 This lesson is on modeling, using separable differential equations, other types of models besides what you did in the last lesson. So let's just go straight to our different types of problems. The first one is an acceleration velocity position type problem. It reads, the acceleration on the moon, A, is negative 18 feet per second squared and at the time t equals zero, the velocity of a ball thrown up into the air is 45 feet per second. And the height at which it is thrown is 6 feet. What is the equation which models the ball's distance from the ground? Now if you've had a good precalculus background, you could figure this out right away and we will talk about that after we finish our total explanation through differential equations. So let's solve this. So we know that the acceleration is equal to negative 18 feet per second squared and we know that the antiderivative of acceleration is velocity. So the velocity equals the integral of the acceleration and of course that's the integral of the negative 18 feet per second. We take these antiderivatives and we get v is equal to negative 18t plus some constant which we will call c1. We have some initial condition here. It says when t is equal to zero, the velocity is 45. So when t is equal to zero, v is 45. So we'll put the 45 in and that will equal c1. The velocity now equals negative 18t plus 45. They ask us for the equation which models the ball's distance from the ground so we know the equation for distance is the integral or antiderivative of velocity. So let's take this antiderivative now and we will get s which is the distance from the ground. So s is now equal to negative 18t squared over 2 so that becomes negative 9t squared plus 45t plus a c sub 2, a new constant. We also know that the ball when it is thrown is 6 feet from the ground so we know s of zero equals 6. So if we put in a zero for t, we get 6 is equal to c sub 2. So our final equation is s is equal to negative 9t squared plus 45t plus 6. Now let's go back and think about what happened in pre-calculus and how this equation read in pre-calculus. It really read something like this, a general form for the equation. s is equal to negative one half gt squared plus v sub zero initial velocity times time plus the initial height from the ground. And if you had memorized this formula you did not need to go through all the differential equations for this. And using it for our problem it would be negative one half times 18t squared plus the initial velocity which is 45 times the t plus the initial height which is 6 feet. And again we get the same equation that we had before. So either way with this type of problem if you can go directly to the final equation, fine. If you need to develop it, that's fine too. Let's go to another type of problem. This one reads, a fish wildlife study shows that the growth of salmon population per month is directly proportional to the square root of the number of salmon. If the original number of salmon is 1500 and two months later there are 2,535, write the equation which models this growth. Well we see that this one is a square root type of differential equation. So let's try to solve that one. So if we use ds dt, the change in population is equal to when we do proportionalities, we're going to put the k in but this time it's proportional to the square root of the number present which we'll call s. We also know that s of 0 equals 1500 and we also know that s of 2 equals 2535. So let's go on and solve the differential equation and substitute in the values that we know. So ds over the square root of s, separate the variables, is equal to k dt. If we integrate both sides we get 2 square roots of s is equal to kt plus c. We can substitute our 0 in for t and find out what c is. So now c is equal to 2 times the square root of 1500 and we can take the square root of the 1500, take out 100 of that so we'll have 20 times the square root of 15 equals c. So this gives us now the equation 2 square roots of s is equal to kt plus 20 square roots of 15. Now we can substitute the 2 in so we can determine our k. So we have s is equal to 2535 and that is equal to k times 2 plus 20 square roots of 15. If we simplify our 2535 it really is a 13 squared times 15. So we can take the 13 squared out and multiply it to the 2 so we get 26 square roots of 15. We can subtract the 20 square roots of 15, divide all that by 2 to get our k. So ultimately we get k is equal to 3 square roots of 15. Put that back into our equation and we get 2 square roots of s is equal to k which is 3 square roots of 15. t plus our c which is 20 square roots of 15. If we want to go on and square this and divide by 2 we will get an answer of s is equal to 135 over 4 t squared plus 450. Again this is slightly different from the ones you have been doing because we have the square root of x as our original change in s over change in t. Let's go to another type of problem. This is a rate in rate out problem. It reads, a pool contains 35 gallons of non-chlorinated water. Water containing a 1% chlorine solution is poured into the pool at a rate of 3 gallons per minute. The chlorine instantly mixes with the non-chlorinated water and the excess is drained out at 3 gallons per minute. Write an equation for the amount of chlorine in the pool at any time t. So we have a lot of things going on here. We have water coming into the pool and then the water with the chlorine in it being drained out of the pool. So let's set up our differential equation for this. We know that there's some sort of change in the amount of chlorine and salt that enters the pool. So that's the sdt. And we have the rate at which it enters the pool which is the concentration times the flow rate. So that's this. And then we have it coming out of the pool and again that's concentration of salt. So we have s pounds of salt out of the 35 gallons that we have in there and then times the 3 gallons per minute. So this becomes our equation. dsdt equals 3 times 0.01 minus 3s over 35. Before we separate variables, let's factor out a negative 3. Remember we never want our variable to be negative on the right hand side. So we'll take out a negative 3 and we'll have s over 35 minus 0.01. Separate our variables. We'll have ds over s over 35 minus 0.01 equals negative 3 dt. So now we integrate and we get 35 times ln of the s over 35 minus 0.01 is equal to negative 3t plus c. We divide both sides by 35, that side, each term on the right hand side. And we get ln of the absolute value of s over 35 minus 0.01 is equal to negative 3t over 35 plus c over 35. Raise e to each power and we get s over 35 minus 0.01 is equal to e to the c over 35 times e to the negative 3t over 35. Substituting our initial condition of s of 0 is equal to 35. We have a 35 here for the s and a 0 here for the t. This becomes 1 and this becomes 1. So both sides, 35 divided by 35 is a 1 and e to the 0 is a 1. So we'll have 1 minus 0.01 and that's 0.99 is equal to e to the c over 35. Putting that back into our equation, we have s over 35 minus 0.01 is equal to 0.99 e to the negative 3t over 35. Multiplying by 35 and putting the 0.01 on the other side, we get s is equal to 0.35 plus 34.65 e to the 3t over 35 with a negative of course in front of the 3t. This is our equation for the flow rate for this particular problem. This concludes our lesson on other types of differential equations.