%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\documentclass[12pt]{article}
\usepackage{amssymb}
\topmargin-3cm
\textheight 25cm
\textwidth18.5cm
\oddsidemargin-.7cm
\evensidemargin-.7cm
%\pagestyle{empty}
\parindent0cm
\parskip 2.5ex
\newcommand{\re}{\rm Re}
\newcommand{\im}{\rm Im}
\newcommand{\C}{\Bbb C}
\newcommand{\Ds}{\displaystyle}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{document}
\begin{center}
{\bf Mathematics 226.02: Differential Equations and Linear Algebra }
\\
{\bf Fall Semester 1997}
\\
{\bf Solutions of Exam 1}
\end{center}
\vspace*{1cm}
{\bf 1.} \begin{minipage}[t]{17.5cm} 
Solve the following initial value problem and determine the interval, where the 
solution is defined.
\[
      (x^2-4)\frac{dy}{dx}+2xy=\frac{2x^2-8}{x}\,,\qquad y(1)=2
\]
{\bf Solution:} This is a linear equation, which can be rewritten as
\[
     \frac{dy}{dx}+\frac{2x}{x^2-4}y=\frac{2}{x}.
\]
The integrating factor is 
\[
    \mu={\rm exp}\int\frac{2x}{x^2-4}dx=x^2-4
\]
and therefore the general solution is given by 
\[
   y=\frac{\int(2x-\frac{8}{x})dx +c}{x^2-4}=\frac{x^2-8\,{\rm ln}\,x+c}{x^2-4}.
\]
From the initial condition one calculates $c=-7$ so that the solution of the 
problem is
\[
 y=\frac{x^2-8\,{\rm ln}\,x-7}{x^2-4}\quad\mbox{\rm defined on the interval }
 0<x<2.
\] 
%\mbox{}\hfill{\bf Answer:} \underline{\mbox{\hspace*{8cm}}}
\end{minipage}

\vspace*{2cm}
{\bf 2.} \begin{minipage}[t]{17.5cm} 
Find the value of $a$, for which the equation is exact, and then solve it using
that value of $a$.
\[
      (y^2+x^3+2x)+(axy+y^2+3)\frac{dy}{dx} =0
\] 
{\bf Solution:} The equation is exact for $a=2$.
\\
For this value, the solution of the differential
equation is obtained as follows: define
\[
   \psi=\int(y^2+x^3+2x)dx +h(y)=xy^2+\frac{1}{4}x^4+x^2+h(y).
\]
Then the partial derivative with respect to $y$ of $\psi$ has to be equal to
$2xy+y^2+3$, thus we get
\[
\psi_y=2xy+h'(y)=2xy+y^2+3,
\]
hence $h'(y)=y^2+3$ and we can take $h(y)=\frac{1}{3}y^3+3y$.
The implicit formula for the solution is
\[
    xy^2+\frac{1}{4}x^4+x^2+\frac{1}{3}y^3+3y=c.
\]
\end{minipage}

\newpage
{\bf 3.} \begin{minipage}[t]{17.5cm} 
(a) Solve the differential equation 
   \[   
         x^3 \frac{dy}{dx} + (y-2)^2 =0
   \] 
{\bf Solution:} Separating variables leads to the expression
\[
   \frac{1}{(y-2)^2}dy+\frac{1}{x^3}dx=0.
\]
Integrating this equation yields
\[
   -\frac{1}{y-2}-\frac{1}{2x^2}=c
\]
which can be solved for $y$:
\[
   y=\frac{2x^2}{-2cx^2-1}+2.
\]
The function $y=2$ is a solution, too.
\\[10ex]
(b) Find an implicit expression for the solution of the differential equation
   \[
          x\frac{dy}{dx} = y+2 x e^{\frac{y}{x}}
   \]  
{\bf Solution:} This is a homogeneous equation
\[
     \frac{dy}{dx}=\frac{y}{x}+2e^{\frac{y}{x}}.
\]
The transformation of variables $v=\frac{y}{x}$ leads to the equation
\[
  \frac{1}{x}dx=\frac{1}{2e^v}dv
\]
which after integration yields
\[
   {\rm ln}\,|x|+c=-\frac{1}{2}e^{-v}
\]
and substituting back $\frac{y}{x}$ for $v$ results in
\[
  2\,{\rm ln}\,|x|+C+e^{\frac{-y}{x}}=0.
\]
\end{minipage}

\newpage
{\bf 4.} \begin{minipage}[t]{17.5cm} 
(a) A skydiver weighing 160 lbs. (including equipment) falls vertically down 
from a certain high altitude.
The parachute opens when the skydiver reaches speed equal to 130 ft/sec.
Assume that $t=0$ when the parachute opens and that the air resistance is
$10|v|$ when the parachute is open. 
Assume also that the gravitational force is $32\ \mbox{\rm ft/sec}^2$.
\\[1.3ex]
(i) Write the initial value problem (i.~e. a differential equation and an 
initial condition) for the speed $v(t)$ at any time $t>0$.
{\bf DO NOT SOLVE IT!}
\\[1.3ex]
(ii) Find the limiting velocity $v_L$ of the skydiver after the parachute opens.
\\[2ex]
{\bf Solution:} 
Newton's law says: mass $\times$ acceleration = external force. 
\\
In the present case the external force is given by 
\[
 \mbox{\rm weight of the skydiver minus air resistance}\; = 160 - 10v
\]
(the weight acts in the direction of the motion, the air resistance acts in 
the opposite direction).
\\
The mass $m$ of the skydiver is related to his/her weight via 
\[
  \mbox{\rm weight}\,=160=m\,\times\,\mbox{\rm gravitational force, thus }
  m=160/32=5.
\]
Therefore we get the equation
$5\frac{dv}{dt}= 160-10v$. 
The motion under consideration starts with time $t=0$ when the parachute opens, 
which is when the speed is 130. Hence we get the initial value problem
\[
    \frac{dv}{dt}=32-2v, \qquad v(0)=130.
\]
The limiting velocity is reached when $v'=0$, thus we have to solve the
equation $32-2v=0$, which leads $v_L=16$.
\\[5ex]
(b) Circle the differential equation whose direction field is shown in the
following picture.
\\[2ex]
\begin{tabular}{lll}
A. $y'=y(x+y)$     & B. $y'=(x-y)(x+y)$\hspace*{.6cm} & \fbox{C. $y'=(2-y)(x+y)$}\\[1.5ex]
D. $y'=(y-2)(x+y)$\hspace*{.6cm} & E. $(x-2)(x+y)$    &  
\end{tabular} 
\end{minipage}


\newpage
{\bf 5.} \begin{minipage}[t]{17.5cm} 
(a) Consider a tank holding initially 500 gallons of a salt solution with
concentration 0.3 lb of salt per gallon.
A solution containing 0.4 lb of salt per gallon is pumped into the tank at a
rate of 3 gallons per minute, and the well-stirred mixture flows out of the 
tank at a rate  of 5 gallons per minute.
Write the initial value problem (i.~e. a differential equation and an initial
condition) needed to find the amount $S(t)$ of salt in the tank at time $t>0$
prior to the instant when the tank is empty.
{\bf DO NOT SOLVE IT!}
\\[1ex]
{\bf Solution:} 
Measure the time in minutes.
\\
Each minute $3\times0.4=1.2$ pounds of salt flow into the tank.
\\
Each minute 3 gallons of solution enter the tank whereas 5 gallons flow out.
As a result,  the tank looses 2 gallons per minute.
Hence , at time $t$ there are $500-2t$ gallons of solution in the tank.
\\
The amount of {\em salt per gallon\/} in the tank at time $t$ is therefore 
given by the ratio $\frac{S(t)}{500-2t}$. 
\\
This means that $5\times \frac{S(t)}{500-2t}$ pounds of salt are leaving the tank
at time $t$. 
\\
At the beginning $500\times0.3=150$ pounds of salt are in the tank.
\\
All this together leads to the initial value problem 
\[
   \frac{dS}{dt}= 1.2-\frac{5}{500-2t}S(t),\quad S(0)=150.
\]
\\[5ex]
(b)
Find the constant (equilibrium) solutions of the differential equation
\[
     \frac{dy}{dt}=-0.002\big(1-\frac{y}{3}\big)\big(1-\frac{y}{7}\big)y
\]
and classify each one as asymptotically stable or unstable.
Sketch the graphs of the four solutions $y_1,\,y_2,\,y_3,\,y_4$, which have initial
values $y_1(0)=3,\,y_2(0)=4,\,y_3(0)=7,\,y_4(0)=1$. 
\\[1ex]
{\bf Solution:} 
The solutions $y=0$ and $y=7$ are asymptotically stable, the solution $y=3$ is 
unstable.
\end{minipage}


\end{document}