>> Welcome to the Cypress College Math Review
on mixture applications of differential equations.
This is an application of systems
of differential equations.
Here we have two tanks; each contains 40 gallons
of a salt solution initially.
The first tank initially contains 10 pounds of salt
and the second tank, five pounds of salt.
A salt solution containing four pounds of salt per gallons runs
into the first tank at a rate of two gallons per minute.
The mixture from the first tank runs into the second tank
at a rate of two gallons per minute.
The mixture in the second tank runs out at the same rate.
Find the amount of salt in each tank as a function of time.
So those are our unknowns.
So let's let X of one of T equal the amount of salt in tank one
at time T. X of two of T then is equal to the amount of salt
in tank two at time T. So initially,
X of one of zero is 10 pounds and X of two
of zero is five pounds.
Alright we know that over here
on the left the solution that's going
into tank one is four pounds per gallon, that's the concentration
of the solution that's going into tank one.
We need to know the concentration
of the solution that's going from tank one into tank two.
We know how fast it's running;
it's going at two gallons per minute.
But we don't know the concentration.
We do know that there are 40 gallons in tank one.
It started out at 40 gallons.
We do know that the solution that's running
into tank one is going in at two gallons per minute.
And the solution that's going out of tank one is going
at two gallons per minute.
So it's going to maintain a volume of 40 gallons.
But how much salt is in tank one?
Oh, that's our unknown, that's X of one.
So there is X of one salt in tank one at time T.
So that's the salt and there's 40 gallons so the concentration
in tank one is X of one over 40.
So X of one over 40 is the concentration
in pounds per gallon of the solution that's going
from tank one to tank two.
Similarly over here on the right, the solution that's going
out of tank two it's concentration is X of two
over 40 pounds per gallon.
The concentration of the solution going
out of tank two would be X of two over 40 pounds per gallon.
So the amount of salt in each tank is changing
with respect to time.
That's the rate in minus the rate out.
I think it's a really good idea to write out the units
as you're going through this.
What you got coming
in the concentration is 4 pounds per gallon;
it's coming in at two gallons per minute.
Take a look at the units, the gallons cancel,
and what about the units that are left?
Pounds per minute.
And look at what you have over on the left hand side.
X of one is in pounds, T is in minutes,
everything makes sense, yes?
Derivatives make sense, the Leibniz's notation
for derivatives; the units
for the numerator should match the units
for the denominator should match.
So, pounds per minute, it's perfect.
Over here the concentration of what's going out,
X of one over 40 pounds per gallon,
two gallons per minute, ok.
Second tank, what's going in the concentration is X of one
over 40 pounds per gallon.
It's going in at two gallons per minute.
What's going out?
The concentration is X of two over 40 pounds per gallon
and the rate that it's going out is two gallons per minute.
Now we're going to write that first one without the units.
We used the prime notation, so X of one prime is equal
to eight minus X of one over 20.
And the second one, X of two prime equals X of one
over 20 minus X of two over 20.
Clean those up a little bit and we have 20 X of one prime plus X
of one equals 160 and 20 X of two prime plus X
of two equals X of one.
And that's the system of differential equations
that we have to solve for this problem.
So here's our first differential equation.
For the homogeneous our auxiliary equation looks
like this.
So the solution is -1/20.
So the complimentary solution is C sub one E
to the -1/20T, or E to the -T/20.
And the particular solution
to the nonhomogeneous would simply be a constant we'll call
it A. Using the method of undetermined coefficients,
we have one derivative.
So we take a derivative of that function and we get zero,
substitute it into the differential equation
and we get zero plus A equal 160,
which tells us that A is 160.
So X of one of T is equal to the complimentary solution
which is C sub one, E
to the -T/20 plus the particular solution to the nonhomogeneous,
which is simply the number 160
which is A. Now we find what C sub one is.
We know that X of one of zero is equal to 10.
Because we know there are ten pounds of salt
in the tank at time zero.
Substituting that in we have 10 equal C sub one times one plus
160, therefore C sub one is equal to -150.
So X of one of T is equal to -150 E
to the power of -T/20 plus 160.
There's the first solution.
Now we look at our second differential equation.
20 X of two prime, plus X of two equals X of one.
We'll go ahead and put X of one in there.
And let's solve the homogeneous first.
Just like above, the homogeneous gives us R is equal to -1/20.
So X of two C is C sub two E to the -1/20 times T,
or I'll write it as -T/20.
Now let's find the particular solution to the nonhomogeneous.
We notice immediately that we have a repetition
between the homogeneous and the nonhomogeneous,
so we have to multiply by T. And then we have a constant there.
Particular solution to the nonhomogeneous,
we take the derivative, we have a product rule.
So we have the derivative of BT is B times this
and then the derivative of E
to the -T/20 would be -1/20 times BT.
And then the derivative of C is simply zero.
So we have our derivative of our particular solution,
now we substitute it into our differential equation.
So we have 20 times our derivative.
[ Silence ]
Plus the function, and that's equal to -150 E
to the -T/20 plus 160.
Distribute the 20, so those cancel.
[ Silence ]
These two terms cancel
out by equating coefficients 20B is equal to -150,
therefore B is equal to -15/2, and C is equal to 160.
Recall that X of two P is equal to BT, E to the -T/20 plus C.
So we get -15/2T, E to the -T/20 plus 160.
Which gives us that X of two is equal to C sub two,
E to the -T/20 minus 15/2T, E to the -T/20 plus 160.
Now we find C sub two.
Recall that X of two of zero, the initial amount of salt
in the tank at time zero was five pounds.
So five is equal to C sub two times one minus zero plus 160.
So C sub two is equal to -155.
So X of two of T is equal to -155 times E to the power
of -T/20 minus 15/2T, E to -T/20 plus 160.
Recall again that X of one of T was -150 E
to the -T/20 plus 160.
And there's your solution to the system of equations.
[ Silence ]
So we have our solutions, S sub one of T,
X sub two of T. These functions give the amount of salt
in each tank at time T. So we're done.
So we solved a system of differential equations
that we came up with and we have our solution.
And we're happy with that, right?
I remember back in an algebra one class,
when I solved an equation my teacher had me check
the solution.
And then in algebra two class, when I learned to solve a system
of equations, they had me check the solution.
I wonder, shouldn't we check our solution here?
I think we should.
Let's check it.
So X of one prime, we'll need the derivatives of course.
So that would be what?
15/2 E to the -T/20, and similarly X
of two prime would be 155/20 E to the -T/20,
minus since I have a product rule going on,
I'll pull out the 15/2.
So then I have E to the -T/20 minus 1/20T, E to the -T/20,
and then we'll distribute so we have 15/40 T, E to the -T/20.
So there's our second derivative.
Now the differential equations that we had
to substitute them into.
We have 20 X of one prime plus X of one equals 160,
was our first differential equation.
So 20 times the derivative
of the first function plus the first function should equal 160.
So we get 150 E to the -T/20 minus 150 E to the -T/20,
plus 160 equals 160, that works.
Now we go to the other equation,
the other differential equation was 20 X of two prime.
Plus X of two should equal X of one.
So we have 20 times 155/20, E to the -T/20, minus 15/2 E
to the -T/20, plus 15/40 T, E to the -T/20,
and then plus X of two.
And I'll put that below there, so we have -155 E
to the -T/20 minus 15/2 T, E to the -T/20, plus 160
and that's supposed to equal X of one, X of one was -150 E
to the -T/20 plus 160.
You can see when you distribute the 20
that you'll get 155 E to the -T/20.
So this term will cancel with this one.
When you distribute the 20 to this term you'll get 15/2
and here you'll have -15/2 of that term.
So on the left hand side you'll be left with 20 times -15/2.
So you'll have -150 E to the -T/20 plus 160 Equals -150 E
to the -T/20 plus 160 and we just checked our solution.
Not a bad idea.