% 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{~~~~~~~~~~} }
\def\EJECT{\SeparatedPages{\vfill Use this page to start your solution.
Attach extra pages as needed, then staple.\eject
\thispagestyle{empty} \hfill  2280 Sample Exam 1 S2015
\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. 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{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
{\bf 6(a)}\newline
Divide the first solution by $2$. Then Euler atom $e^{-x}$ is a solution, which implies
that $r=-1$ is a root of the characteristic equation. Subtract $y_1=e^{-x}$ and
$y_2=e^{-x}-e^{2x/3}$ to justify that $y=y_1-y_2=e^{2x/3}$ is a solution. It is an Euler atom
corresponding to root $r=2/3$. Then the characteristic equation should be $(r-(-1))(r-2/3)=0$,
or $3r^2+r-2=0$. The differential equation is $3y''+y'-2y=0$.

{\bf 6(b)}\newline
The characteristic equation factors into $r^4(r^2+4r+4)=0$ with roots $r=0,0,0,0,-2, -2$.
Then $y$ is a linear combination of the Euler atoms $1,x,x^2,x^3,e^{-2x},xe^{-2x}$.

{\bf 6(c)}\newline
The roots of the fully factored equation $ r^4(r+2)^4(r-2)^3((r+1)^2+4)=0$ are $$r=0,0,0,0, -2,-2,-2,-2, 2,2,2, -1\pm 2i.$$ The solution $y$ is a linear combination of the Euler atoms
$$1,x,x^2,x^3; \quad
e^{-2x},xe^{-2x},x^2e^{-2x},x^3e^{-2x};\quad
e^{2x},xe^{2x},x^2e^{2x}; \quad
e^{-x}\cos(2x), e^{-x}\sin(2x).$$

{\bf 6(d)}\newline
Use $4r^2+4r+65=0$ and the quadratic formula to obtain roots
$r=-1/2+4i,-1/2-4i$. Case 2 of the recipe gives $y=(c_1\cos 4t+c_2\sin 4t)e^{-t/2}$.
This is under-damped (it oscillates). The illustration shows a spring,
dashpot 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 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^{(4)} + y^{(2)} = x +
2e^{x}+3\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}{7} ~\newline
{\bf 7(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 7(b)}. The homogeneous equation $ y^{(4)} + y^{(2)}=0$ has solution $y_h=c_1+c_2x+c_3\cos x+c_4\sin x$,
because the characteristic polynomial has roots $0$, $0$, $i$,
$-i$.

{\bf \fbox{1}~} Rule I constructs an initial trial solution $y$ from the list of independent Euler 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 $F(x) =  x +
2e^{x}+3\sin x$, 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, $x=xe^{0x}$ 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}
$$\begin{array}{lcl}
y &=& y_1+y_2+y_3+y_4,\\
 y_1 &=& d_1e^{x},\\
 y_2 &=& x^2(d_2+d_3x),\\
 y_3 &=& x(d_4\cos x),\\
 y_4 &=& x(d_5\sin x).
\end{array}
$$
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. The factor used 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 $y_h$. The atom in
$y_1$ is not a solution of the homogeneous equation, therefore $y_1$ is unaltered.



\end{solution}
\fi

\end{exercises}
\EJECTEND

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\endinput
