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CLEVE MOLER: Hello.

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I'm Cleve Moler,
one of the founders

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and chief mathematician
at the MathWorks.

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This series of videos is about
solving ordinary differential

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equations in MATLAB.

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We can begin by recalling
the definition of derivative.

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The derivative of a
function at a point

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is the slope of the
tangent line to the graph

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of the function at that point.

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Our numerical
approximations will

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rely upon the slope of
the secant to the graph.

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That's a line through two points
separated by a distance h.

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We'll have a lot to say about
the step size h as we go along.

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What's important to realize
is that as h goes to 0,

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the slope of the
secant approaches

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the slope of tangent.

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The wiggly equals sign means
approximately equal to.

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T0 is the point where we are
finding the approximation.

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The value of the
derivative at t0

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is approximately equal to
the slope of the secant.

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The slope of the
secant is the change

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in the y value over the
change in the t value.

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The change in y value
is the difference

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between the two values of y.

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The change in the t
value is the step size h.

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If we rewrite this,
we get the value of y

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at the point t0 plus
h is approximately

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equal to the value of
y at t0 plus h times

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the value of y prime at t0.

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This is the basis for our
first numerical method,

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Euler's method.

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Leonhard Euler was a 18th
century Swiss mathematician.

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Probably the most influential
mathematician of his era.

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He made important
contributions to a wide range

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of fields of mathematics,
physics, and astronomy.

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He invented the notion of
a function, for example.

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The differential equation
is given by this function

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f of two variables t and y.

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And the task in general
is to find the function y

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whose derivative is equal to f.

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Now, there's lots of functions y
whose derivative is equal to f.

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And so there's an initial
condition, a point t

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naught or t0, and a value y0.

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And the initial
condition is that y at t0

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should be equal to y0.

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Here's some examples.

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The compound interest problem is
just the interest rate times y.

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Here, the function of t and y
doesn't actually depend upon t.

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And it's linear in y.

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The initial condition
is at time 0,

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there's a specified
value of y, like $100.

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That's the compound
interest problem.

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Here's the logistic equation.

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Nonlinear equation.

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Here, f of t and y, again,
doesn't depend upon t.

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And it's a constant times y
minus another constant times

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y squared.

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That's the logistic equation.

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And again, the value
is specified at 0.

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Let's say y at 0 is equal to 1.

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Here's another
nonlinear equation.

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F of t and y is t
squared plus y squared.

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It's not possible to
find an analytic solution

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to this equation.

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We'll use these numerical
methods to find a solution

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to this equation.

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Initial condition y
at 0 is equal to 0.

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That's an example of
a function of t and y.

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Euler's method actually isn't
a practical numerical method

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in general.

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We're just using it
to get us started

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thinking about the ideas
underlying numerical methods.

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Euler's method involves
a sequence of points

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t sub n separated by
a fixed step size h.

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And then y sub n is an
approximation to the value

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of the solution at t sub n.

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The approximation comes from
the slope of the secant.

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The ratio of the difference
of the values of y

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and to the step size h.

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The differential equation
says that this ratio

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should be the value of
the function at t sub n.

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And if we rearrange
this equation,

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we get Euler's method.

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That yn plus 1 is yn plus
h times the function f

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evaluated at t
sub n and y sub n.

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This is Euler's method.

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We're now ready for our
first MATLAB program, ODE1.

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It's called ODE1 because
it's our first program

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and because it
evaluates the function f

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that defines the differential
equation once per step.

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There are five input arguments.

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The first is f, a function
that defines the differential

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equation.

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This is something called
an anonymous function.

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I'll talk more about
that in a moment.

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The other four are
scalar numerical values.

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The first three define the
interval of integration.

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We're going from t0 in
steps of h to t final.

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The fifth input argument
is y0, the initial value.

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The output is a vector.

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Vector y out is the
values of the solution

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at the points in the interval.

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We start by putting y0,
the initial value, into y

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and then putting y
into the output vector.

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The body of the
function is a four loop,

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t goes from T0 not in steps of H
up to one step short of t final

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and then the final passage
through the body of the code

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takes t up to t final.

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We evaluate the
function f at t and y.

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That gives us a slope
s, s is for slope.

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Here's the Euler step.

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Take the current value of
y, add h, times the slope.

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That gives us this
new value of y.

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And then y is appended to y out.

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This MATLAB construction
with the square brackets

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takes a vector y, adds
another value to it,

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making it one element longer
and puts the resulting y out

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back in y out.

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This is the entire code.

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This is it.

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This is ODE1 that
implements Euler's method.

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The first argument to any
of the MATLAB ODE solvers

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is the name of a function that
specifies the differential

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equation.

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This is known as
a function handle.

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The easiest way to
get a function handle

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is to make use of an
anonymous function created

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with the ampersand or at sign.

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All of the
differential equations

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involve anonymous functions
of two variables, t and y.

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And so we have f equals
at parenthesis t comma y

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closed parenthesis.

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This is followed by any
expression involving

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either t or y.

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And many of them
don't depend upon t.

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So here is an anonymous function
defining our interest problem.

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And we can just evaluate this
like any ordinary function.

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When y is equal to
1, f of 1 is 0.06.

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Here's an example of a function
that depends upon both t and y.

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The functions can involve
constants that have values.

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So here, we can
define two constants.

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And then we can use
those two constants

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to define the logistic
equation f of a times

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y minus b times y squared.

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Again, this is an autonomous
equation that doesn't actually

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depend upon t.

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Let's see how Euler's
method and ODE1

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work on this simple example.

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Y prime equals 2y with the
initial condition y of 0

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equals 10 on the interval
t between 0 and 3.

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We define the anonymous function
f of t and y is equal to 2y.

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The initial condition
is t0 equals 0.

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We're going to take
a step size of 1.

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Go to t final equals
3 starting at y0

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equals 10 and here's
our call to ODE1.

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We have an animation
that shows these steps.

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Start at t0 equals
0 and y0 equals 10.

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Here's our first point.

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We evaluate the function there.

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We get a slope of 20.

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That's 2 times 10.

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We take an Euler step of
length 1 across the first step.

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That brings us to
the second point, 30.

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Evaluate the function there.

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Two times 30 is 60.

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That's our slope.

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Take the second
step to get to y2.

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Y2 is 90.

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Evaluate the function there.

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Get 2 times 90 is 180.

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That gives us a slope.

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Take a step across the
interval with that slope

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would get us to a third point.

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The third point
is 270 and that's

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the end of the integration.

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So that's three Euler steps
to get from t0 to t final.

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Euler's method is
actually the same

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as computing compound interest.

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So let's do a compound
interest problem.

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Define the interest rate.

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Define our anonymous function
using that interest rate.

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Start at time 0.

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Go in steps of a month.

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Go for 10 years.

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Start with $100.

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And here's our
result of using ODE1

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to compute compound interest.

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That's 121 numbers.

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MATLAB actually has a format for
looking at dollars and cents.

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And so here they are as dollars
and cents starting with $100

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and compounding every month.

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We get up to just over $180.

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I'm going to plot that.

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So I want a time vector months.

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And I actually want to compare
it with simple interest.

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So here's how you
compute simple interest.

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$0.50 a month.

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And now let's plot those two.

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So the straight line is simple
interest getting up to $160.

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And the blue line is
the compound interest.

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There is a slight
upward curvature,

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getting us up to $180.

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There's a dot every
month here as we

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show the results of Euler's
method, which as I said

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is the same thing as
computing compound interest.

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Finally, here's an exercise.

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Find the differential equation
that produces linear growth

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and rerun this example
using ODE1 twice,

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once to compute the
compound interest

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and once to compute
the simple interest.