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For the following problems, set up and solve the differential equations.A 1000 -liter tank contains pure water and a solution of 0.2 kg saltL is pumped into the tank at a rate of 1 $\mathrm{L} /$ min and is drained at the same rate. Solve for total amount of salt in the tank at time $t$ .
$u(t)=200\left(1-e^{-\frac{t}{1000}}\right)$
Calculus 2 / BC
Chapter 4
Introduction to Differential Equations
Section 5
First-order Linear Equations
Differential Equations
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So we're going to start this problem by setting up a formula for the rate of change of the amount of salt in the tank with respect to time. So we know that this race is going to be given to us by the rate of salt going in minus the rate of salt going out. So that's what you tell us that the S d T will be equal to. We know that one of the solutions in Compton is going to a pump insult at a rate of 0.2 killer grounds for leader times 20 liters per minute. Well, give us kilograms per minute. Plus, our other solution is pumped in iterated 0.5 kilograms for leader times five liters per minute will again give us units of leaders per minute or accuse me kilograms for a minute in minus. Uh, the salt out will be. We know we're pumping it out at a rate of 25 liters per minute. But the concentration of the assault that's being pumped out is going to be equal to the amount of salt in the tank. I'm divided by the volume of tanks and so when we simplify this, Um, we can actually go ahead and substitute in, um, our salt the amount of salt in the tank is equal to the concentration times the volume because that's going to give us units of kilograms. So the rate of change of salt in the tank with respect to time, um, is going to be equal to 0.0 4 to 5, um, minus, uh, 25 times concentration times the volume over the volume so we can see that this is going to cancel. And we know that the rate of change of the concentration with time is going to be equal to the amount of salt divided by the volume. So essentially, we're just going to divide every term here by the volume. So when we do that, we get the CD T is equal to 0.0 for 4 to 5. Should have been not for five to minus 25 times a concentration over the volume. And, um, if we are to calculate the rate of change of volume with respect to time, we're going to find that the volume is actually a constant 1000 leaders. You can calculate this on your own if you want to check me. But I've already calculated it. Um And so essentially, we're just going to plug in 1000 for V. So when we do that, um, we get DC DT is equal to I'm in this step, actually already divided by 1000. I'm sorry if your number was a little bit off there, That would explain why that was my bad. I actually already divided that I'm so that's equal to 0.425 minus, um, 0.25 c so we can go ahead and simple for this to DC DT is going to be equal to a 0.17 minus C over 40. And so I'll go ahead and start a new page here, but we can see that we're going to be able to use our separation of variables here to integrate. Um, And so when we separate our variables, we're going to end up with one over 10.17 minus C. D. C is equal to won over 40 duty. And so when we integrate both sides of our equation here we get the negative natural log of 0.17 minus seat is equal to one over 40 teeth. And so when we solve for C, you will find that C is equal to 0.17 minus a e to the negative one over 40 teeth. And so to solve for our constant A we're going to plug in or initial condition. We know that at times zero see is going to be equal Thio 10 kilograms divided by 1000 liters. So that's going to be, um, 0.1 And so when we plug this in, we get the point. No one is equal to 0.17 minus a E to zero power, and we solved for A and get that A's equal 2.16 So our final equation is our concentration is equal to 0.17 minus 0.16 a e to the negative or he's not a a 0.16 e to the negative one over 40 teeth, Um, and that's given to us in kilograms per liter.
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