% 2019 Differential Equations 2280-1
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\usepackage{linalg}
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\begin{document}
\thispagestyle{empty}
%{\large\bf Name}
%\underline{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}\hfill
%{{\bf Time of your class}~\underline{~~~~~~~~~~} }
\def\EJECT{\SeparatedPages
{\vfill Use this page to start your solution.
Attach extra pages as needed, then staple.\newpage
\thispagestyle{empty} \hfill  2280 Sample Exam 2 S2019
%\newline{\large\bf Name}.
%\underline{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}\hfill
}
}
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\begin{center}\Large\bf
Differential Equations 2280 \\
\ifsolutions
\small Sample Midterm Exam 2 with Solutions\\
\else
\small Sample Midterm Exam 2 Problems Only\\
\fi
\small Exam Date: 5 April 2019 at 7:30am
\end{center}

{\bf Instructions}: This in-class exam is 80 minutes. No
calculators, notes, tables or books. No answer check is expected.
Details count 3/4, answers count 1/4. Problems below cover the possibilities,
but the exam day content will be much less, as was the case for Exam 1. Exam 2 covers only problems 1-7, which is
Chapters 1 to 5 and 7 in the textbook. Chapter 6 (Problem 8)
is moved to the final exam.

%The first 5 problems are from
%a midterm exam in 2009, solutions appended to this PDF.
%The last two problems
%have solutions immediately after the problem statement. Last edit 23 Feb.

\begin{exercises}
\prob{3.}{(Laplace Theory)} ~\newline
\BF{(a)} [50\%] Solve by Laplace's method
$x'' + 2x' + x = e^t$, $x(0)=x'(0)=0$.

\BF{(b)} [25\%] Assume $f(t)$ is of exponential order. Find $f(t)$ in the relation
$$\dd\left.\frac{d}{ds}\L(f(t))\right|_{s \to (s-3)}=\L(f(t)-t).$$

\BF{(c)} [25\%] Derive an integral formula for $y(t)$ by Laplace transform
methods, explicitly using the Convolution Theorem, for the problem
$$y''(t)+4y'(t)+4y(t)=f(t),\quad y(0)=y'(0)=0.$$
This is similar to a required homework problem from Chapter 7.

\Answer{~\newline
\BF{(a)}  \newline
$x \left( t \right) =-1/4\,{{\rm e}^{-t}}-1/2\,{{\rm e}^{-t}}t+1/4\,{
{\rm e}^{t}}$
\\
An intermediate step is $\L(x(t)) = \dd\frac{1}{(s-1)(s+1)^2}$. The solution uses partial fractions $\dd\frac{1}{(s-1)(s+1)^2} = \frac{A}{s-1}+\frac{B}{s+1}+\frac{C}{(s+1)^2}$, with answers
$A=1/4$, $B=-1/4$, $C=-1/2$.
\\
\BF{(b)}  \newline
Replace by the shift theorem and the $s$-differentiation theorem the given equation by
$$\dd\L\left((-t)f(t)e^{3t}\right)=\L(f(t)-t).$$
Then Lerch's theorem cancels $\L$ to give $-te^{3t}f(t) = f(t)-t$. Solve for
$$f(t) = \dd\frac{t}{1+te^{3t}}.$$
\\
\BF{(c)}  \newline
The main steps are:
\\
$\dd (s^2+4s+4)\L(y(t)) = \L(f(t)),$ \newline
$\dd \L(y(t)) = \frac{1}{(s+2)^2}\L(f(t))$, \newline
$\dd \L(y(t)) = \L(te^{-2t})\L(f(t))$, by the first shifting theorem,\newline
$\dd \L(y(t)) = \L(\mbox{convolution of $te^{-2t}$ and $f(t)$})$, by the Convolution Theorem, \newline
$\dd \L(y(t)) = \L\left(\int_0^t xe^{-2x}f(t-x)dx\right)$, insert definition of convolution, \newline
$y(t) =\dd\int_0^t xe^{-2x}f(t-x)dx$, by Lerch's Theorem.
}
\end{exercises}
\EJECT
\begin{exercises}
\prob{4.}{(Laplace Theory)} ~\newline
  (4a) [20\%] Explain Laplace's Method, as applied to the differential
 equation $x'(t) + 2x(t) = e^t$, $x(0)=1$. Reference only. Not to appear on any exam.

  (4b) [15\%] Solve $\L(f(t))=\dd\frac{100}{(s^2+1)(s^2+4)}$ 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 $f(t)=(-t)e^{2t}\sin(3t)$.

  (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
(4a) 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 evaluate $\L(e^t)$. Then 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 plus the backward Laplace table 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 $a=1/3$, $b=2/3$.
\\
(4b) $\L(f)=\frac{100}{(u+1)(u+4)}=\frac{100/3}{u+1}+\frac{-100/3}{u+4}$ 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 $f(t)=\frac{100}{3}(\sin t- \frac12\sin 2t)$.
\\
(4c) $\L(f)=\frac{a}{s}+\frac{b}{s^2}+\frac{c}{s+3}=\L(a+bt+ce^{-3t})$ implies $f(t)=a+bt+ce^{-3t}$.
The constants, by Heaviside coverup, are $a=-1/9$, $b=1/3$, $c=1/9$.
\\
(4d) $\L(f)=\frac{d}{ds}\L(e^{2t}\sin 3t)$ by the $s$-differentiation theorem.
The first shifting theorem implies $\dd\L(e^{2t}\sin 3t)=\left.\L(\sin 3t)\right|_{s\to(s-2)}$. Finally, the forward table
implies $\dd\L(f)=\frac{d}{ds}\left(\frac{1}{(s-2)^2+9}\right)=\frac{-2(s-2)}{((s-2)^2+9)^2}$.
\\
(4e) The answer is $x(t)=1$, by guessing, then checking the answer. The Laplace details jump through hoops to arrive at
$\dd (s^3+s^2)\L(x(t))=s^2+s$, or simply $\L(x(t))=1/s$. Then $x(t)=1$.
% de:=diff(x(t),t,t,t)+diff(x(t),t,t)=0;ic:=x(0)=1,D(x)(0)=0,D(D(x))(0)=0;dsolve({de,ic},x(t));
\\
(4f) The transformed system is
$$\begin{array}{rcrcl}
(s-1)\L(x) &+& (-1)\L(y) &=& 0,\\
(-1)\L(x) &+& (s+1)\L(y) &=& \L(2).
\end{array}
$$
Then $\L(2)=2/s$ and Cramer's Rule gives the formulas
$$\dd \L(x) = \frac{2}{s(s^2-2)},\quad \L(y) = \frac{2(s-1)}{s(s^2-2)}.
$$
% des:=diff(x(t),t)=x(t)+y(t), diff(y(t),t)=x(t)-y(t)+2;ic:=x(0)=0,y(0)=0;
% dsolve({des,ic},{x(t),y(t)})
% # x = \cosh(\sqrt2 t)-1, y = \sqrt2\sinh(\sqrt2 t) - \cosh(\sqrt2 t)
After partial fractions and the backward table,
$$ x = -1+\cosh(\sqrt2 t),\quad y = \sqrt2\sinh(\sqrt2 t) - \cosh(\sqrt2 t)+1.$$
}
\end{exercises}
\EJECT
\begin{exercises}
\prob{5.}{(Laplace Theory)} ~\newline
  (a) [30\%] Solve $\L(f(t))=\dd\frac{1}{(s^2+s)(s^2-s)}$ for $f(t)$.
  \newline
  (b) [20\%] Solve for $f(t)$ in the equation
  $\L(f(t))= \dd \frac{s+1}{s^2+4s+5}$.
  \newline
   (c) [20\%] Let $u(t)$ denote the unit step. Solve for $f(t)$ in the
   relation
  $$\L(f(t))=\dd\frac{d}{ds}\L(u(t-1)\sin 2t)$$
Remark: This is not a second shifting theorem problem.\newline
   (d) [30\%] Compute $\L(e^{2t}f(t))$ for
  $$f(t)=\dd\frac{e^t-e^{-t}}{t}.$$

\Answer{~\newline
(a) $f(t) = \sinh(t)-t = \frac12 e^t -\frac12e^{-t} - t$
% F:=1/((s^2+s)*(s^2-s));invlaplace(F,s,t);
\\
(b) $f(t) = e^{-2t}(\cos(t)-\sin(t))$
% F:=(s+1)/(s^2+4*s+5);invlaplace(F,s,t);
\\
(c) Replace $d/ds$ by factor $(-t)$ in the Laplace integrand:
$$\L(f(t))=\dd\L((-t)\sin (2t)u(t-1))$$
Apply Lerch's theorem to cancel $\L$ on each side, obtaining the answer
$$f(t))=(-t)\sin (2t)u(t-1).$$
(d) The first shifting theorem reduces the problem to computing $\L(f(t))$.
\\
$\dd \L(tf(t))=\L(e^t-e^{-t}) = \frac{1}{s-1}-\frac{1}{s+1}$
\\
$\dd -\frac{d}{ds}\L(f(t)) = \frac{1}{s-1}-\frac{1}{s+1}$, by the $s$-differentiation theorem,
\\
Then $F(s)=\L(f(t))$ satisfies a first order quadrature equation $F'(s) = h(s)$ with solution
$F(s)=\ln|s+1|-\ln|s-1|+c=\ln\left| \frac{s+1}{s-1}\right| + c$ for some constant $c$. Because $F=0$
at $s=\infty$ (a basic theorem for functions of exponential order) and $\ln|1|=0$, then $c=0$ and
$\L(f(t))=F(s)=\ln|s+1|-\ln|s-1|=\ln\left| \frac{s+1}{s-1}\right| $. Then the shifting theorem implies
$$ \L\left(e^{2t}f(t)\right) = \left.\L(f(t))\right|_{s:=s-2} = \ln\left| \frac{s-1}{s-3}\right|.$$
}
\end{exercises}
\EJECTEND

\begin{exercises}
\prob{6.}{(Systems of Differential Equations)} ~\newline
 The eigenanalysis method says that, for a $3\times 3$ system
$\.x' = A\.x$, the general solution is $\.x(t)=c_1\.v_1e^{\lambda_1
t}+c_2\.v_2e^{\lambda_2 t}+c_3\.v_3e^{\lambda_3 t}$. In the
solution formula, $(\lambda_i,\.v_i)$, $i=1,2,3$, is an eigenpair
of $A$.  Given
$$A=\left[\begin{array}{rrr}
 4 & 1 &  1  \\
 1 & 4 &  1  \\
 0 & 0 &  4
 \end{array}\right],$$
 then \newline
 (a) [75\%] Display eigenanalysis details for $A$. \newline
 (b) [25\%] Display the solution $\.x(t)$ of $\.x'(t)=A\.x(t)$.
\newline
 (c) Repeat (a), (b) for the matrix
$A=\left[\begin{array}{rrr}
 5 & 1 &  1  \\
 1 & 5 &  1  \\
 0 & 0 &  7
 \end{array}\right]$.

\Answer{~\newline
(a): The details should solve the equation $|A-\lambda I|=0$ for three values $\lambda=5,4,3$. Then solve the three systems $(A-\lambda I)\vec{v}=\vec{0}$ for eigenvector $\vec{v}$, for $\lambda=5,4,3$.
\\
The eigenpairs are
$$
 5,  \colvector{1\\ 1\\ 0};\quad 4, \colvector{-1\\ -1\\ 1};\quad
  3, \colvector{1\\ -1\\ 0}.
$$
(b): The eigenanalysis method implies
$$
\.x(t)=c_1e^{5t}  \colvector{1\\ 1\\ 0}+ c_2e^{ 4t} \colvector{-1\\ -1\\
1}+ c_3e^{3t} \colvector{1\\ -1\\ 0}.
$$
(c):  The eigenpairs are
$$
 6,  \colvector{1\\ 1\\ 0};\quad 7, \colvector{1\\ 1\\ 1};\quad
  4, \colvector{1\\ -1\\ 0}.
$$
and
the eigenanalysis method implies
$$
\.x(t)=c_1e^{6t}  \colvector{1\\ 1\\ 0}+ c_2e^{ 7t} \colvector{1\\ 1\\
1}+ c_3e^{4t} \colvector{1\\ -1\\ 0}.
$$
%A:=<5,1,1|1,5,1|0,0,7>^+;LinearAlgebra[Eigenvectors](A);
%P:=<1,1,0|-1,-1,1|1,-1,0>;DD:=<5,0,0|0,4,0|0,0,3>;A:=P.DD.P^(-1);
%# A.P=P.DD
}

\end{exercises}
\EJECT

\begin{exercises}
\prob{7.}{(Systems of Differential Equations)} ~\newline
(a) [30\%]  Find the eigenvalues of the matrix
$A=\left[\begin{array}{rrrr}
 4 & 1 & -1 & 0 \\
 1 & 4 & -2 & 1 \\
 0 & 0 &  2 & 0 \\
 0 & 0 &  2 & 4
 \end{array}\right]$.
%with(LinearAlgebra):
%B:=Matrix([[4,1,-1,0],[1,4,-2,1],[0,0,2,0],[0,0,2,4]]);
%Eigenvalues(B);
%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)=4*y(x)+(-exp(-x)+1)*exp(3*x),y(0)=0},y(x));
%dsolve({diff(y(x),x)=5*y(x)+(1/2*exp(-2*x)-exp(-x)+1/2)*exp(4*x),y(0)=0},y(x));

(b) [20\%] Justify that eigenvectors of $A$ corresponding to the eigenvalues in increasing order are the four vectors
$$
\colvector{ 1\\ -5\\ -3 \\ 3},\quad
\colvector{-1\\ 1\\ 0 \\ 0},\quad
\colvector{-1\\ 0\\ 0 \\ 1},\quad
\colvector{1 \\ 1 \\ 0 \\ 0}.
$$


(c) [50\%] Display the general solution of $\.u'=A\.u$ according to the Eigenanalysis method.


\Answer{~\newline
(a) Eigenvalues are $\lambda=2,3,4,5$. \\Define
$$A=\left[\begin{array}{rrrr}
 4 & 1 & -1 & 0 \\
 1 & 4 & -2 & 1 \\
 0 & 0 &  2 & 0 \\
 0 & 0 &  2 & 4
 \end{array}\right].$$
Subtract $\lambda$ from the diagonal elements of $A$ and expand
the determinant $\det(A-\lambda I)$ to obtain the characteristic
polynomial $(2-\lambda)(3-\lambda)(4-\lambda)(5-\lambda)=0$. The
eigenvalues are the roots: $\lambda=2,3,4,5$. Used here was the
{\em cofactor rule} for determinants. Also possible is the special result for block matrices, $\left|\begin{array}{cc} B_1 & 0\\ C & B_2\end{array}\right|=|B_1||B_2|$. Sarrus' rule does not apply
for $4\times 4$ determinants (an error) and the triangular rule
likewise does not directly apply (another error).
\\
(b)
To be justified is $AP=PD$ where $D=\diag(2,3,4,5)$ is the diagonal matrix of eigenvalues (see part (a)) and $P$ is the augmented matrix of eigenvectors supplied above. Matrix multiply can check the answer, by expanding each side of $AP=PD$. \\
Alternative method:\\
 Solve $(A-\lambda I)\vec{v}=\vec{0}$ four times, for $\lambda=2,3,4,5$. Each is a homogeneous system of linear algebraic equations, reduced to RREF by swap, combo, multiply. Each eigenvector answer is Strang's Special Solution.
\\
(c)
Because the eigenvalues are $\lambda=2,3,4,5$, then the solution is a vector linear combination of the Euler solution atoms $e^{2t}, e^{3t}, e^{4t}, e^{5t}$:
$$\.u(t) = \vec{d}_1 e^{2t} + \vec{d}_2 e^{3t} +\vec{d}_3 e^{4t} +\vec{d}_4 e^{5t}.$$
According to the theory, $\vec{d}_j=c_j\vec{v}_j$, where $(\lambda_1,\vec{v}_1)$, \ldots, $(\lambda_4,\vec{v}_4)$ are the eigenpairs of $A$ and $c_1,c_2,c_3,c_4$ are invented symbols representing real, arbitrary constants. Then
$$
\vec{u}=c_1e^{2t}\colvector{ 1\\ -5\\ -3 \\ 3}+
c_2e^{3t}\colvector{-1\\ 1\\ 0 \\ 0}+
c_3e^{4t}\colvector{-1\\ 0\\ 0 \\ 1}+
c_4e^{5t}\colvector{1 \\ 1 \\ 0 \\ 0}.
$$
%de:=diff(r1(t),t)=2*r1(t),
%diff(r2(t),t)=3*r2(t)+r1(t),
%diff(r3(t),t)=4*r3(t)+r2(t),
%diff(r4(t),t)=5*r4(t)+r3(t);
%ic:=r1(0)=1,r2(0)=0,r3(0)=0,r4(0)=0;
%dsolve([de,ic]);
}
\end{exercises}
\EJECT

\begin{exercises}
\prob{8.}{(Systems of Differential Equations)}
\newline
\BF{(a)} [100\%]  The eigenvalues are $3$, $5$ for the matrix
$A=\left[\begin{array}{rrrr}
 4 & 1   \\
 1 & 4
 \end{array}\right]$.
%with(linalg):
%C:=matrix([[4,1],[1,4]]); eigenvals(C);

Display the general solution of $\.u'=A\.u$ according to the Cayley-Hamilton-Ziebur shortcut (textbook chapters 4,5).
 Assume initial condition $\vec{u}_0 = \colvector{1\\ -1}$.

\Answer{~\newline
(a)
{\bf Cayley-Hamilton Ziebur Shortcut}. The method says that the components $x(t),y(t)$ of the solution to the system
$$\vec{u}'=A\vec{u}, \quad\vec{u}(0)=\colvector{1\\ -1}$$
with
$A=\left(\begin{array}{rrrr}
 4 & 1   \\
 1 & 4
 \end{array}\right)$ and $\vec{u}=\colvector{x(t)\\ y(t)}$ are linear combinations of the Euler atoms found from the roots of the characteristic equation $|A-rI|=0$. The roots are $r=3,5$ and the atoms are $e^{3t},e^{5t}$. The scalar system is
$$\left\{\begin{array}{lcl}
x'(t) &=& 4x(t) + y(t), \\
y'(t) &=& x(t) + 4y(t), \\
x(0) &=& 1, \\
y(0) &=& -1.
\end{array}\right.
$$
The C-H-Z method implies $x(t)=c_1e^{3t}+c_2e^{5t}$, but $c_1,c_2$ are not arbitrary constants: they are determined by the initial conditions $x(0)=1,y(0)=-1$. Then $x'=4x+y$ can be solved for $y$ to obtain $y(t)=x'(t) -4x(t)$. Substitute expression $x(t)=c_1e^{3t}+c_2e^{5t}$ into $y(t)=x'(t) -4x(t)$ to obtain
$$y(t)=x'(t) -4x(t) = 3c_1e^{3t}+5c_2e^{5t} - 4(c_1e^{3t}+c_2e^{5t}) = -c_1e^{3t}+c_2e^{5t}.$$
Then
\begin{equation}\label{scalar.answer}
\left\{\begin{array}{lcrcl}
x(t)&=&c_1e^{3t}&+&c_2e^{5t},\\
y(t) &=& -c_1e^{3t}&+&c_2e^{5t}.
\end{array}\right.\end{equation}
Initial data $x(0)=1,y(0)=-1$ are used in the last step, to evaluate $c_1,c_2$. Inserting these conditions produces a $2\times 2$ linear system for $c_1,c_2$
$$\left\{\begin{array}{lcrcl}
1&=&c_1e^{0}&+&c_2e^{0},\\
-1 &=& -c_1e^{0}&+&c_2e^{0}.
\end{array}\right.$$
The solution is $c_1=1$ and $c_2=0$, which implies the final answer $x(t)=e^{3t}$, $y(t)=-e^{3t}$.
\newline
\rule{0.2\linewidth}{0.05em}
\newline
{\bf Remark on Fundamental Matrices}. The answer prior to evaluation of $c_1,c_2$ can be written as
$$\colvector{x(t)\\ y(t)} = \twobytwo{e^{3t}&e^{5t}}{-e^{3t}&e^{5t}}\colvector{c_1\\ c_2}.$$
The matrix $\Phi(t)=\twobytwo{e^{3t}&e^{5t}}{-e^{3t}&e^{5t}}$ is called a {\bf fundamental matrix}, because it is nonsingular and satisfies $\Phi'=A\Phi$ (its columns are solutions of $\vec{u}'=A\vec{u}$). In terms of $\Phi$,
$$e^{At} = \Phi(t)\Phi^{-1}(0).$$
This formula gives an alternative way to compute $e^{At}$ by using the Cayley-Hamilton-Ziebur shortcut. Please observe that the columns of $\Phi$ are the formal partial derivatives of the vector solution $\vec{u}$ on the symbols $c_1,c_2$. Partial derivatives on symbols is a general method for discovering basis vectors. Therefore, $\Phi$ can be written directly from equations (\ref{scalar.answer}).
}
\end{exercises}
\EJECT


\subsection*{Chapters 4 and 5}
\begin{exercises}
\prob{6.}{(Systems of Differential Equations)} ~\newline
\BF{Background}.  Let $A$ be a real $3\times 3$ matrix with eigenpairs
 $(\lambda_1,\.v_1)$,  $(\lambda_2,\.v_2)$,  $(\lambda_3,\.v_3)$.
 The eigenanalysis method says that the $3\times 3$ system
$\.x' = A\.x$ has general solution
$$\.x(t)=c_1\.v_1e^{\lambda_1
t}+c_2\.v_2e^{\lambda_2 t}+c_3\.v_3e^{\lambda_3 t}.$$
\BF{Background}. Let $A$ be an $n\times n$ real matrix. The method called \BF{Cayley-Hamilton-Ziebur} is based upon the result
\begin{quote}\sf
The components of solution $\.x$ of $\.x'(t)=A\.x(t)$ are linear combinations of Euler solution atoms obtained from the roots of the characteristic equation $|A-\lambda I|=0$.
\end{quote}
\BF{Background}. Let $A$ be an $n\times n$ real matrix. An augmented matrix $\Phi(t)$ of $n$ independent solutions of $\.x'(t)=A\.x(t)$ is called a
 \BF{fundamental matrix}. It is known that the general solution is $\.x(t)=\Phi(t)\.c$, where $\.c$ is a column vector of arbitrary constants $c_1, \ldots, c_n$. An alternate and widely used definition of fundamental matrix is  $\Phi'(t) = A\Phi(t)$, $|\Phi(0)| \ne 0$.

\bigskip

 \BF{(a)}  Display eigenanalysis details for the $3\times 3$ matrix
$$A=\left(\begin{array}{rrr}
 4 & 1 &  1  \\
 1 & 4 &  1  \\
 0 & 0 &  4
 \end{array}\right),$$
 then display the general solution $\.x(t)$ of $\.x'(t)=A\.x(t)$.

\BF{(b)}  The $3\times 3$ triangular matrix
$$A=\left(\begin{array}{rrr}
 3 & 1 &  1  \\
 0 & 4 &  1  \\
 0 & 0 &  5
 \end{array}\right),$$
 represents a linear cascade, such as found in brine tank models.
 Using the linear integrating factor method, starting with component $x_3(t)$, find the vector general solution $\.x(t)$ of $\.x'(t)=A\.x(t)$.

 \BF{(c)}  The exponential matrix $e^{At}$ is defined to be a fundamental matrix $\Psi(t)$ selected such that $\Psi(0)=I$, the $n\times n$ identity matrix. Justify the formula $e^{At} = \Phi(t)\Phi(0)^{-1}$, valid for {\em any} fundamental matrix $\Phi(t)$.

  \BF{(d)} Let $A$ denote a $2\times 2$ matrix. Assume  $\.x'(t)=A\.x(t)$ has scalar general solution $x_1=c_1e^t+c_2e^{2t}$, $x_2=(c_1-c_2)e^t + 2c_1+c_2)e^{2t}$, where $c_1,c_2$ are arbitrary constants. Find a fundamental matrix $\Phi(t)$ and then go on to find $e^{At}$
  from the formula in part (c) above.

  \BF{(e)} Let $A$ denote a $2\times 2$ matrix and consider the system $\.x'(t)=A\.x(t)$. Assume
   fundamental matrix
    $\dd\Phi(t) = \twobytwo{e^t & e^{2t}}{2e^t & -e^{2t}}$.
 Find the $2\times 2$ matrix $A$.

   \BF{(f)}  The Cayley-Hamilton-Ziebur shortcut  applies especially to the system
   $$ x' = 3x + y,\quad y' = -x + 3y,$$
   which has complex eigenvalues $\lambda = 3\pm i$. Show the details of the method, then go on to report a fundamental matrix $\Phi(t)$.\\
    \BF{Remark}. The vector general solution is $\.x(t)=\Phi(t)\.c$, which
 contains no complex numbers. Reference: 4.1, Examples 6,7,8.

\ifsolutions
\vfill\eject
\Answer{~\newline
\BF{Part (a)}~~ The details should solve the equation $|A-\lambda I|=0$ for the three eigenvalues $\lambda=5,4,3$. Then solve the three systems $(A-\lambda I)\vec{v}=\vec{0}$ for eigenvector $\vec{v}$, for $\lambda=5,4,3$.
\\
The eigenpairs are
$$
 5,  \colvector{1\\ 1\\ 0};\quad 4, \colvector{-1\\ -1\\ 1};\quad
  3, \colvector{1\\ -1\\ 0}.
$$
The eigenanalysis method implies
$$
\.x(t)=c_1e^{5t}  \colvector{1\\ 1\\ 0}+ c_2e^{ 4t} \colvector{-1\\ -1\\
1}+ c_3e^{3t} \colvector{1\\ -1\\ 0}.
$$
%A:=<5,1,1|1,5,1|0,0,7>^+;LinearAlgebra[Eigenvectors](A);
%P:=<1,1,0|-1,-1,1|1,-1,0>;DD:=<5,0,0|0,4,0|0,0,3>;A:=P.DD.P^(-1);
%# A.P=P.DD
\\[1ex]
\BF{Part (b)}~~ Write the system in scalar form
$$\begin{array}{lcl}
x' &=& 3x + y + z, \\
y' &=& 4y+z,\\
z' &=& 5z.
\end{array}
$$
Solve the last equation as \\
$z=\frac{\mbox{constant}}{\mbox{integrating factor}} = c_3e^{5t}$. \\
\fbox{$z = c_3e^{5t}$}\\
The second equation is\\
$y' = 4y + c_3e^{5t}$\\
The linear integrating factor method applies.\\
$y' - 4y = c_3e^{-5t}$\\
$\dd\frac{(Wy)'}{W} = c_3e^{5t}$, where $W=e^{-4t}$,\\
$(Wy)' = c_3We^{5t}$\\
$(e^{-4t}y)' = c_3e^{-4t}e^{5t}$\\
$e^{-4t}y = c_3e^t + c_2.$\\
\fbox{$y = c_3e^{5t}+c_2e^{4t}$}\\
Stuff these two expressions into the first differential equation:\\
$x' = 3x + y + z = 3x + 2c_3e^{5t}+c_2e^{4t}$\\
Then solve with the linear integrating factor method. \\
$x' - 3x = 2c_3e^{5t}+c_2e^{4t}$\\
$\dd\frac{(Wx)'}{W} = 2c_3e^{5t}+c_2e^{4t}$, where $W=e^{-3t}$. Cross-multiply:\\
$(e^{-3t}x)' = 2c_3e^{5t}e^{-3t}+c_2e^{4t}e^{-3t}$, then integrate:\\
$e^{-3t}x = c_3e^{2t}+c_2e^{t}+c_1$\\
$e^{-3t}x = c_3e^{2t}+c_2e^{t}+c_1$, divide by $e^{-3t}$:\\
\fbox{$x = c_3e^{5t}+c_2e^{4t}+c_1e^{3t}$}\\
\\[1ex]
\BF{Part (c)}~~
The question reduces to showing that $e^{At}$ and $\Phi(t)\Phi(0)^{-1}$ have equal columns. This is done by showing that the matching columns are solutions of $\vec{u}\, ' = A\vec{u}$ with the same initial condition $\vec{u}(0)$, then apply Picard's theorem on uniqueness of initial value problems.
\\[1ex]
\BF{Part (d)}~~
Take partial derivatives on the symbols $c_1,c_2$ to find vector solutions $\vec{v}_1(t)$, $\vec{v}_2(t)$. Define $\Phi(t)$ to be the augmented matrix of $\vec{v}_1(t)$, $\vec{v}_2(t)$. Compute $\Phi(0)^{-1}$, then multiply on the right of $\Phi(t)$ to obtain $e^{At}=\Phi(t)\Phi(0)^{-1}$. Check the answer in a computer algebra system or using Putzer's formula.
\\[1ex]
\BF{Part (e)}~~
The equation $\Phi'(t)=A\Phi(t)$ holds for every $t$. Choose $t=0$ and then solve for $A=\Phi'(0)\Phi(0)^{-1}$.
\\[1ex]
\BF{Part (f)}~~
By C-H-Z, $x=c_1e^{3t}\cos(t)+c_2e^{3t}\sin(t)$. Isolate $y$ from the first differential equation $x'=3x+y$, obtaining the formula $y=x'-3x=-c_1e^{3t}\sin(t)+c_2e^{3t}\cos(t)$. A fundamental matrix is found by taking partial derivatives on the symbols $c_1,c_2$. The answer is exactly the Jacobian matrix of $\colvector{x\\ y}$ with respect to variables $c_1,c_2$.\\
$\dd\Phi(t)=\twobytwo{e^{3t}\cos(t) & e^{3t}\sin(t)}{-e^{3t}\sin(t) & e^{3t}\cos(t)}$.
}
\fi
\end{exercises}
\EJECT
\subsection*{Chapter 6}

\begin{exercises}
\prob{1}{(Linear and Nonlinear Dynamical Systems)}
\newline
\BF{(a)}
 Determine whether the unique equilibrium $\vec{u}=\vec{0}$ is stable or
unstable. Then classify the equilibrium point $\vec{u}=\vec{0}$ as a
saddle, center, spiral or node. %Sub-classification into improper or proper node is not required.
 $$\vec{u}'=\twobytwo{3&4}{-2&-1}\vec{u}$$
\BF{(b)}
 Determine whether the unique equilibrium $\vec{u}=\vec{0}$ is stable or
unstable. Then classify the equilibrium point $\vec{u}=\vec{0}$ as a
saddle, center, spiral or node. %Sub-classification into improper or proper node is not required.
 $$\vec{u}'=\twobytwo{-3&2}{-4&1}\vec{u}$$
\BF{(c)} Consider the nonlinear dynamical system
  $$\begin{array}{lcl}
  x'&=& x-2y^2-y+32, \\
   y'&=& 2x^2-2xy.
  \end{array}
 $$
 An equilibrium point is $x=4$, $y=4$. Compute
 the Jacobian matrix $A=J(4,4)$ of the linearized system at this
 equilibrium point.
 \\[2ex]
 \BF{(d)} Consider the nonlinear dynamical system
  $$\begin{array}{lcl}
  x'&=& -x-2y^2-y+32, \\
   y'&=& 2x^2+2xy.
  \end{array}
 $$
 % solve({-x-2*y^2-y+32=0, 2*x^2+2*x*y=0},{x,y});
 % {x = -4, y = 4}, {x = 4, y = -4}, {x = 0, y = RootOf(2*_Z^2+_Z-32)}
 An equilibrium point is $x=-4$, $y=4$. Compute
 the Jacobian matrix $A=J(-4,4)$ of the linearized system at this
 equilibrium point.
\\[2ex]
\BF{(e)} Consider the nonlinear dynamical system
  $\dd\left\{\begin{array}{lcl}
  x'&=& -4x+4y+9-x^2, \\
   y'&=& 3x-3y.
  \end{array}\right.$\\
 At equilibrium point $x=3$, $y=3$, the Jacobian matrix is
 %$A=J(3,3)=\left.\twobytwo{-4-2x&4}{3&-3}\right|_{x=3}=\twobytwo{-10&4}{3&-3}$.
 $\dd A=J(3,3)=\twobytwo{-10&4}{3&-3}$.
 \begin{quote}
   {\bf (1)}
  Determine the stability at $t=\infty$ and the phase portrait
  classification saddle, center, spiral
  or node at $\vec{u}=\vec{0}$ for the linear system $\frac{d}{dt}\vec{u}=A\vec{u}$.

   {\bf (2)}
  Apply the Pasting Theorem to classify $x=3$, $y=3$ as a saddle, center, spiral
  or node for the {\bf nonlinear dynamical system}. Discuss all details of the application of the theorem.
  {\em Details count 75\%}.
  \end{quote}
\BF{(f)} Consider the nonlinear dynamical system
 $\dd\left\{
  \begin{array}{lcl}
  x'&=& -4x-4y+9-x^2, \\
   y'&=& 3x+3y.
  \end{array}\right.$\\
 At equilibrium point $x=3$, $y=-3$, the Jacobian matrix is
 %$A=J(3,3)=\left.\twobytwo{-4-2x&-4}{3&3}\right|_{x=3}=\twobytwo{-10&-4}{3&3}$.
 $\dd A=J(3,-3)=\twobytwo{-10&-4}{3&3}$.
 \begin{quote}
  \BF{Linearization}.~ Determine the stability at $t=\infty$ and the phase portrait
  classification saddle, center, spiral
  or node at $\vec{u}=\vec{0}$ for the \BF{linear dynamical system} $\dd\frac{d}{dt}\vec{u}=A\vec{u}$.

  \BF{Nonlinear System}.~ Apply the Pasting Theorem to classify $x=3$, $y=-3$ as a saddle, center, spiral
  or node for the {\bf nonlinear dynamical system}.
  Discuss all details of the application of the theorem.
  {\em Details count 75\%}.
  \end{quote}



\bigskip
\ifsolutions
\vfill\eject
\Answer{~\\
\BF{Part (a)}~~ It is an unstable spiral. Details: The eigenvalues of $A$ are roots of $r^2-2r+5=(r-1)^2+4=0$, which are complex conjugate roots $1\pm 2i$.
Rotation eliminates the saddle and node.
Finally, the atoms $e^{t}\cos 2t$, $e^{t}\sin 2t$ have limit zero at $t=-\infty$, therefore the system is stable at $t=-\infty$ and unstable at $t=\infty$.
So it must be a spiral [centers have no exponentials]. Report: {\bf unstable spiral}.
\\[1ex]
\BF{Part (b)}~~
%A:=<-3,-4|2,1>;
%linalg[eigenvects](A);
%   # [-1+2*I, 1, {Vector[row](2, {(1) = 1, (2) = 1+I})}],
% # [-1-2*I, 1, {Vector[row](2, {(1) = 1, (2) = 1-I})}]
% DD:=matrix([[-1,-2],[2,-1]]); P:=matrix([[1,2],[1,1]]);A:=evalm(inverse(P)&*DD&*P);
% DD:=matrix([[1,-2],[2,1]]); P:=matrix([[1,2],[1,0]]);
% A:=evalm(inverse(P)&*DD&*P);
% A:=Matrix([[3, 4], [-2, -1]]);
 % Answer: unstable spiral
% A == matrix([[5, 10], [-4, -7]])
 % Answer: stable spiral
It is a stable spiral. Details: The eigenvalues of $A$ are roots of $r^2+2r+5=(r+1)^2+4=0$, which are complex conjugate roots $-1\pm 2i$.
Rotation eliminates the saddle and node.
Finally, the atoms $e^{-t}\cos 2t$, $e^{-t}\sin 2t$ have limit zero at $t=\infty$, therefore the system is stable at $t=\infty$ and unstable at $t=-\infty$.
So it must be a spiral [centers have no exponentials]. Report: {\bf stable spiral}.
\\[1ex]
\BF{Part (c)}~~
The Jacobian is $J(x,y)=\twobytwo{1 & -4y-1}{4x-2y&-2x}$. Then $A=J(4,4)=\twobytwo{1 & -17}{8&-8}$.
\\[1ex]
\BF{Part (d)}~~
The Jacobian is $J(x,y)=\twobytwo{-1 & -4y-1}{4x+2y&2x}$. Then $A=J(-4,4)=\twobytwo{-1 & -17}{-8&-8}$.
\\[1ex]
\BF{Part (e)}~~
(1) The Jacobian is $J(x,y)=\twobytwo{-4-2x& 4}{3&-3}$. Then
$A=J(3,3)=\twobytwo{-10 & 4}{3&-3}$. The eigenvalues of $A$ are found
from $r^2+13r+18=0$, giving distinct real negative roots
$-\frac{13}{2}\pm(\frac12)\sqrt{97}$. Because there are no trig functions in the Euler solution aistoms,
then no rotation happens, and the classification must be a saddle or node. The Euler solution
atoms limit to zero at $t=\infty$, therefore it is a node and we report
a {\bf stable node} for the linear problem $\vec{u}'=A\vec{u}$ at
equilibrium $\vec{u}=\vec{0}$.
\\
(2) Theorem 2 in Edwards-Penney section 6.2 applies to say that
the same is true for the nonlinear system: {\bf stable node} at $x=3$,
$y=3$. The exceptional case in Theorem 2 is a proper node, having characteristic equation roots that are equal.
Stability is always preserved for nodes.
% DD:=matrix([[-1,-2],[2,-1]]); P:=matrix([[1,2],[1,1]]);A:=evalm(inverse(P)&*DD&*P);
% DD:=matrix([[1,-2],[2,1]]); P:=matrix([[1,2],[1,0]]);
% A:=evalm(inverse(P)&*DD&*P);
% A:=Matrix([[3, 4], [-2, -1]]);
 % Answer: unstable spiral
% A == matrix([[5, 10], [-4, -7]])
 % Answer: stable spiral
\\[1ex]
\BF{Part (f)}~~\\
% Answer: (1) u'=Au, A:=matrix([[-10,-4],[3,3]]);
%  (2) stable node  (3) Thm 2 in E+P ==> stable node
\BF{Linearization}. The Jacobian is $J(x,y)=\twobytwo{-4-2x& -4}{3&3}$. Then
$A=J(3,3)=\twobytwo{-10 & -4}{3&3}$. The eigenvalues of $A$ are found
from $r^2+7r-18=0$, giving distinct real roots
$2,-9$. Because there are no trig functions in the Euler solution atoms $e^{2t}, e^{-9t}$,
then no rotation happens, and the classification must be a saddle or node. The Euler solution
atoms do not limit to zero at $t=\infty$ or $t=-\infty$, therefore it is a saddle and we report
a {\bf unstable saddle} for the linear problem $\vec{u}'=A\vec{u}$ at
equilibrium $\vec{u}=\vec{0}$. \\
\BF{Nonlinear System}. Theorem 2 in Edwards-Penney section 6.2 applies to say that
the same is true for the nonlinear system: {\bf unstable saddle} at $x=3$,
$y=3-$.
}
\fi
\end{exercises}
\EJECT
\subsection*{Final Exam Problems}

{\bf Chapter 3}: Linear Constant Equations of Order $n$.\\[1ex]
\BF{(a)}  Find by variation of parameters a particular solution $y_p$ for the equation
$y''=1-x$. Show all steps in variation of parameters. Check the answer
by quadrature.
\\[1ex]
\BF{ (b) }  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 $x_{\mbox{ss}}$.
\\[1ex]
 \BF{(c)}  A fourth order linear homogeneous differential equation with
constant coefficients has two particular solutions $2e^{3x}+4x$ and
$xe^{3x}$. Write a formula for the general solution.
\\[1ex]
\BF{(d)} Find the {\bf Beats} solution for
the forced undamped spring-mass problem
$$ x'' + 64x = 40\cos(4t),\quad x(0)=x'(0)=0.$$
It is known that this solution is the sum of two harmonic oscillations of different frequencies.
{\bf To save time, don't convert to phase-amplitude form}.
\\[1ex]
 \BF{(e)}
 Write the solution $x(t)$ of
 $$x''(t)+25x(t) = 180\sin (4t), \quad x(0)=x'(0)=0,$$ as the sum
 of two harmonic oscillations of different natural frequencies.\newline
 {\bf To save time, don't convert to phase-amplitude form}.
\\[1ex]
 \BF{(f)}  Find the steady-state periodic
 solution for the forced spring-mass system $ x'' + 2x' + 2x =
5\sin(t)$.
\\[1ex]
 \BF{(g)} Given $5x''(t) + 2x'(t) + 4x(t) = 0$, which represents a
 damped spring-mass system with $m=5$, $c=2$, $k=4$, determine if the equation is
 over-damped , critically damped  or
 under-damped.\\
 {\bf To save time, do not solve for $x(t)$}!
\\[1ex]
 \BF{(h)}
Determine the practical resonance frequency $\omega$ for the electric
current equation
$$2I''+7I'+50I = 100\omega\cos(\omega t).$$
 \BF{(i)} Given the forced spring-mass system $ x'' + 2x' + 17x =
82\sin(5t)$, find the steady-state periodic  solution.
\\[1ex]
 \BF{ (j)} Let $f(x)=x^3e^{1.2x}+x^2e^{-x}\sin(x)$. Find the
characteristic polynomial of a constant-coefficient
linear homogeneous differential equation of least order which has $f(x)$ as a
solution. To save time, do not expand the polynomial and do not find the differential equation.

\bigskip



\BF{Chapter 5}.  Solve a homogeneous system $u'=Au$, $u(0)=\colvector{1\\ 2}$,
             $A=\twobytwo{2&3}{0&4}$ using the matrix exponential,
             Zeibur's method, Laplace resolvent and eigenanalysis.

 \bigskip

\BF{Chapter 5}.  Solve a non-homogeneous system $u'=Au+F(t)$, $u(0)=\colvector{0\\ 0}$,
             $A=\twobytwo{2&3}{0&4}$, $F(t)=\colvector{3\\ 1}$ using variation
             of parameters.


\bigskip

\ifsolutions
\vfill\eject
\Answer{
\BF{Chapter 3 final exam sample solutions}.\\
\BF{Part (a)} ~~ Answer: $\dd y_p=\frac{x^2}{2} - \frac{x^3}{6}$.
\\
\BF{Variation of Parameters}.\newline
 Solve $y''=0$ to get $y_h=c_1y_1+c_2y_2$,
$y_1=1$, $y_2=x$.  Compute the
Wronskian $W=y_1y_2'-y_1'y_2=1$. Then for
$f(t)=1-x$,
\\
$\dd
 y_p=y_1\int y_2\frac{-f}{W}dx+y_2\int y_1\frac{f}{W}dx,
 $
\\
 $\dd
 y_p=1\int -x(1-x)dx+x\int 1(1-x)dx,
 $
\\
  $\dd
 y_p=-1(x^2/2-x^3/3)+x(x-x^2/2),
 $
\\
  $\dd
 y_p=x^2/2 - x^3/6.
 $
\\
This answer is checked by quadrature, applied twice to $y'' = 1-x$ with initial conditions zero.
\\[1ex]
\BF{Part (b)} ~~ It has to be 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 $.
\\[1ex]
\BF{Part (c)}~~
In order for $xe^{3x}$ to be a solution, the general solution must have Euler atoms $e^{3x}, xe^{3x}$. Then the first solution $2e^{3x}+4x$ minus $2$ times the Euler atom $e^{3x}$ must be a solution, therefore $x$ is a solution, resulting in Euler atoms $1,x$. The general solution is then a linear combination of the four Euler atoms: $y = c_1(1)+c_2(x)+c_3\left(e^{3x}\right)+c_4\left(xe^{3x}\right)$.
\\[1ex]
\BF{ Part (d)}~~
% solve(r^2+64=0,r);
% de:=diff(x(t),t,t)+64*x(t)=40*cos(4*t);dsolve(de,x(t));
% dsolve({de,x(0)=0,D(x)(0)=0},x(t));
Use undetermined coefficients trial solution $x=d_1\cos 4t + d_2 \sin 4t$. Then $d_1=5/6$, $d_2=0$, and finally $x_p(t)=(5/6)\cos(4t)$. The characteristic equation $r^2+64=0$ has roots $\pm 8i$ with corresponding Euler solution atoms $\cos(8t), \sin(8t)$. Then $x_h(t)=c_1\cos(8t)+c_2\sin(8t)$. The general solution is $x=x_h+x_p$. Now use $x(0)=x'(0)=0$ to determine $c_1=-5/6, c_2=0$, which implies the particular solution $x(t)=-\frac56\cos(8t)+\frac56\cos(4t)$.
\\[1ex]
{\bf Part (e)}~~  The answer is $x(t)= -16\sin(5t)+20\sin(4t)$ by the method of undetermined coefficients.\\
Rule I: $x=d_1\cos(4t)+d_2\sin(4t)$. Rule II does not apply due to natural frequency $\sqrt{25}=5$
not equal to the frequency of the trial solution (no conflict).  Substitute the trial solution into
$x''(t)+25x(t) = 180\sin (4t)$
to get $9d_1\cos(4t)+9d_2\sin(4t) = 180\sin(4t)$. Match coefficients, to arrive at the equations
$9d_1=0$, $9d_2=180$. Then $d_1=0$, $d_2=20$ and $x_p(t)=20\sin(4t)$. Lastly, add the homogeneous solution to obtain $x(t)=x_h+x_p=c_1\cos(5t)+c_2\sin(5t)+20\sin(4t)$. Use the initial condition relations $x(0)=0, x'(0)=0$ to obtain the equations
$\cos(0)c_1+\sin(0)c_2+20\sin(0)=0$, $-5\sin(0)c_1 + 5\cos(0)c_2 +80\cos(0)=0$. Solve for the coefficients $c_1=0$, $c_2=-16$
% de:=diff(y(x),x,x)+25*y(x)=180*sin(4*x);dsolve(de,y(x));
% subs(y(x)=d1*cos(4*x)+d2*sin(4*x),de);collect(%,{cos(4*x),sin(4*x)});
% ic:=x(0)=0,D(x)(0)=0;dsolve([de,ic],y(x)); #y(x) = -16*sin(5*x)+20*sin(4*x) % F2010
\\[1ex]
\BF{Part (f)}~~
% de:=diff(y(x),x,x)+2*diff(y(x),x)+2*y(x)=5*sin(x);dsolve(de,y(x));
% subs(y(x)=d1*cos(x)+d2*sin(x),de);collect(%,{cos(x),sin(x)});
The answer is $x=\sin t - 2\cos t$ by the method of undetermined coefficients.\\
Rule I: the trial solution $x(t)$ is a linear combination of the Euler atoms found in $f(x)=5\sin(t)$. Then $x(t)=d_1\cos(t) + d_2\sin(t)$. Rule II does not apply, because solutions of the homogeneous problem contain negative exponential factors (no conflict). Substitute the trial solution into
$ x'' + 2x' + 2x = 5\sin(t)$ to get
$(-2d_1+d_2)\sin(t)+(d_1+2d_2)\cos(t) = 5\sin(t)$. Match coefficients to find the system of equations $(-2d_1+d_2)=5$, $(d_1+2d_2)=0$. Solve for the coefficients $d_1=-2$, $d_2=1$.
\\[1ex]
\BF{Part (g)}~~Use the quadratic formula to decide. The number under the radical sign in the formula,
called the discriminant, is $b^2-4ac=2^2-4(5)(4)=(19)(-4)$, therefore there are two complex conjugate roots
and the equation is {\bf under-damped}. Alternatively, factor $5r^2+2r+4$ to obtain roots $(-1\pm\sqrt19 i)/5$
and then classify as {\bf under-damped}.
\\[1ex]
\BF{Part (h)}~~ The resonant frequency is $\omega= 1/\sqrt{LC} = 1/\sqrt{2/50}=\sqrt{25}=5$. The solution uses the theory in the textbook, section 3.7, which says that electrical resonance occurs for $\omega= 1/\sqrt{LC}$.
\\[1ex]
\vfill\eject
\BF{Part (i)}~~
% de:=diff(x(t),t,t)+2*diff(x(t),t)+17*x(t)=82*sin(5*t);dsolve(de,x(t));
% yp:=A*cos(5*t)+B*sin(5*t);eval(subs(x(t)=yp,de));
% solve({-8*A+10*B=0, -10*A-8*B=82});
The answer is $x(t)=-5\cos(5t)-4\sin(5t)$ by undetermined coefficients.\\
 Rule I: The trial solution is  $x_p(t)=A\cos(5t)+B\sin(5t)$. Rule II:  because
the homogeneous solution $x_h(t)$ has limit zero at $t=\infty$, then Rule II does not apply (no conflict). Substitute the trial solution into the differential equation. Then $-8A\cos(5t)-8B\sin(5t)-10A\sin(5t)+10B\cos(5t) = 82\sin(5t)$. Matching coefficients of sine and cosine gives the equations $-8A+10B=0$, $-10A-8B=82$. Solving, $A=-5, B=-4$. Then $x_p(t)=-5\cos(5t)-4\sin(5t)$ is the unique periodic steady-state solution.
\\[1ex]
\BF{Part (j)}~~ The characteristic polynomial is the expansion $(r-1.2)^4((r+1)^2+1)^3$. Because
$x^3e^{ax}$ is an Euler solution atom for the differential equation if and only if
$e^{ax},xe^{ax},x^2e^{ax},x^3e^{ax}$ are Euler solution
atoms, then the characteristic equation must have roots
$1.2,1.2,1.2,1.2$, listing according to multiplicity. Similarly, $x^2e^{-x}\sin(x)$ is an Euler solution
 atom for the differential equation
if and only if $-1\pm i,-1\pm i,-1\pm i$ are roots of the characteristic
equation. There is a total of 10 roots with product of the
factors $(r-1)^4((r+1)^2+1)^3$ equal to the 10th degree characteristic polynomial.
}
\fi
\ifsolutions
\BF{Chapter 5 Final Exam Sample Solutions}. Presently, there are no solutions available for the two sample problems. If you solve one, then kindly email your solution to post.
\fi
\end{document}
\endinput
