WEBVTT

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>> Welcome to the Cypress
College Math Review

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on mixture applications
of differential equations.

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This is an application
of systems

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of differential equations.

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Here we have two tanks;
each contains 40 gallons

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of a salt solution initially.

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The first tank initially
contains 10 pounds of salt

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and the second tank,
five pounds of salt.

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A salt solution containing four
pounds of salt per gallons runs

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into the first tank at a rate
of two gallons per minute.

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The mixture from the first
tank runs into the second tank

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at a rate of two
gallons per minute.

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The mixture in the second tank
runs out at the same rate.

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Find the amount of salt in each
tank as a function of time.

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So those are our unknowns.

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So let's let X of one of T equal
the amount of salt in tank one

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at time T. X of two of T then
is equal to the amount of salt

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in tank two at time
T. So initially,

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X of one of zero is
10 pounds and X of two

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of zero is five pounds.

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Alright we know that over here

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on the left the solution
that's going

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into tank one is four pounds per
gallon, that's the concentration

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of the solution that's
going into tank one.

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We need to know the
concentration

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of the solution that's going
from tank one into tank two.

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We know how fast it's running;

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it's going at two
gallons per minute.

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But we don't know
the concentration.

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We do know that there are
40 gallons in tank one.

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It started out at 40 gallons.

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We do know that the
solution that's running

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into tank one is going in
at two gallons per minute.

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And the solution that's going
out of tank one is going

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at two gallons per minute.

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So it's going to maintain
a volume of 40 gallons.

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But how much salt
is in tank one?

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Oh, that's our unknown,
that's X of one.

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So there is X of one salt
in tank one at time T.

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So that's the salt and there's
40 gallons so the concentration

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in tank one is X of one over 40.

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So X of one over 40
is the concentration

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in pounds per gallon of
the solution that's going

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from tank one to tank two.

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Similarly over here on the
right, the solution that's going

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out of tank two it's
concentration is X of two

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over 40 pounds per gallon.

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The concentration of
the solution going

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out of tank two would be X of
two over 40 pounds per gallon.

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So the amount of salt
in each tank is changing

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with respect to time.

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That's the rate in
minus the rate out.

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I think it's a really good
idea to write out the units

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as you're going through this.

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What you got coming

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in the concentration
is 4 pounds per gallon;

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it's coming in at two
gallons per minute.

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Take a look at the
units, the gallons cancel,

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and what about the
units that are left?

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Pounds per minute.

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And look at what you have
over on the left hand side.

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X of one is in pounds,
T is in minutes,

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everything makes sense, yes?

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Derivatives make sense,
the Leibniz's notation

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for derivatives; the units

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for the numerator
should match the units

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for the denominator
should match.

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So, pounds per minute,
it's perfect.

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Over here the concentration
of what's going out,

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X of one over 40
pounds per gallon,

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two gallons per minute, ok.

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Second tank, what's going in
the concentration is X of one

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over 40 pounds per gallon.

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It's going in at two
gallons per minute.

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What's going out?

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The concentration is X of
two over 40 pounds per gallon

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and the rate that it's going
out is two gallons per minute.

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Now we're going to write that
first one without the units.

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We used the prime notation,
so X of one prime is equal

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to eight minus X of one over 20.

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And the second one, X of
two prime equals X of one

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over 20 minus X of two over 20.

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Clean those up a little bit and
we have 20 X of one prime plus X

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of one equals 160 and
20 X of two prime plus X

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of two equals X of one.

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And that's the system of
differential equations

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that we have to solve
for this problem.

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So here's our first
differential equation.

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For the homogeneous our
auxiliary equation looks

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like this.

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So the solution is -1/20.

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So the complimentary
solution is C sub one E

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to the -1/20T, or
E to the -T/20.

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And the particular solution

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to the nonhomogeneous would
simply be a constant we'll call

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it A. Using the method of
undetermined coefficients,

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we have one derivative.

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So we take a derivative of
that function and we get zero,

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substitute it into the
differential equation

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and we get zero plus
A equal 160,

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which tells us that A is 160.

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So X of one of T is equal to
the complimentary solution

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which is C sub one, E

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to the -T/20 plus the particular
solution to the nonhomogeneous,

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which is simply the number 160

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which is A. Now we
find what C sub one is.

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We know that X of one
of zero is equal to 10.

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Because we know there
are ten pounds of salt

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in the tank at time zero.

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Substituting that in we have 10
equal C sub one times one plus

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160, therefore C sub
one is equal to -150.

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So X of one of T
is equal to -150 E

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to the power of -T/20 plus 160.

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There's the first solution.

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Now we look at our second
differential equation.

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20 X of two prime, plus
X of two equals X of one.

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We'll go ahead and
put X of one in there.

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And let's solve the
homogeneous first.

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Just like above, the homogeneous
gives us R is equal to -1/20.

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So X of two C is C sub two
E to the -1/20 times T,

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or I'll write it as -T/20.

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Now let's find the particular
solution to the nonhomogeneous.

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We notice immediately
that we have a repetition

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between the homogeneous
and the nonhomogeneous,

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so we have to multiply by T. And
then we have a constant there.

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Particular solution
to the nonhomogeneous,

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we take the derivative,
we have a product rule.

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So we have the derivative
of BT is B times this

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and then the derivative of E

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to the -T/20 would
be -1/20 times BT.

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And then the derivative
of C is simply zero.

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So we have our derivative
of our particular solution,

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now we substitute it into
our differential equation.

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So we have 20 times
our derivative.

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[ Silence ]

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Plus the function, and
that's equal to -150 E

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to the -T/20 plus 160.

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Distribute the 20,
so those cancel.

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[ Silence ]

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These two terms cancel

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out by equating coefficients
20B is equal to -150,

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therefore B is equal to
-15/2, and C is equal to 160.

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Recall that X of two P is equal
to BT, E to the -T/20 plus C.

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So we get -15/2T, E
to the -T/20 plus 160.

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Which gives us that X of
two is equal to C sub two,

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E to the -T/20 minus 15/2T,
E to the -T/20 plus 160.

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Now we find C sub two.

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Recall that X of two of zero,
the initial amount of salt

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in the tank at time
zero was five pounds.

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So five is equal to C sub two
times one minus zero plus 160.

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So C sub two is equal to -155.

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So X of two of T is equal
to -155 times E to the power

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of -T/20 minus 15/2T,
E to -T/20 plus 160.

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Recall again that X
of one of T was -150 E

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to the -T/20 plus 160.

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And there's your solution
to the system of equations.

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[ Silence ]

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So we have our solutions,
S sub one of T,

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X sub two of T. These functions
give the amount of salt

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in each tank at time
T. So we're done.

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So we solved a system of
differential equations

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that we came up with and
we have our solution.

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And we're happy with
that, right?

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I remember back in
an algebra one class,

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when I solved an equation
my teacher had me check

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the solution.

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And then in algebra two class,
when I learned to solve a system

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of equations, they had
me check the solution.

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I wonder, shouldn't we
check our solution here?

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I think we should.

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Let's check it.

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So X of one prime, we'll need
the derivatives of course.

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So that would be what?

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15/2 E to the -T/20,
and similarly X

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of two prime would be
155/20 E to the -T/20,

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minus since I have a
product rule going on,

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I'll pull out the 15/2.

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So then I have E to the -T/20
minus 1/20T, E to the -T/20,

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and then we'll distribute so we
have 15/40 T, E to the -T/20.

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So there's our second
derivative.

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Now the differential
equations that we had

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to substitute them into.

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We have 20 X of one prime
plus X of one equals 160,

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was our first differential
equation.

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So 20 times the derivative

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of the first function plus the
first function should equal 160.

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So we get 150 E to the -T/20
minus 150 E to the -T/20,

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plus 160 equals 160, that works.

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Now we go to the other equation,

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the other differential
equation was 20 X of two prime.

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Plus X of two should
equal X of one.

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So we have 20 times 155/20,
E to the -T/20, minus 15/2 E

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to the -T/20, plus
15/40 T, E to the -T/20,

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and then plus X of two.

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And I'll put that below
there, so we have -155 E

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to the -T/20 minus 15/2 T,
E to the -T/20, plus 160

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and that's supposed to equal
X of one, X of one was -150 E

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to the -T/20 plus 160.

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You can see when you
distribute the 20

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that you'll get 155
E to the -T/20.

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So this term will
cancel with this one.

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When you distribute the 20
to this term you'll get 15/2

00:21:25.306 --> 00:21:27.966 A:middle
and here you'll have
-15/2 of that term.

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So on the left hand side you'll
be left with 20 times -15/2.

00:21:35.246 --> 00:21:48.766 A:middle
So you'll have -150 E to the
-T/20 plus 160 Equals -150 E

00:21:48.766 --> 00:21:54.816 A:middle
to the -T/20 plus 160 and we
just checked our solution.

00:21:57.666 --> 00:21:58.746 A:middle
Not a bad idea.