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\underline{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}\hfill\fi
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\thispagestyle{empty} \hfill  2280 Sample Exam 1 S2019
\newline{\large\bf Name}.
\underline{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}\hfill}}
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\begin{center}\Large\bf
Differential Equations 2280 \\
\small Sample Midterm Exam 1\\
\small Exam Date: Friday, 22 February 2019 at 7:30am in LCB 215
\end{center}

{\bf Instructions}: This in-class exam is designed for 80 minutes. Maximum of seven (7) problems. Expect dramatically fewer parts to each problem. No
calculators, notes, tables or books. No answer check is expected.
Details count 3/4, answers count 1/4.

\begin{exercises}
\prob{1.}{(Quadrature Equations)} ~\newline
     \BF{(a)} [40\%] Solve $y'=\dd\frac{2x^3}{1+x^2}$. \newline
%     \BF{(b)} [25\%] Solve $y'=(2\sin x + \cos x)(\sin x - 2\cos x)$.\newline
%     \BF{(c)} [25\%] Solve $y'=\dd \frac{x\tan\left(\ln(1+x^2)\right)}{1+x^2}$, $y(0)=2$.\newline
     \BF{(b)} [60\%] Find the position $x(t)$ from the velocity model $\dd\frac{d}{dt}(e^{-t}v(t))= 2\,e^{t}$, $v(0)=5$ and the position model $\dd\frac{dx}{dt}=v(t)$, $x(2)=2$.
% F:=(2*x^3)/(1+x^2); int(F,x);
% x^2-ln(x^2+1)
%
%de:=diff(exp(-t)*v(t),t)=2*exp(t);
%dsolve([de,v(0)=5],v(t));
%vv:=rhs(%);
%dsolve([D(x)(t)=vv,x(0)=2],x(t))
%                                 / d      \
%         -exp(-t) v(t) + exp(-t) |--- v(t)| = 2 exp(t)
%                                 \ dt     /
%                  v(t) = (2 exp(t) + 3) exp(t)
%                     (2 exp(t) + 3) exp(t)
%                 x(t) = exp(2 t) + 3 exp(t) - 2
\ifsolutions
\begin{solution}{1} ~\newline
 \BF{(a)} Answer $y = {x}^{2}-\ln  \left( {x}^{2}+1 \right)  + c$.
 The integral of $F(x) = \frac{2x^3}{1+x^2}$ is found by substitution $u=1+x^2$, resulting in
 the new integration problem $\int F\,dx = \int \frac{u-1}{u}du= \int (1)du - \int \frac{du}{u}$.


 \BF{(b)} Velocity $v(t) = 2e^{2t} + 3e^{t}$ by quadrature. Integrate $x'(t) =2e^{2t} + 3e^{t}$ with
 $x(0)=2$ to obtain position $x(t) = e^{2t} + 3e^{t} -2$.


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

\iffalse
\begin{exercises}
\prob{2.}{(Classification of Equations)} ~\newline
 The differential equation $y'=f(x,y)$ is defined to be {\bf separable} provided
$f(x,y)=F(x)G(y)$ for some functions $F$ and $G$.

% \BF{(a)} [40\%]  Check (\fbox{X}) the problems that can be put into separable form.
% No details expected.
%\begin{center}\large\begin{tabular}{|lp{0.3\linewidth}|lp{0.4\linewidth}|}
%\hline
% \fbox{$\phantom{X}$} & $ y'+xy=y(2x+e^x)+x^2y$ & \fbox{$\phantom{X}$} &   $y'=(x-1)(y+1) + (1-x)y$ \\[2ex]  \hline
%  \fbox{$\phantom{X}$} &  $y'=2e^{2x-y}e^{3y}+3e^{3x+2y}$ & \fbox{$\phantom{X}$} &  $\dd y'+x^2e^y=xy$ \\[2ex] \hline
%  \end{tabular}
%\end{center}

 \BF{(a)} [40\%] The equation $y'+x(y+3)=ye^x+3x$ is separable. Provide formulas for $F(x)$ and $G(y)$.

 %\BF{(c)} [10\%] Give an example of $y'=f(x,y)$ which is separable and linear but not quadrature. No details expected.

  \BF{(b)} [60\%] Apply partial derivative tests to show that $y'=x + y$ is linear but not separable. Supply all details.

\ifsolutions
\begin{solution}{2} ~\newline
 \BF{(a)} The equation is $y'=ye^x-xy=(e^x-x)y$. Then $F(x)=e^x-x$, $G(y)=y$.

 \BF{(b)} Let $f(x,y)=x+y$. Then $\partial f/\partial y = 1$, which is independent of $y$, hence the equation $y'=f(x,y)$ is linear. The negative test is $\frac{\partial f/\partial y}{f}$ depends on $x$. In this case, the fraction is
 $$\frac{\partial f/\partial y}{f}=\frac{1}{f}= \frac{1}{x+y}.$$
  At $y=0$, this reduces to $1/x$, which depends on $x$, therefore the equation $y'=f(x,y)$ is not separable. Symmetrically, the test $f_x/f$ depends on $y$ implies $y'=f(x,y)$ is not separable.

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

\begin{exercises}
\prob{2.}{(Solve a Separable Equation)} ~\newline
 % u:=x*exp(-x)+sin(x)*cos(x); int(u,x);
 % # -x*exp(-x)-exp(-x)+(1/2)*sin(x)^2
 % w:=(5*y+10)/(y^2+3*y-4);int(w,y);
 % # ans=2*ln(y+4)+3*ln(y-1)
 Given
   $\dd (5y+10)y' = \left( xe^{-x} +
\sin(x)\cos(x) \right)(y^2+3y-4)$.


  Find a non-equilibrium solution in implicit form. ~\newline
 To save time, {\bf do not solve} for $y$ explicitly and {\bf do not
 solve} for equilibrium solutions.

\ifsolutions
\begin{solution}{2} ~\newline
The solution by separation of variables identifies the separated equation $y'=F(x)G(y)$ using
$$F(x)= xe^{-x} + \sin(x)\cos(x),\quad G(y) = \frac{y^2+3y-4}{5y+10}.$$
The integral of $F$ is done by parts and also by $u$-substitution.
$$
\begin{array}{lcl}
\int Fdx &=& \int xe^{-x}\, dx + \int \sin(x)\cos(x)\, dx\\[2ex]
         &=& I_1 + I_2 .\\[3ex]
I_1 &=& \int xe^{-x}\, dx \\[2ex]
         &=& -xe^{-x} - \int e^{-x}\, dx,\quad \mbox{parts}\quad u=x, dv=e^{-x}dx,\\
         &=& -xe^{-x} - e^{-x} + c_1 \\
I_2 &=& \int \sin(x)\cos(x)\, dx\\[2ex]
         &=& \int u\,du,\quad u=\sin(x), du=\cos(x)dx,\\
         &=& u^2/2+c_2 \\
         &=& \frac12\sin^2(x)+c_2
\end{array}
$$
Then
$\dd\int Fdx = -xe^{-x} - e^{-x} + \frac12\sin^2(x)+c_3$.

The integral of $1/G(y)$ requires partial fractions. The details:
$$
\begin{array}{lcl}
\dd\int\frac{dx}{G(y(x))} &=& \dd\int \frac{5u+10}{u^2+3u-4}\, du,\quad u=y(x), du=y'(x)dx,\\[2ex]
         &=& \dd\int \frac{5u+10}{(u+4)(u-1)}\, du\\[2ex]
         &=& \dd\int \frac{A}{u+4} + \frac{B}{u-1}\, du,\quad A,B\quad\mbox{determined later},\\[2ex]
         &=& A\ln|u+4|+B\ln|u-1|+c_4
\end{array}
$$
The partial fraction problem
$$\frac{5u+10}{(u+4)(u-1)} = \frac{A}{u+4} + \frac{B}{u-1}$$
can be solved in a variety of ways, with answer $A=\frac{-20+10}{-5}=2$ and $B=\frac{15}{5}=3$.
The final implicit solution is obtained from $\int \frac{dx}{G(y(x))} = \int F(x)dx$, which gives the equation
$$2\ln|y+4|+3\ln|y-1| = -xe^{-x} - e^{-x} + \frac12\sin^2(x)+c.$$
\end{solution}
\fi
\end{exercises}
\EJECT

\begin{exercises}
\prob{3.}{(Linear Equations)} ~\newline
% v:='v':de:=2*diff(v(t),t)=-64+(10/(3*t+2))*v(t); ic:=v(0)=32;
% dsolve({de,ic},v(t));
% #                        v(t) = 32+48*t

 \BF{(a)} [60\%] Solve the linear model $\dd 2x'(t)= - 64 + \frac{10}{3t+2}x(t)$, $x(0)=32$. Show
 all integrating factor steps.
 % int( (5*u+10)/(u^2+3*u-4),u); int(x*exp(-x)+sin(x)*cos(x),x);
 \newline
 \BF{(b)} [20\%]  Solve  $\dd \frac{dy}{dx}-(\cos(x))y=0$ using the homogeneous linear equation shortcut.
\newline
 \BF{(c)} [20\%] Solve $\dd 5\frac{dy}{dx}-7y=10$ using the superposition principle $y=y_h+y_p$ shortcut. Expected are answers for $y_h$ and $y_p$.

\ifsolutions
\begin{solution}{3} ~\newline
 \BF{(a)} The answer is $v(t)=32+48t$. The details:
 $$\begin{array}{l}
 v'(t) = -32 + \dd\frac{5}{3t+2}\,v(t),\\[2ex]
 v'(t) + \dd\frac{-5}{3t+2}\, v(t) = -32,\quad\mbox{standard form $v'+p(t)v=q(t)$}\\[2ex]
 p(t) = \dd\frac{-5}{3t+2},\\[2ex]
 W = e^{\int p\,dt},\quad \mbox{integrating factor} \\[2ex]
 W = e^u,\quad u = \int p\,dt = -\frac{5}{3}\ln|3t+2|= \ln\left(|3t+2|^{-5/3}\right)\\[2ex]
 W = (3t+2)^{-5/3},\quad\mbox{Final choice for $W$}.
 \end{array}
 $$
 Then replace the left side of $v'+pv=q$ by $(vW)'/W$ to obtain
 $$\begin{array}{l}
 v'(t) + \dd\frac{-5}{3t+2}\, v(t) = -32,\quad\mbox{standard form $v'+p(t)v=q(t)$}\\[2ex]
\dd\frac{(vW)'}{W} = -32,\quad \mbox{Replace left side by quotient $(vW)'/W$}\\[2ex]
 (vW)' = -32W,\quad\mbox{cross-multiply}\\[2ex]
 vW = -32\int Wdt,\quad\mbox{quadrature step.}
 \end{array}
 $$
The evaluation of the integral is from the power rule:
$$\int -32W\,dt = -32\int (3t+2)^{-5/3}dt=-32\frac{(3t+2)^{-2/3}}{(-2/3)(3)} + c.$$
Division by $W=(3t+2)^{-5/3}$ then gives the general solution
$$v(t) = \frac{c}{W} -\frac{32}{-2}(3t+2)^{-2/3}(3t+2)^{5/3}.$$
Constant $c$ evaluates to $c=0$ because of initial condition $v(0)=32$. Then
$$v(t) =\frac{32}{-2}(3t+2)^{-2/3}(3t+2)^{5/3}=16(3t+2)^{-\frac23+\frac53} = 16(3t+2).$$

 \BF{(b)} The answer is $y=$ constant divided by the integrating factor: $y=\dd\frac{c}{W}$.
 Because $W=e^u$ where $u=\int -\cos(x)dx = -\sin x$, then $y=ce^{\sin x}$.
 % dsolve(diff(y(x),x)-cos(x)*y(x)=0,y(x));

 \BF{(c)} The equilibrium solution (a constant solution) is $y_p = -\frac{10}{7}$. The homogeneous solution is $y_h = c e^{7x/5}$ = constant divided by the integrating factor.
 Then $y=y_p+y_h = -\frac{10}{7} + ce^{7x/5}$.
 % dsolve(5*diff(y(x),x)-7*y(x)=10,y(x));
\end{solution}
\fi
\end{exercises}



\EJECT

\begin{exercises}
\prob{4.}{(Stability)} ~\newline
 %\BF{(a)} [50\%] Draw a phase line diagram for the differential equation
% $$\frac{dx}{dt} = \left(\ln(1+5x^2)\right)^{1/5}\left(|2x-1|-3\right)^3(2+x)^2(4-x^2)(1-x^2)^{3}e^{\cos x}.$$
% Expected in the phase line diagram are equilibrium
% points and signs of $dx/dt$.
% \newline
% \rule{\textwidth}{0.01in}

 %\vspace*{1.5in}

        %\BF{(b)} [50\%]
        Assume an autonomous equation $x'(t)=f(x(t))$. Draw a phase portrait with at least 12 threaded curves, using the phase line diagram given below. Add these labels as appropriate: funnel, spout, node [neither spout nor funnel], stable, unstable.

 \def\AA{\rule{5ex}{0.02in}}
 \def\BB{\rule[-0.1in]{0.4ex}{0.2in}}
 \def\PLUS{\mbox{\Large$+$}}
 \def\MINUS{\mbox{\Large$-$}}
 \begin{center}\begin{tabular}{*{11}{@{}c}}
 \PLUS & & \MINUS & & \PLUS & & \MINUS & & \PLUS & & \PLUS \\
\AA\AA & \AA & \AA\AA\AA & \AA & \AA & \AA & \AA & \AA & \AA & \AA& \AA\AA\AA \mbox{\Large ~$x$} \\[-3ex]
          & \BB &           & \BB &     & \BB &     & \BB &     & \BB&  \\[1ex]
    & $-10$   &     & $-5$   &     & $-3$    &     & $0$    &     & $3$    &
 \end{tabular}\end{center}
 \rule{\textwidth}{0.01in}

 \bigskip
 \ifsolutions
 \begin{solution}{4} ~\newline
 The graphic is drawn using increasing and decreasing curves, which may or may not be depicted with turning points. The rules:

 \begin{quote}\sf
 1. A curve drawn between equilibria is increasing if the sign is PLUS.

 2. A curve drawn between equilibria is decreasing if the sign is MINUS.

 3. Label: FUNNEL, STABLE \newline
 \hspace*{5pc} The signs left to right are PLUS MINUS crossing the equilibrium point.

 4. Label: SPOUT, UNSTABLE \newline
 \hspace*{5pc} The signs left to right are MINUS PLUS crossing the equilibrium point.

 5. Label: NODE, UNSTABLE \newline
 \hspace*{5pc} The signs left to right are PLUS PLUS crossing the equilibrium point, or \newline
 \hspace*{5pc} The signs left to right are MINUS MINUS crossing the equilibrium point.
 \end{quote}

 The answer:

  $x = -10$: FUNNEL, STABLE \newline
  $x = -5$: SPOUT, UNSTABLE \newline
  $x = -3$: FUNNEL, STABLE \newline
  $x = 0$: SPOUT, UNSTABLE \newline
  $x = 3$: NODE, UNSTABLE

\end{solution}

 \fi
\end{exercises}
\EJECT
\begin{exercises}
\prob{5.}{(Chapter 3: Linear $n$th Order DE)} ~\newline
Using Euler's theorem on Euler solution atoms and the characteristic equation for
higher order constant-coefficient differential equations, solve
(a), (b), (c).

(a) [40\%] Find a constant coefficient differential equation $ay'' + by' + cy=0$ which has
particular solutions $-5e^{-x}+xe^{-x}$, $10e^{-x}+xe^{-x}$.
%
%(b) [25\%] Solve $y^{(6)} + 4y^{(5)}+ 4y^{(4)} = 0$.

(b) [30\%] Given characteristic equation
 $ r(r-2)(r^3+4r)(r^2+2r+37)=0$, solve the differential equation.

(c) [30\%] Given $mx''(t) + cx'(t) + kx(t) = 0$, which
represents an unforced damped spring-mass system. Assume $m=4$, $c=4$, $k=129$.
Classify the equation as
over-damped, critically damped or under-damped. Illustrate
in a spring-mass-dashpot drawing the assignment of physical
constants $m$, $c$, $k$ and the initial conditions $x(0)=1$, $x'(0)=0$.


% ((r+1/2)^2+32=0 -->4 4r^2+4r+129=0
\ifsolutions
\begin{solution}{5} ~\newline
{\bf 5(a)}\newline
Multiply the first solution by $2$ and add it to the second solution. Then Euler atom $xe^{-x}$ is a solution, which implies
that $r=-1$ is a double root of the characteristic equation. Then the characteristic equation should be
$(r-(-1))(r-(-1))=0$,
or $r^2+2r+1=0$. The differential equation is $y''+2y'+y=0$.

{\bf 5(b)}\newline
The characteristic equation factors into $r^2(r-2)(r^2+4)((r+1)^2+36)=0$ with roots $r=0,0;  2; \pm 2i; -1\pm 6i $.
Then $y$ is a linear combination of the Euler solution atoms:

 $1,x,e^{2x},\cos(2x), \sin(2x); e^{-x}\cos(6x),e^{-x}\sin(6x)$.

{\bf 5(c)}\newline
Use $4r^2+4r+129=0$ and the quadratic formula to obtain roots
$r=-1/2+4\sqrt2 i,-1/2-4\sqrt2 i$ and Euler solution atoms $e^{-x/2}\cos 4\sqrt2 t, e^{-x/2}\sin 4\sqrt2 t$. Then $y$ is a linear combination of these two solution atoms, and it oscillates, therefore the classification is
{\bf under-damped}. The illustration shows a spring, a
dashpot and a mass with labels $k$, $c$, $m$. Initial conditions mean mass elongation $x=1$, at rest.

A {\bf dashpot} is represented as a cylinder and piston with rod, the rod attached to the mass. Variable $x$ is positive in the down direction and negative in the up direction. The equilibrium position is $x=0$.

\end{solution}
\fi

\end{exercises}
\EJECT

\begin{exercises}
\prob{6.}{(Chapter 3: Linear $n$th Order DE)} ~\newline
(a) [25\%] The trial solution $y$ with fewest Euler solution atoms, according to the
method of undetermined coefficients, contains no solution of the
homogeneous equation. Explain why, using the example $y'' = 1+x$.

(b) [75\%]
Determine for $ y^{(3)} + y^{(2)} = x +
2e^{-x}+\sin x$ the corrected trial solution for $y_p$ according
to the method of undetermined coefficients.
{\bf Do not evaluate the
undetermined coefficients}! The trial solution should be the one with
fewest Euler solution atoms.


\ifsolutions
\begin{solution}{6} ~\newline
{\bf (a)}. Rule I says the trial solution is $y=d_1+d_2x$. Rule II says to multiply
by $x$ until no atom is a solution of $y''=0$. Then $y=d_1x^2+d_2x^3$ contains no terms of the homogeneous solution $y_h=c_1+c_2x$.

{\bf (b)}.
The homogeneous equation $ y^{(3)} + y^{(2)}=0$ has solution $y_h=c_1+c_2x+c_3e^{-x}$,
because the characteristic polynomial has roots $0$, $0$, $-1$.

{\bf \fbox{1}~} Rule I constructs an initial trial solution $y$ from the list of independent Euler solution atoms
$$e^{-x},\quad 1,\quad x,\quad  \cos x,\quad \sin x.$$
Linear combinations of these atoms are supposed
to reproduce, by assignment of constants, all derivatives of $f(x) =  x +
2e^{-x}+\sin x$, which is the right side of the differential equation.
Each of $y_1$ to $y_4$ in the display below is
constructed to have the same {\bf base atom}, which is the Euler atom
obtained by stripping the power of $x$. For example, Euler solution atom $xe^{0x}$ (or $x$, because $e^{0x}=1$) strips to base atom $e^{0x}$ or $1$.
$$\begin{array}{lcl}
 y &=& y_1+y_2+y_3+y_4,\\
 y_1 &=& d_1e^{-x},\\
 y_2 &=& d_2+d_3x,\\
 y_3 &=& d_4\cos x,\\
 y_4 &=& d_5\sin x.
\end{array}
$$

{\bf \fbox{2}~} Rule II is applied individually to each of
$y_1$, $y_2$, $y_3$, $y_4$ to give the {\bf corrected trial
solution}
$$\begin{array}{lcl}
y &=& y_1+y_2+y_3+y_4,\\
 y_1 &=& d_1xe^{-x},\\
 y_2 &=& x^2(d_2+d_3x),\\
 y_3 &=& d_4\cos x,\\
 y_4 &=& d_5\sin x.
\end{array}
$$
The powers of $x$ multiplied in each case are selected to
eliminate terms in the initial trial solution which duplicate
homogeneous equation Euler solution atoms. For instance, $y_1=d_1e^{-x}$ is {\bf in conflict} with the homogeneous equation, because $e^{-x}$ is a common Euler atom of both $y_1$ and the homogeneous solution ($y_h=c_1+c_2x+c_3e^{-x}$). Then Rule II multiplies $y_1$ by $x$ to obtain the replacement $y_1=d_1xe^{-x}$. This new term is again subjected to the Rule II test: Does $y_1$ contain an Euler atom of the homogeneous equation? The answer is NO, so the $x$-multiplication stops and the term $y_1$ is finished. We go on to the remaining terms, in the same way. Term $y_2$ needs two $x$-multiplications. The factor used after so many $x$-multiplications is
exactly $x^s$ of the Edwards-Penney table, where $s$ is the
multiplicity of the characteristic equation root $r$ that produced
the conflicting atom in the homogeneous solution $y_h$. The atoms in terms
$y_3, y_4$ are not solutions of the homogeneous equation, therefore $y_3, y_4$ are unaltered by Rule II.



\end{solution}
\fi

\end{exercises}

\EJECT





\begin{exercises}
\prob{7.}{(Chapters 1 and 3)} ~\newline
% de:=2*diff(v(t),t)=-8+(2/(2*t+1))*v(t); ic:=v(0)=-4; dsolve({de,ic},v(t));
 \BF{(a)} [20\%] Solve $\dd 2v'(t)=-8 + \frac{2}{2t+1}v(t)$, $v(0)=-4$. Show
 all integrating factor steps.

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

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

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

 \BF{(e)} [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 $x_{\mbox{ss}}$.

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

\BF{(g)} [50\%] Find by any applicable method
the steady-state periodic solution for the current equation $
I'' + 2I' + 5I = -10\sin(t)$.
\newline
\BF{(h)} [50\%] 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.

\ifsolutions
\begin{solution}{7}

 \BF{(a)} $v(t) = -4 -8t$
 \newline
 \BF{(b)} $r^2+4r+6=0$, $y=c_1y_1+c_2y_2$, $y_1=e^{-2x}\cos \sqrt2 x$, $y_2=e^{-2x}\sin \sqrt2 x$.
\newline
 \BF{(c)} 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
 \BF{(d)} The atoms that appear in $y(x)$ are $x, \sin\sqrt2x, e^{-x}\cos 3x$. Derivatives of these atoms create a longer list:
 $1,x, \cos\sqrt2x, \sin\sqrt2x, e^{-x}\cos 3x, e^{-x}\sin 3x$. These atoms correspond to characteristic equation roots $0,0$; $\sqrt2 i$, $-\sqrt2 i$, $-1+3i$, $-1-3i$. Then the characteristic equation has factors $r, r$; $x^2+2$; $((r+1)^2+9)$. The product of these factors is the correct characteristic equation, which corresponds to the differential equation of least order such that $y(x)$ is a solution. Report $r^6+2r^5+12r^4+4r^3+20r^2=0$ as the characteristic equation or $y^{(6)}+2y^{(5)}+12y^{(4)}+4y'''+20y''=0$ as the differential equation.
\newline
 \BF{(e)} It has to equal 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
 \BF{(f)} 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 Rule \fbox{1}~} An initial trial solution $y$ is constructed for
atoms $1$, $x$, $x^2$, $\cos x$, $\sin x$ giving 3 groups, each group having the same base atom:
$$\begin{array}{lcl}
 y &=& y_1+y_2+y_3,\\•••••••••
 y_1 &=& d_1+d_2x+d_3x^2,\\
 y_2 &=& d_4\cos x,\\
 y_3 &=& d_5\sin x.
\end{array}
$$
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 Rule \fbox{2}~} The correction rule is applied individually to each of
$y_1$, $y_2$, $y_3$.\newline
Multiplication by $x$ is done repeatedly, until the replacement atoms do not appear in
atom list for the homogeneous differential equation. The result is the {\bf shortest trial
solution}
$$ y = y_1+y_2+y_3=(d_1x^2+d_2x^3+d_3x^4) + (d_4\cos x) + (d_5\sin x). $$
Some facts: (1) If an Euler solution atom of the homogeneous equation appears in a group, then it is removed because of
$x$-multiplication, but replaced by a new atom having the same base atom. (2) The number of terms in each of $y_1$ to $y_3$ is unchanged from Rule I to Rule II.

\BF{(g)} ~~ Answer: $ \dd I_{\mbox{ss}}(t) = \cos t-2\sin t.$
\begin{quote}
\BF{Variation of Parameters}.\newline
This is the same problem as $x''+2x'+5x=-10\sin(t)$.\\
 Solve $x''+2x'+5x=0$ to get $x_h=c_1x_1+c_2x_2$,
$x_1=e^{-t}\cos 2 t$, $x_2=e^{-t}\sin 2 t$.  Compute the
Wronskian $W=x_1x_2'-x_1'x_2=4e^{-2t}$. Then for
$f(t)=-10\sin(t)$,
$$
 x_p=x_1\int x_2\frac{-f}{W}dt+x_2\int x_1\frac{f}{W}dt.
 $$
 The integrations are too difficult, so the
 method won't be pursued.
\\
 \BF{Undetermined Coefficients}.
\\
The trial solution by Rule I is $I=d_1\cos t + d_2\sin t$. The homogeneous solutions
have exponential factors, therefore the Euler solution atoms in the trial solution cannot be solutions of the homogeneous problem, hence Rule II does not apply.
\\
Substitute the
trial solution into the non-homogeneous equation to obtain the answers $d_1=1$, $d_2=-2$. The
unique periodic solution $I_{\mbox{ss}}$ is extracted from the
general solution $I=I_h+I_p$ by crossing out all negative
exponential terms (terms which limit to zero at infinity). Because
$I_p=d_1\cos t + d_2\sin t = \cos t - 2\sin t$ and the homogeneous solution $x_h$ has negative exponential terms, then
$$
I_{\mbox{ss}}=\cos t - 2\sin t.
$$
\BF{Laplace Theory}.\newline
   Plan: Find the general solution, then extract the steady-state solution by dropping negative exponential terms. The computation can use initial data $I(0)=I'(0)=0$, because every particular solution contains the terms of the steady-state solution. Some details:
\\
   $\dd (s^2+2s+5)\L(I) = \frac{-10}{s^2+1}$ \newline
   $\dd \L(I) = \frac{-10}{(s^2+1)(s^2+2s+5)}$ \newline
   $\dd \L(I) = \frac{-10}{(s^2+1)((s+1)^2+4)}$ \newline
   $\dd \L(I) = \frac {s-2}{s^2+1}-\frac {s}{(s+1)^2+4}$ \newline
   $\dd \L(I) = \frac {s}{s^2+1}-2\frac{1}{s^2+1}-\frac {s+1}{(s+1)^2+4} + \frac12\frac {2}{(s+1)^2+4}$ \newline
   $\dd \L(I) = \L(\cos t)-2\L(\sin t)-\L(e^{-t}\cos 2t)+\frac12\L(e^{-t}\sin 2t)$ \newline
 $\dd I(t) = \cos t-2\sin t-e^{-t}\cos 2t+\frac12e^{-t}\sin 2t$, by Lerch's Theorem.
 \\
 Dropping the negative exponential terms gives the steady-state solution
 $ \dd I_{\mbox{ss}}(t) = \cos t-2\sin t.$
\end{quote}
\BF{(h)} ~~ Answer: $\dd y_p=\frac{x^2}{2} - \frac{x^3}{6}$.
\begin{quote}
\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.
\end{quote}

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

\end{document}
\endinput
