\documentclass{article}
\input{../macros.tex}

\begin{document}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{Lecture 1}

\subsection*{ Welcome to 18.03SC}

Here's a list of some MIT courses that list 18.03 Differential Equations as a prerequisite
or co-requisite.

2.001    Mechanics and Materials I

2.003    Dynamics and Vibrations

2.005    Thermal-Fluids Engineering

2.016    Hydrodynamics

3.23     Electrical, Optical, and Magnetic Properties of Materials

6.002    Circuits and Electronics

6.021    Quantitative Physiology: Cells and Tissues

8.04     Quantum Physics I

10.301   Fluid Mechanics

12.005   Applications of Continuum Mechanics to Earth, Atmospheric, and
Planetary Sciences

16.01    Unified Engineering

18.100   Analysis I

18.330   Introduction to Numerical Analysis

20.309  Biological Engineering II

22.05   Neutron Science and Reactor Physics

Altogether 140 courses at MIT list 18.03 as a prerequisite or a co-requisite.


\subsection*{What and why}

 A {\em differential equation} is a relation between a function and its
derivatives.

Differential equations form the language in which the basic laws of
science are expressed. The science tells us how the system at hand
changes "from one instant to the next." The challenge addressed by the
theory of differential equations is to take this short-term
information and obtain information about long-term overall behavior.
So the art and practice of differential equations involves the following
sequence of steps: one "models" a system (physical, chemical, biological,
economic, or even mathematical) by means of a differential equation;
one then attempts to gain information about solutions of this equation;
and one then translates this mathematical information back into the
scientific context.

\pagebreak

\begin{verbatim}
                              Solve
Differential Equation:   _______________\    Behavior over time
Short term information                  /

            /\
             \                                   /
              \                                 /
     Model     \                               /  Interpret
                \                             /
                 \                           /
                                            \/

                       Physical World
\end{verbatim}


A basic example is given by Newton's law,  $F = ma$.  Here $a =$ acceleration,
the second derivative of  $x =$ position. Forces don't affect  $x$  directly,
but only through its derivatives.  This is a second order ODE, and we
will study second order ODEs extensively later in the course.

\subsection*{Separable equations}
 In this first Unit we will study ODEs involving only the first derivative;
first order:  $y' = F(x,y)$ .

\bigskip

{\bf Example 1:}  $y' = 2x$.  Solution by integrating:  $y = x^2 + c$.

Notice that there are many solutions.

An expression like this, involving a constant ($c$ here) is called the
{\em general solution}. The constant is a "constant of integration."

\bigskip

{\bf Example 2:}  $y' = ky$.  General Solution:  $y = Ce^{kx}$.  

MEMORIZE THIS -- it's the central example in this course.

(In fact, a good  definition of the exponential function  $e^x$  is that it is
the solution of the differential equation  $y' = y$  such that  $y(0) = 1$ .)

\bigskip

{\bf Question 1:}  What is the general solution to the ODE  $dy/dx = 2y+1$ ?

\begin{enumerate}
\item $y = Ce^{2x} - 1$

\item $y = Ce^{x/2} - 2$

\item $x = y^2 + y + c$

\item $y = e^{x/2} + C$

\item $y = Ce^{2x} - 1$

\item \label{q1ans6} $y = Ce^{2x} - 1/2$

\item $y = e^{2x} + c$

\item  None of the above

\item Don't know
\end{enumerate}

The method: ``separation of variables.'' 

Recall the method: Put all the $x$'s on one side and $y$'s on the other
(if possible):
$$dy/(2y+1) = dx$$

Integrate both sides:
$$(1/2) ln|2y+1| + c1 = x + c2$$

Amalgamate the constants and (if possible) solve for y in terms of x:
$$ln|2y+1| = 2x + c  ,   |2y+1| = e^c e^{2x}  ,  2y+1 = C e^{2x}$$
$$y = C e^{2x} - 1/2$$

So the answer is: \ref{q1ans6}.

We can check this!:  $ y' = 2 C e^{2x} = 2(y + 1/2) = 2y+1$.

It's a nice feature of differential equations in general: it's easy to
check your answer!

\bigskip

{\bf Question 2:} Is  $y' + xy = x$  separable?

\begin{itemize}
\item Yes

\item No

\item don't know.
\end{itemize}

Well,  $y' = x - xy = x(1-y)$  so  $dy/(1-y) = x dx$  :  YES.

We could go on to solve this, but you can do that on your own.

We will see many other methods of solving various types of equations.
Unfortunately, most real life equations aren't explicitly solvable,
and often you don't actually care as much about the explicit solution
as about the general properties. That's one place where today's topic
is helpful.


\subsection*{Graphical approach}

The ODE  $y' = F(x,y)$  specifies a derivative -- that is, a slope -- at every
point in the plane. This is a {\em direction field} or {\em slope field}.

\bigskip

{\bf Example:}  $y' = 2x$. 

\begin{center}FIGURE NEEDED\end{center}


 Notice that the slope
$F(x,y)$  does not depend on $y$ here: It is invariant under vertical
translation.

A {\em solution} of the differential equation is a function whose graph has the
given slope at every point it goes through. I drew some. The graphs of
solutions are {\em integral curves}. They are vertically nested parabolas.
The translation invariance of the direction field is reflected in the
fact that a vertical translate of a solution is another solution.

To specify a particular solution of an ODE you have to give an {\em initial condition}: when  $x$  takes on a certain value,  $y$  takes on a specified
value.


{\bf Example:} $y' = y$.

\begin{center}FIGURE NEEDED\end{center}

Notice that the slope
$F(x,y)$  does not depend on $x$ here: it is invariant under horizontal
translation.

Graphs of solutions are now horizontal translates of each other.

{\bf Example:} $y' = y^2 - x$.

This equation does not admit solutions in elementary functions.
Nevertheless we can say interesting things about its solutions.

To draw the direction field, find where  $F(x,y)$  is constant, say  $F(x,y) = m$.
This is an {\em isocline}. For example, Figure ??? shows that $x = y^2$ is the isocline for $m=0$.

\begin{center}FIGURE NEEDED\end{center}

When $m = 1$ the isocline is $x = y^2 - 1$.

When $m = -1$ the isocline is $x = y^2 + 1$ .

The Mathlet Isoclines illustrates more examples.

We have seen in action the
{\em existence and uniqueness theorem for ODE's}:
$y' = F(x,y)$  has exactly one solution such that  $y(a) = b$ ,
for any  $(a,b)$  in the region where  $F$  is defined.
(The solution  $y(x)$  may only exist for  $x$  near to  $a$ .)

(You also have to put some technical conditions on $F$.)

The E. and U. theorem says that there is just one integral curve through
each point:\\
EVERY POINT LIES ON JUST ONE INTEGRAL CURVE: NO CROSSING ALLOWED.\\
The applet makes it look like many solutions coalesce, but this is just
a pixel problem. In reality they are separate, but very close to each other.


Many seem to bunch up along the bottom
branch of the parabola. Can we explain this?

I cleared the solutions and drew in just the isoclines $m = -1$ and $m = 0$.
Once a solution gets between these two parabolas, it can never escape.
The poor thing can't cross the  $m = -1$ parabola because it would have
to have slope greater than $-1$ when it does; and it can't cross the  $m = 0$
parabola because it would have to have slope less than zero when it does.
This is a {\em funnel}.

So solutions in that region stay in that region: they are trapped
between those two parabolas, which are asymptotic as  $x \rightarrow \infty$.
All these solutions become very close to the function  $- sqrt(x)$
for large  $x$ .  This is an ideal situation!  We know, approximately,
but with increasing accuracy, about the long term behavior of these
solutions, and the answer doesn't depend on initial conditions (as long
as you are in this range). This is "stability." (Of course if the solution
doesn't get trapped, it's a different story.)


Direction fields let you visualize the qualitative behavior of solutions
to differential equations, and this is often what you want to know.
But we also want to be able to solve ODEs "analytically," that is,
using formulas.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\end{document}
