% 2015 Differential Equations 2280-2
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\def\BF#1{{\large\bf #1}}
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\begin{document}
\thispagestyle{empty} {\large\bf Name}
\underline{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}\hfill
%{{\bf Time of your class}~\underline{~~~~~~~~~~} }
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Attach extra pages as needed, then staple.\eject
\thispagestyle{empty} \hfill  2280 Sample Exam 1 S2015
[8:35]\newline{\large\bf Name}.
\underline{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}\hfill}}
%%% {\bf Time of your class}~\underline{~~~~~~~~~~} }
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\begin{center}\Large\bf
Differential Equations 2280 \\
\small Sample Midterm Exam 1\\
\small Exam Date: Friday, 27 February 2015 at 12:50pm
\end{center}

{\bf Instructions}: This in-class exam is 50 minutes. 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)} [25\%] Solve $y'=\dd\frac{3+x^2}{1+x^2}$. \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{(d)} [25\%] Find the position $x(t)$ from the velocity model $\dd\frac{d}{dt}(t^2v(t))= 0$, $v(2)=10$ and the position model $\dd\frac{dx}{dt}=v(t)$, $x(2)=-20$.

\iftrue {\small [Integral tables will be supplied
 for anything other than basic formulas. This sample problem
 would require no integral table. The exam problem will be
 shorter.]}\fi

\end{exercises}
\EJECT


\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{(b)} [10\%] Is $y'+x(y+1)=ye^x+x$ separable? No details expected.

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

  \BF{(d)} [40\%] Apply tests to show that $y'=x + e^y$ is not separable and not linear. Supply all details.

\end{exercises}
\EJECT

\begin{exercises}
\prob{3.}{(Solve a Separable Equation)} ~\newline
 % u:=(2*x^2+3)/(x+2); int(u,x);
 % # 1/3*ln(y-2)+2/3*ln(y+1)= -exp(2-x)+x^2-4*x+11*ln(x+2)+C
 Given
   $\dd (x+3)(y+1)y' = \left( (x+3)e^{-x+2} +
3 x^2+2 \right)(y-1)(y+2)$.


  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.

\end{exercises}
\EJECT

\begin{exercises}
\prob{4.}{(Linear Equations)} ~\newline
% v:='v':de:=5*diff(v(t),t)=-160+(25/(2*t+3))*v(t); ic:=v(0)=32; dsolve({de,ic},v(t));
 \BF{(a)} [60\%] Solve the linear model $\dd 5 x'(t)= - 160 + \frac{25}{2t+3}x(t)$, $x(0)=32$. Show
 all integrating factor steps.\newline
 \BF{(b)} [20\%]  Solve the homogeneous equation $\dd \frac{dy}{dx}-(2x)y=0$.
\newline
 \BF{(c)} [20\%] Solve $\dd 5\frac{dy}{dx}+10y=7$ using the superposition principle $y=y_h+y_p$. Expected are answers for $y_h$ and $y_p$.

\end{exercises}



\EJECT

\begin{exercises}
\prob{5.}{(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 diagram 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 & & \PLUS & & \PLUS & & \MINUS & & \PLUS & & \MINUS \\
\AA\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]
    & $-9$   &     & $-5$   &     & $-3$    &     & $0$    &     & $3$    &
 \end{tabular}\end{center}
 \rule{\textwidth}{0.01in}
\end{exercises}
\EJECT
\begin{exercises}
\prob{6.}{(ch3)} ~\newline
Using Euler's theorem on atoms and the characteristic equation for
higher order constant-coefficient differential equations, solve
(a), (b), (c) and (d).

(a) [25\%] Find a differential equation $ay'' + by' + cy=0$ with
           solutions $2e^{-x}$, $e^{-x}-e^{2x/3}$.

(b) [25\%] Solve $y^{(6)} + 4y^{(5)}+ 4y^{(4)} = 0$.

(c) [25\%] Given characteristic equation 
 $ r(r+2)(r^3-4r)^3(r^2+2r+5)=0$, solve the differential equation.

(d) [25\%] Given $4x''(t) + 4x'(t) + 65x(t) = 0$, which
represents an unforced damped spring-mass system with $m=4$, $c=4$, $k=65$.
 Solve the differential equation [15\%]. Classify the answer as
over-damped, critically damped or under-damped [5\%]. Illustrate
in a drawing of the physical model the meaning of constants $m$, $c$, $k$
[5\%].

% ((r+1/2)^2+16=0 -->4 r^2+4r+1+64=0
\ifsolutions
\begin{solution}{6} ~\newline
Part (a)\newline
 {\bf 1}: $r^2+r+1=0$, $y=c_1y_1+c_2y_2$, $y_1=e^{-x/2}\cos(\sqrt3
x/2)$, $y_2=e^{-x/2}\sin(\sqrt3 x/2)$.\newline
 {\bf 2}: $r^{iv}+4r^2=0$, roots $r=0,0,2i,-2i$. Then
 $y=(c_1+c_2x)e^{0x}+c_3\cos 2x + c_4\sin 2x$. \newline
 {\bf 3}: Write as $(r-a)^2(r+a)^2(r^2+16)^3=0$ where $a=\sqrt3$.
 Then $y=u_1e^{ax}+u_2e^{-ax}+u_3\cos 4x + u_5\sin 3x$. The
 polynomials are $u_1=c_1+c_2x$, $u_2=c_3+c_4x$,
 $u_3=c_5+c_6x+c_7x^2$, $u_4=c_8+c_9x+c_{10}x^2$.

Part (b)\newline
Use $4r^2+4r+1=0$ and the quadratic formula to obtain roots
$r=-1/2,-1/2$. Case 2 of the recipe gives $y=(c_1+c_2t)e^{-t/2}$.
This is critically damped. The illustration shows a spring,
dampener and mass with labels $k$, $c$, $m$, $x$ and the
equilibrium position of the mass.
\end{solution}
\fi

\end{exercises}
\EJECT

\begin{exercises}
\prob{7.}{(ch3)} ~\newline
(a) [25\%] The trial solution $y$ with fewest 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^{(4)} + y^{(2)} = x +
2e^{x}+3\sin x$ the corrected trial solution for $y_p$ according
to the method of undetermined coefficients. 
Do not evaluate the
undetermined coefficients! The trial solution should be the one with
fewest atoms.



%(b) [40\%] The corrected trial solution in undetermined coefficients for
%$y^{(4)}(t)+y''(t)=t- \sin t$ can be constructed from $T(s)\L(t - s\sin
%t)$ where $T(s)$ is the transfer function. Show the Laplace and partial
%fraction details necessary to obtain the corrected trial solution. 
%
%Note 1. Don't find $y_p(t)$ -- find the trial solution with symbols
%$d_1$, $d_2$, \ldots.
%
%Note 2. Certain partial fraction terms are removed before forming the trial
%solution. Document where it happens, but don't explain why.

\ifsolutions
\begin{solution}{7} ~\newline
The homogeneous solution is $y_h=c_1+c_2x+c_3e^{3x}+c_4e^{-3x}$,
because the characteristic polynomial has roots $0$, $0$, $3$,
$-3$.

{\bf \fbox{1}~} An initial trial solution $y$ is constructed for
atoms $1$, $x$, $e^{3x}$, $e^{-3x}$, $\cos x$, $\sin x$ giving
$$\begin{array}{lcl}
 y &=& y_1+y_2+y_3+y_4,\\
 y_1 &=& (d_1+d_2x)e^{3x},\\
 y_2 &=& d_3+d_4x+d_5x^2+d_6x^3,\\
 y_3 &=& d_7e^{-3x},\\
 y_4 &=& d_8\cos x + d_9\sin x.
\end{array}
$$
Linear combinations of the listed independent atoms are supposed
to reproduce, by assignment of constants, all derivatives of the
right side of the differential equation.

{\bf \fbox{2}~} The fixup rule is applied individually to each of
$y_1$, $y_2$, $y_3$, $y_4$ to give the \BF{corrected trial
solution}
$$\begin{array}{lcl}
y &=& y_1+y_2+y_3,\\
 y_1 &=& x(d_1+d_2x)e^{3x},\\
 y_2 &=& x^2(d_3+d_4x+d_5x^2+d_6x^3),\\
 y_3 &=& x(d_7e^{-3x}),\\
 y_4 &=& d_8\cos x + d_9\sin x.
\end{array}
$$
The powers of $x$ multiplied in each case are designed to
eliminate terms in the initial trial solution which duplicate
atoms appearing in the homogeneous solution $y_h$. The factor is
exactly $x^s$ of the Edwards-Penney table, where $s$ is the
multiplicity of the characteristic equation root $r$ that produced
the related atom in the homogeneous solution $y_h$. By design,
unrelated atoms are unaffected by the fixup rule, and that is why
$y_4$ was unaltered.

{\bf \fbox{3}~} Undetermined coefficient step skipped, according
to the problem statement.

{\bf \fbox{4}~} Undetermined coefficient step skipped, according
to the problem statement.


\end{solution}
\fi

\end{exercises}
\EJECTEND

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