Okay, so this is the second mixing problem I'm going to do. We still have two tanks, tank A and
B, and we still have a constant volume of solution of salt water in each tank. This time, the rate
of flow from tank A to B and B to A is going to be different. To account for a constant volume still,
we're going to have a spigot that pours pure water into tank A to maintain a constant volume
and we're also going to have a drain on tank B that exits salt water solution. That's kind of
my mediocre drawing of it. We have from B to A is 4 liters per minute. Drawing the picture can
probably help you figure out what's going in and what's going out. And the second one goes nine.
Let me draw arrows to show the directions. Again, it's the same idea. The first thing we're going
to do is write out the differential equations. Let's just call this X1 as well and X2. X1 prime,
you want to think of it as the rate in minus the rate out. And the idea of rate times concentration
is the same because we're looking for the amount of salt present in the tanks. The rate in of salt
is X2 over 360 because the volume this time is 360 liters. That's the concentration. And the rate in
is 4. The rate out is going to be 9. Nine times X1 over 360. That's the first one. You notice we
have 5 liters per minute of pure water going into tank A but we're not going to include that in our
differential equation because we're not worried about how much pure water is in the tank. We're
looking for an equation to measure salt but we will have salt in the solution leaving tank B, so
we do include that in our differential equation of X2 or tank B. Let's do that now. X prime 2 again
is rate in minus rate out. That's going to be: the rate going in is X1 over 360 times 9 minus...The
rate going out here is 4 times X2 over 360. Initially, this is just pure water but eventually
salt solution will come in and will leave again. And then we also have to consider--this is what
makes it different from the last problem--we also have to consider the rate going out from
the drain. That's going to be--solution drains at a rate of 5. That makes sense to keep a constant
volume. It's going to be 5 over X2 over 360. All right. Not too different. You can simplify
this to be 9 over 360 times X1 minus 9 over 360 times X2. And now we're ready to solve and find
the eigenvalues and eigenvectors. Again, we're just going to write out the constant coefficient
matrix. That's going to be -9 over 360, because we're looking at X1 first, plus 4 over 360. Then
we're going to do the second equation. That's 9 over 360 minus 9 over 360. In this case, we don't
have a symmetric matrix like last time. Times X1, X2 equals to X prime of T. We're going to write
out the characteristic equation again. That's lambda squared minus T lambda plus D equals 0. T
is going to be -18 over 360. That simplifies as -1 over 20. D is going to be this diagonal multiplied
minus this diagonal. And that's going to be 1 over 2880. Our characteristic equation is lambda
squared plus 1 over 20 plus 1 over 2880. Using the quadratic equation, we're going to get lambda
equal to -1 over 20 plus or minus square root of 1 over 20 squared minus 4 times 1 times 1 over
2880 all over 2. This will give us the two values of lambda. That simplifies to -1 over 20 plus or
minus 1 over 30 over 2. The two values are going to be -1 over 24 and 1 over 20. Now we have lambda
1 is equal to -1 over 24 and lambda 2. Now we can plug in the values to find the eigenvector. That's
going to be -9 over 360. Just the A matrix again. Minus the identity matrix times lambda times V1,
V2. If you simplify this out, that gives you 1 over 60, 1 over 90, 1 over 40, and 1 over 60,
V1, V2 equal to 0. That's going to give you 1 over 60 V1 plus 1 over 90 V2 is equal to 0.
For lambda equal to -1 over 24, we're going to get the vector 1 over -32. For lambda 2, we're
going to repeat the same process, just plugging in 1 over 20. And I'm going to skip showing that,
but it's basically the same thing and you're going to get 1 over 1 and 3 halves. Okay, so I
just realized I cannot read my own handwriting. This obviously is wrong, it's supposed to be
-1 over 120, so it's not 1 over 20. Okay I did this problem on a sheet of paper and I couldn't
read my own handwriting. If you do lambda 2 is equal to -1 over 120 you get 1 and 3 halves as
the vector. Okay, that was annoying. Anyway, now you can write the solution. The general solution,
and this is an IVP problem, so we're going to eventually find the coefficients. But we have
C1 e to the--let's do this one first. Negative 1 20, 1, 3/2, plus C2 e the -1 over 24.
And that's 1, -3/2. And we have that at X of 0, I think it was 60 and 0. Let me just check. That
was the last line. Yep, 60 tank A and zero and tank B. This is our IVP problem. Again,
we plug in 0. That's going to give us C1, 3/2 C1 plus C2, -3/2 C2. And X of 0 is 60 and 0. That's
going to give us 60 is equal to C1 plus C2 and 0 is equal to 3/2 C1 minus 3/2 C2. Again we have
that C1 is equal to C2. That means C1 is equal to C2 is equal to 30. Our final solution is X
of T is equal to 30 e to the -1 over 120, 1, 3/2 plus 30 e to the -1 over 24, 1, -3/2. And if you
can follow along with these two problems, then you should be able to do these problems pretty
easily. I think that the application of this is just being able to write the correct differential
equation and then you already know all the techniques to solving differential equations,
so that should not be the hard part. It's just making sure you understand that, if it asks you
for salt, then it does matter how much water is pouring in and you just want to think about the
concentrations and the rates of flow. It's not that difficult but yeah, those were two examples.