00:01
So here we're going to look at an example of a one -gallon mixing problem.
00:07
And the goal is to work out y, which is the pounds or mass of salt in the tank as a function of time.
00:25
So the way you do these types of rate problems is we recognize that that amount is changing in time because of the inflow of raw.
00:35
And the outflow of brine.
00:37
So we want to write the rate of change of why is the rate of stuff coming in minus the rate going out.
00:53
That should work.
00:54
Difference in the rates.
00:56
So if we look at the rate in, we can use what information we're given that we're told there are two pounds of salt dissolved per gallon in the stuff coming in.
01:10
With three gallon per minute flow.
01:13
So we're going to simply multiply those together and we get the rate in is two pounds per minute.
01:21
I'm sorry, six pounds per minute.
01:23
Don't forget the three, six pounds per minute.
01:31
On the outlet side, okay, that's a little bit harder because we do have the volume rate coming in, coming out as well as coming in, of course.
01:46
But we do not have the amount of salt per gallon coming out.
01:52
And so what we need to do is use the y in the tank, which is a function of time, and the volume in the tank as a function of time, is the concentration of salt in the brink coming out, and then we have two gallons per minute.
02:15
And that's essentially what we have done with the rate, in with the variables put in where we don't know the pounds per gallon coming out.
02:29
Okay, but it should simplify down two pounds per minute if we knew those amounts.
02:37
Now, here's the thing is there's no problem with the y in that situation.
02:45
The problem is we have a new variable volume as a function of time.
02:50
And we need to know what that is in order to write this as a single differential equation.
02:56
So what we'll have to do is also write an equation for the rate of change of volume, which is the rate of the volume change in minus the rate of the volume change coming out.
03:15
And those we do know.
03:16
So we know that there's a rate of three gallons per minute coming in.
03:23
Notice i'm keeping my units so we can see what we're quantity we're looking at a rate of change in.
03:31
And so that difference is simply equal to one.
03:35
We could drop the unit.
03:39
And now we can just simply separate and integrate.
03:44
That's always a good thing to do.
03:52
And that produces basically volume equals.
03:57
The initial volume was 100 gallons.
04:00
And then we have a single time.
04:02
Time there.
04:05
So this is what we can substitute for our variable here.
04:12
And now we just have time in there.
04:18
Putting it all together, we finally have a differential equation that we can look at for the rate of change of salt in the tank.
04:31
It is six.
04:33
We'll drop the units.
04:36
And we have 2y divided by 100.
04:40
Plus t, that is not separable.
04:53
And what makes it non -separable is simply the function of time that is sitting in front of the y.
05:08
So there is a technique that you can use to make this equation separable, and that is called the integrating factor approach or technique integrating factor.
05:30
So we'll take a little side trip down the integrating factor lane.
05:35
I'm going to write the y stuff together on one side, though, all the y and t stuff together.
05:45
And what the integrating factor will do for us is to allow us to write everything in the left hand side as a single exact differential, which we'll see it a little bit.
06:00
So the integrating factor, you take e raise the integral of that stuff that's sitting at front of the y.
06:14
So that is 2 over 100 plus tdt.
06:21
And doing that integral, yeah, this is really nice.
06:25
It comes out to be a natural logarithm.
06:30
Perfect.
06:32
We can furthermore, the exponent comes out as a multiplying factor, we'll put it in as an exponent in front of 100 plus 2 square, sorry, that should be 100 plus t, too excited about that too.
06:55
And what's nice about these is that the exponential cancels the logarithmic function.
07:04
They are inverses of each other.
07:07
This is not always true with the integrating factor approach, but for this particular example, it is.
07:17
And so the integrating factor is 100 plus t squared.
07:28
Now what do you do with the integrating factor? you multiply every single term in the equation by that integrating factor...