=A\.u$ with a saddle at equilibrium point $x=y=0$, and $A$ is not triangular.
 \newline
 (c) [25\%] Give an example of a nonlinear 2-dimensional predator-prey system with
  exactly four equilibria.
 \newline
 (d) [25\%] Display a formula for the general solution of the equation
 $\.u'=\twobytwo{1 & 1}{-1 & 1}\.u$. Then explain why the system has a
 spiral at $(0,0)$.
 \newline
 (e) [25\%] Is the origin an isolated equilibrium point of the linear system
 $\.u'=\twobytwo{1 & 1}{1 & 1}\.u$? Explain your answer.

\Answer{~\newline
(a) All except the center, which is stable but not asymptotically stable. All the others correspond to a general solution which can have an exponential factor $e^{kt}$ in each term. If $k>0$, then the solution cannot approach the origin at $t=\infty$.
\newline
(b) Required are characteristic roots like $1$, $-1$. Let $B=\twobytwo{-1 & 0}{0 & 1}$. Define $A=PBP^{-1}$ where $P=\twobytwo{1 & 1}{1&2}$. Then $\.u'=A\.u$ has a saddle at the origin, because
the characteristic roots of $A$ are still $1$, $-1$. And $A=\twobytwo{-3&2}{-4&3}$ is not triangular.
\newline
(c)
{\bf Example}: The nonlinear predator-prey system  $x'=(x+y-4)x$, $y'=(-x+2y-2)y$ has exactly four equilibrium points $(0,0)$, $(4,0)$, $(0,1)$, $(2,2)$.
\newline
(d) The characteristic equation $\det(A-\lambda I)=0$ is $(1-\lambda)^2+1=0$ with complex roots $1\pm i$
and corresponding atoms $e^t\cos t$, $e^t\sin t$. Then the Cayley-Hamilton-Ziebur Method implies
$$\vec{\.u}(t) = e^t\cos t \vec{\.c}_1 + e^t\sin t \vec{\.c}_2.$$
{\bf Explanation}, why the classification is a spiral. Such solutions containing sine and cosine factors wrap around the origin. This makes it a spiral or a center. Because of the exponential factor $e^t$, it is asymptotically stable at $t=-\infty$, which disallows a center, so it is a spiral.
\newline
(e) No, because $\det(A)=0$. In this case, $A\.u=\.0$ has infinitely many solutions, describing a
line of equilibria through the origin. This implies the equilibrium
point $(0,0)$ is not isolated [you cannot draw a circle about $(0,0)$
which contains no other equilibrium point].
}
\end{exercises}
\EJECT
\begin{exercises}
\prob{16.}{(ch7)} ~\\[2ex]
 (a) Define the direct Laplace Transform.

 (b) Define Heaviside's unit step function.

 (c) Derive a Laplace integral formula for Heaviside's unit step
 function.

 (d) Explain all the steps in Laplace's Method, as applied to the differential
 equation $x'(t) + 2x(t) = e^t$, $x(0)=1$.

\Answer{~\newline
(a) Definition of Direct Laplace Transform
$$\L(f(t))=\int_0^\infty f(t)e^{-st}dt.$$
(b) Definition of the Heaviside unit step\newline
$$u(t-a) = \left\{ \begin{array}{ll} 1 & t \ge a, \\ 0 & t < a. \end{array}\right. $$
(c) Derivation\newline
We prove the second shifting theorem $\L(u(t-a)f(t-a))=e^{-as}\L(f(t))$, which includes an integral formula for the Heaviside function.
$$
 \begin{array}{lcl}
   \L(u(t-a)f(t-a))
        &=& \int_0^\infty u(t-a)f(t-a)e^{-st}dt \\
        &=& \int_0^a (\mbox{integrand})dt + \int_a^\infty (\mbox{integrand})dt \\
        &=& 0 + \int_a^\infty f(t-a)e^{-st}dt \\
        &=& \int_0^\infty f(u)e^{-s(a+u)}du \\
        &=& e^{-sa}\int_0^\infty f(u)e^{-su}du \\
        &=& e^{-as}\L(f(t))
 \end{array}
$$
Used in the derivation is a change of variable $u=t-a$, $du=dt$. Line 3 uses $u(t-a)=0$ on the interval $0\le t \le a$ and $u(t-a)=1$ on $a \le t < \infty$, which simplifies each integrand.
Line 5 observes that factor $e^{-sa}$ in the integrand is a constant relative to $u$-integration, therefore it can move through the integral sign.
\newline
(d) 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 $a=1/3$, $b=2/3$.
}
\end{exercises}
 \EJECT
\begin{exercises}
\prob{17.}{(ch7)} ~\\[2ex]
  (a) Solve $\L(f(t))=\dd\frac{100}{(s^2+1)(s^2+4)}$ for $f(t)$.

  (b) Solve for $f(t)$ in the equation
  $\L(f(t))=\dd\frac{1}{s^2(s-3)}$.

  (c) Find $\L(f)$ given $f(t)=(-t)e^{2t}\sin(3t)$.

  (d) Find $\L(f)$ where $f(t)$ is the periodic function of period
  $2$ equal to $t/2$ on $0\le t \le 2$ (sawtooth wave).


\Answer{~\newline
(a) $\L(f)=\frac{100}{(u+1)(u+4)}=\frac{100/3}{u+1}+\frac{-100/3}{u+4}$ 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
(b) $\L(f)=\frac{a}{s}+\frac{b}{s^2}+\frac{c}{s-3}=\L(a+bt+ce^{3t})$ implies $f(t)=a+bt+ce^{3t}$.
The constants, by Heaviside coverup, are $a=-1/9$, $b=-1/3$, $c=1/9$.
\newline
(c) $\L(f)=\frac{d}{ds}\L(e^{2t}\sin 3t)$ 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}

 \EJECT
\begin{exercises}
\prob{18.}{(ch7)} ~\\[2ex]
 (a) Solve $y''+4y'+4y=t^2$, $y(0)=y'(0)=0$ by Laplace's Method.

 (b) Solve $x'''+x''-6x'=0$, $x(0)=x'(0)=0$, $x''(0)=1$ by
 Laplace's Method.

 (c) Solve the system $x'=x+y$, $y'=x-y+e^t$, $x(0)=0$, $y(0)=0$
 by Laplace's Method.


\Answer{~\newline
(a) Transform to get $\L(x)=\frac{\L(t^2)}{s^2+4s+4}$. Then $\L(x)=\frac{1}{s^3(s+2)^2}
=\frac{a}{s}+\frac{b}{s^2}+\frac{c}{s^3}+\frac{d}{s+2}+\frac{f}{(s+2)^2}=\L(a+bt+ct^2+de^{-2t}+fte^{-2t})$. The answer is $x(t)=a+bt++ct^2+de^{-2t}+fte^{-2t}$. The partial fraction constants are $a=3/16$, $b=-1/4$, $c=1/4$, $d=-3/16$, $f=-1/8$.
% 3/16/s-1/(4*s^2)+1/(4*s^3)-3/16/(s+2)-1/(8*(s+2)^2)
\newline
(b) Transform to get $\L(x)=\frac{1}{s^3+s^2-6s}=\frac{1}{s(s-2)(s+3)}
=\frac{a}{s}+\frac{b}{s-2}+\frac{c}{s+3}=\L(a+be^{2t}+ce^{-3t})$. Then the answer is $x(t)=a+be^{2t}+ce^{-3t}$. The partial fraction constants are $a=-1/6, b=1/10, c=1/15$.
}
\end{exercises}

 \EJECT
\begin{exercises}
\prob{19.}{(ch7)} ~\newline
(a) [25\%] Solve by Laplace's method
$x'' + x = \cos t$, $x(0)=x'(0)=0$.

(b) [10\%] Does there exist $f(t)$ of exponential order such that
$\L(f(t))=\dd\frac{s}{s+1}$? Details required.

(c) [15\%] Linearity $\L(c_1f+c_2g)=c_1\L(f)+c_2\L(g)$ is one Laplace
rule. State four other Laplace rules. Forward and backward table entries
are not rules, which means $\L(1)=1/s$ doesn't count.

(d) [25\%] Solve by Laplace's resolvent method
$$\begin{array}{lcl}
 x'(t) &=& \phantom{2}x(t) + y(t), \\
 y'(t) &=& 2x(t),
\end{array}
$$
with initial conditions $x(0)=-1$, $y(0)=2$.

(e) [25\%] Derive $y(t)=\dd\int_0^t \sin(t-u)f(u)du$ by Laplace transform
methods from the forced oscillator problem
$$y''(t)+y(t)=f(t),\quad y(0)=y'(0)=0.$$

\Answer{~\newline
(a) Transform to obtain $\L(x)=\frac{s}{(s^2+1)^2}$.
\newline
{\bf Calculus method}. Observe that $\frac{d}{ds}\frac{1}{s^2+1}=\frac{-2s}{(s^2+1)^2}$. Then
$\L(x)=-\frac12\frac{d}{ds}\frac{1}{s^2+1}=-\frac12\frac{d}{ds}\L(\sin t)=-\frac12\L((-t)\sin t)$ by the $s$-differentiation theorem. Finally, $x(t)=\frac12 t\sin t$.
\newline
{\bf Convolution method}. Write $\L(x)=\L(\sin t)\L(\cos t)$. Apply the convolution theorem to obtain $x(t)=\int_0^t \sin u \cos (t-u)du=\frac12t\sin t$. A maple answer check is
\centerline{\fbox{\tt int(sin(u)*cos(t-u),u=0..t);}.}
Hand integration uses the trigonometric identity $2\sin(a)\cos(b)=\cos(a-b)-\cos(a+b)$.
\newline
(b) No. The limit of the Laplace transform of a function of exponential order is zero as $s\to\infty$.
\newline
(c) The possible rules: Linearity, Lerch's cancelation law, parts formula, $s$-differentiation, first shift theorem, second shift theorem, periodic function formula, convolution theorem, delta function formula, integral theorem.
\newline
(d) The resolvent formula $(sI-A)\L(\vec{\.u})=\vec{\.u}_0$ becomes the $2\times $ system of equations
$$ \twobytwo{ s-1 & -1}{ -2 & s-0}\colvector{\L(x)\\ \L(y)} = \colvector{-1 \\ 2}.$$
Multiply by the inverse matrix of $(sI-A)$ on the left to obtain
$$ \colvector{\L(x)\\ \L(y)} = \twobytwo{ s-1 & -1}{ -2 & s-0}^{-1}\colvector{-1 \\ 2}
=\frac{1}{\Delta}\twobytwo{ s-0 & 1}{ 2 & s-1}\colvector{-1 \\ 2},
$$
where $\Delta=\det(sI-A)=(s+1)(s-2)$. Then $\L(x)=\frac{2-s}{\Delta}=\frac{-1}{s+1}$, $\L(y)=\frac{2s}{\Delta}=\frac{2s-4}{(s+1)(s-2)}=\frac{2}{s+1}$.
Then $x(t)=-e^{-t}$, $y(t)=2e^{-t}$.
\newline
(e) Derive $y(t)=\dd\int_0^t \sin(t-u)f(u)du$\newline
Transform $y''+y=f$ to get the transfer function relation
$$\L(y(t)) = \frac{1}{s^2+1}\L(f(t))=\L(\sin t)\L(f(t)).$$
The convolution theorem implies the right side of the equation is
$\L(\int_0^t \sin(t-u)f(u)ud)$. Lerch's cancelation law implies
$y(t)=\int_0^t \sin(t-u)f(u)du$.
}
\end{exercises}
\EJECT

\begin{exercises}
\prob{20.}{(ch7)} ~\newline
  (a) [25\%] Solve $\L(f(t))=\dd\frac{10}{(s^2+8)(s^2+4)}$ for $f(t)$.
  \newline
  (b) [25\%] Solve for $f(t)$ in the equation
  $\L(f(t))=\dd\frac{s+1}{s^2(s+2)}$.
  \newline
  (c) [20\%] Solve for $f(t)$ in the equation
  $\L(f(t))= \dd \frac{s-1}{s^2+2s+5}$.
  \newline
   (d) [10\%] Solve for $f(t)$ in the relation
  $$\L(f)=\dd\frac{d}{ds}\L(t^2\sin 3t)$$
   (e) [10\%] Solve for $f(t)$ in the relation
  $$\L(f)=\dd\left.\left(\L\left(t^3e^{9t}\cos 8t\right)\right)
   \right|_{s\to s+3}.$$

\Answer{~\newline
(a) $\L(f(t))=\frac{10}{u+8}{u+4}$ where $u=s^2$. Use Heaviside's coverup method to find the partial fraction expansion
$$\frac{10}{u+8}{u+4} = \frac{-5/2}{u+8}+\frac{5/2}{u+4} = \frac{-5/2}{s^2+8} + \frac{5/2}{s^2+4}.$$
Then $\L(f(t))=\L\left(-\frac52\frac{\sin \sqrt8 t}{\sqrt8} + \frac52 \frac{\sin 2t}{2}\right)$ implies by Lerch's theorem
$$f(t)=-\frac52\frac{\sin \sqrt8 t}{\sqrt8} + \frac52 \frac{\sin 2t}{2}.$$
(b) Expand the fraction into partial fractions as follows:
$$\L(f)=\frac{s+1}{s^2(s+2)}=\frac{a}{s}+\frac{b}{s^2}+\frac{c}{s+2}=\L(a+bt+ce^{-2t}).$$
Then Lerch's theorem implies $f(t)=a+bt+ce^{-2t}$. The partial fraction constants are
$a=1/4, b=1/2, c=-1/4$.
\newline
(d) Because $\frac{d}{ds}\L(g(t))=\L((-t)g(t))$, then $\L(f)=\L((-t)t^2\sin 3t)$. Lerch's theorem implies $f(t)=-t^3\sin 3t$.
\newline
(e) The shifting theorem $\left.\L(g(t))\right|_{s\to(s-a)}=\L(e^{at}g(t))$ is applied to remove the shift on the outside and put $e^{-3t}$ into the Laplace integrand. Then $\L(f(t))=\L(e^{-3t}t^3e^{9t}\cos 8t)$. Lerch's theorem implies $f(t)=t^3e^{6t}\cos 8t$.
}
\end{exercises}
\begin{exercises}
\prob{21.}{(ch9)} ~\newline
 (a) Find the Fourier sine and cosine coefficients for the $2$-periodic
 function $f(t)$ equal to $t/2$ on $0\le t \le 2$.
 \newline
 (b) State Fourier's convergence theorem.
 \newline
 (c) State the results for term-by-term integration and differentiation
 of Fourier series.

\Answer{~\newline
}

\end{exercises}

 \EJECT
\begin{exercises}
\prob{22.}{(ch9)} ~\newline
 (a) Find a steady-state periodic solution by Fourier's method for
 $x''+x=F(t)$, where $F(t)$ is $2$-periodic and equal to $10$ on
 $0<t<1$, equal to $-10$ on $1 < t < 2$.

 (b) Display Fourier's Model for the solution to the heat problem
 $u_t=u_{xx}$, $u(0,t)=u(1,t)=0$, $u(x,0)=f(x)$ on $0\le x \le 1$, $t\ge
 0$.

 (c) Solve $u_t=u_{xx}$, $u(0,t)=u(\pi,t)=0$, $u(x,0)=80\sin^3 x$ on
 $0\le x \le \pi$, $t\ge 0$.

\Answer{~\newline
(a)\\
Because $F(t)$ equals $10$ times the square wave on $-1<t<1$, then  a standard Fourier series table can be used to write a Fourier series for $F(t)$. The method of undetermined coefficients supplies the answer for $x(t)$.\\
{\bf Some details}. Only basic definitions and results for Fourier series will be used. Because $F(t)$ is odd and $L=1$ in the standard Fourier coefficient formulas, then all cosine coefficients are zero and the sine coefficients $b_n$ are given by
$$b_n=\frac{1}{L}\int_{-L}^L F(t)\sin\left(n\pi \frac{t}{L}\right)dt = \frac{1}{1}\left(2\int_0^1 10\sin(n\pi t)dt \right) = \frac{2(1-(-1)^n)}{n\pi}.$$
Undetermined coefficients assumes the output is
$$x(t) = \sum_{k=1}^\infty c_n\sin(n\pi t).$$
Substitution into the differential equation, using the rules for differentiation of Fourier series, gives the identity
$$\sum_{k=1}^\infty -k^2\pi^2\,c_k\sin(k\pi t) + \sum_{k=1}^\infty c_k\sin(k\pi t) = \sum_{n=1}^\infty b_n\sin(n\pi t).$$
Matching coefficients of the sine terms left and right implies $(1-k^2\pi^2)c_k=b_k$. Then
$$x(t) = \sum_{k=1}^\infty \frac{b_k}{1-k^2\pi^2}\,\sin(k\pi t).$$
(b)\\
The eigenpairs for the problem are $(\lambda_n,f_n)$, where $\lambda_n = (n\pi)^2$, $f_n(x)=\sin(\sqrt{\lambda_n}x)$. Write $f(x)=\sum_{n=1}^\infty b_n\sin(n\pi x)$. Then $u(x,t)$ is obtained from $f(x)$ by insertion of the factor $e^{-\lambda_n t}$ after the sine factor:
$$f(x)=\sum_{n=1}^\infty b_n\sin(n\pi x)\,e^{-n^2\pi^2 t}.$$
(c)\\
The trick is to apply part (b), after writing $f(x)$ as a sine series. This uses trig identities, as follows.\\
$\sin^3(x)=(1-\cos^2(x))\sin(x)=\sin(x)-\frac12\cos(x)(2\sin(x)\cos(x))$\\[1ex]
$\phantom{\sin^3(x)} = \sin(x)- \frac12\cos(x)\sin(2x)$\\[1ex]
$\phantom{\sin^3(x)} = \sin(x)- \frac12\frac12\left(\sin(2x+x)+\sin(2x-x)\right)$\\[1ex]
$\phantom{\sin^3(x)} = \sin(x) -\frac14\sin(3x)+\frac14\sin(x)$\\[1ex]
$\phantom{\sin^3(x)} = \frac34\sin(x) -\frac14\sin(3x)$\\[1ex]
Then
$$f(x) = 80\left(\frac34\sin(x) -\frac14\sin(3x)\right)=60\sin(x)-20\sin(3x).$$
Fourier's method then gives
$$u(x,t) = 60\sin(x)e^{-t}-20\sin(3x)e^{-9t}.$$
}
\end{exercises}
 \EJECT

\begin{exercises}
\prob{23.}{(Vibration of a Finite String)} ~\newline
The {\bf normal modes} for the string equation $u_{tt}=c^2 u_{xx}$ are
given by the functions
$$
\sin\left( \frac{n\pi x}{L} \right)
\cos\left( \frac{n\pi ct}{L} \right),\quad
\sin\left( \frac{n\pi x}{L} \right)
\sin\left( \frac{n\pi ct}{L} \right)
.$$
It is known that each normal mode is a solution of the string equation
and that the problem below has solution $u(x,t)$ equal to an infinite
series of constants times normal modes.

 Solve the finite string vibration problem on $0 \le x \le 2$, $t>0$,

$$
 \begin{array}{lcl}
  u_{tt} &=& c^2 u_{xx}, \\
  u(0,t) &=& 0, \\
  u(2,t) &=& 0, \\
  u(x,0) &=& 0, \\
  u_t(x,0) &=& -11\sin(5\pi x).
 \end{array}
$$
\end{exercises}
\Answer{Because the wave initial shape is zero, then the only normal
modes are sine times sine. The initial wave velocity is already a
Fourier series, using orthogonal set $\{ \sin(n\pi
x/2)\}_{n=1}^\infty$. The 1-term Fourier series $-11\sin(5\pi x)$ can
be modified into a solution by inserting the missing sine factor present
in the corresponding normal mode. Then $u(x,t)=-11\sin(5\pi x)\sin(5\pi
ct)/(5\pi)$. We check it is a solution.
}

\EJECT


\begin{exercises}
\prob{24.}{(Periodic Functions)} ~\newline

(a) [30\%] Find the period of $f(x)=\sin(x) \cos(2x) + \sin (2x)\cos(x)$.

(b) [40\%] Let $p=5$. If $f(x)$ is the odd $2p$-periodic extension to
       $(-\infty,\infty)$ of the function
           $f_0(x)=100x\,e^{10 x}$ on $0\le x \le p$, then find
           $f(11.3)$.
           The answer is not to be simplified or evaluated to a decimal.

(c) [30\%] Mark the expressions which are periodic with letter {\large\bf P},
those odd with {\large\bf O} and those even with {\large\bf E}.
 $$\begin{array}{llll}\sin(\cos(2x))\quad\quad & \ln|2+\sin(x)| \quad\quad & \sin(2x)\cos(x)\quad\quad
& \dd\frac{1+\sin(x)}{2+\cos(x)}
\end{array}$$
\end{exercises}
\Answer{
 (a) $f(x)=\sin(x+2x)$ by a trig identity. Then period = $2\pi/3$.
 \newline
 (b) $f(11.3)=f(11.3-p-p)=f(1.3)=f_0(1.3)=130 e^{13}$.
 \newline
 (c) All are periodic of period $2\pi$, satisfying $f(x+2\pi)=f(x)$.
The first is even and the third is odd.
}
\EJECT

\begin{exercises}
\prob{25.}{(Fourier Series)} ~\newline
Let $f_0(x)=x$ on the interval $0 < x < 2$, $f_0(x)= -x$ on $-2 < x <
0$, $f_0(x)=0$ for $x=0$, $f_0(x)=2$ at $x=\pm 2$. Let $f(x)$ be the
periodic extension of $f_0$ to the whole real line, of period $4$.

(a) [80\%] Compute the Fourier coefficients of $f(x)$ (defined above) for the terms $\sin (67\pi x)$
   and $\cos(2\pi x)$. Leave tedious integrations in integral form, but
   evaluate the easy ones like the integral of the square of sine or cosine.

(b) [20\%] Which values of $x$ in $|x|<12$ might exhibit Gibb's
over-shoot?


\end{exercises}
\Answer{
 (a) Because $f_0(x)$ is even, then $f(x)$ is even. Then the coefficient
 of $\sin(67\pi x)$ is zero, without computation, because all sine terms in
 the Fourier series of $f$ have zero coefficient. The coefficient of
 $\cos(n\pi x/2)$ for $n>0$ is given by the formula
 $$a_n=\frac{1}{2}\int_{-2}^{2} f_0(x)\cos(n\pi x/2)dx
 =\int_0^2 x\cos(n\pi x/2)dx.$$
For $\cos(2\pi x)$, we select $n\pi x/2=2\pi x$, or index $n=4$.
\\[2ex]
 (b) There are no jump discontinuities, $f$ is continuous, so no Gibbs
 overshoot.
}
\EJECT

\begin{exercises}
\prob{26.}{(Cosine and Sine Series)} ~\newline
Find the first nonzero term in the sine series expansion of
 $f(x)$, formed as the odd $2\pi$-periodic extension
of the function $\sin(x)\cos(x)$ on $0 < x < \pi$. Leave the Fourier
coefficient in integral form, unevaluated, unless you can compute
the value in a minute or two.
\end{exercises}
\Answer{
 Because $\sin(x)\cos(x)=(1/2)\sin(2x)$ is odd and $2\pi$-periodic,
 this is the Fourier series of $f$. This term is for coefficient $b_2$,
 so $b_2=1/2$ is the first nonzero Fourier coefficient. The first
 nonzero term is $(1/2)\sin(2x)$.
}
\EJECT

\begin{exercises}
\prob{27.}{(Convergence of Fourier Series)} ~\newline

(a) [30\%] Dirichlet's kernel formula can be used to evaluate the sum
$\cos(2x)+\cos(4x)+\cos(6x)+\cos(8x)$. Report its value according to
that formula. [Not on the final exam 2015]

(b) [40\%] The Fourier Convergence Theorem for piecewise smooth
functions applies to continuously differentiable functions of period
$p$. Re-state the Fourier Convergence Theorem for the special case of a $p$-periodic continuously differentiable function.
It is necessary to translate the results for interval  $-\pi \le x \le \pi$
to the interval $-p\le x \le p$ and simplify the value to which the Fourier series converges.

(c) [30\%] Give an example of a function $f(x)$ periodic of period 2
that has a Gibb's over-shoot at the integers $x=0,\pm 2, \pm 4,
\ldots$, (all $\pm 2n$) and nowhere else.

\end{exercises}
\Answer{
 (a) Dirichlet's kernel identity $\dd\frac12 + \cos (t) + \cdots +
 \cos(nt)=\frac{\sin(nt+t/2)}{2\sin(t/2)}$ is used with $t=2x$ and $n=4$ to obtain the answer $\dd\frac{\sin(8x+x)}{2\sin(x)}-\frac12$.
 \newline
 (b) Let $f$ be a $p$-periodic smooth function on
 $(-\infty,\infty)$. Then for all values of $x$,
$$f(x) = a_0 + \sum_{n=1}^\infty (a_n\cos (n\pi x/p) + b_n\sin
(n\pi x/p),$$
where the Fourier coefficients $a_0,a_n,b_n$ are given by the Euler
formulas:
$$a_0=\frac{1}{2p}\int_{-p}^{p} f(x)dx,\quad
a_n=\frac{1}{p}\int_{-p}^{p} f(x)\cos(n\pi x/p)dx,$$
$$b_n=\frac{1}{p}\int_{-p}^{p} f(x)\sin(n\pi x/p)dx.$$
\newline
 (c) Any $2$-periodic continuous function $f$ will work, if we alter the
 values of $f$ at the desired points to produce a jump discontinuity.
 For example, define $f(x)=\sin(\pi x)$ except at the points $\pm 2n$,
 where $f(2n)=2$ ($n=0,1,2,3,\ldots$).
}
\EJECTEND

\end{document}
\endinput
