 This lesson is on modeling using separable differential equations, growth and decay models. You have learned how to separate differential equations and solve them. Now we are going to do word problems or modeling. Our first example is a basic one. It reads, the population of a small town is growing at a rate proportional to the number of people in the town. If there were 2,000 people in the town in 1990 and 2,800 people in the town in 1998, write an equation which models the growth in population for the town. While there are a couple of things here that you have to really understand, it says the population is growing at a rate proportional to the number of people in the town. So we have some sort of change in Y over change in X but we'll call it change in Q over change in T. And proportional to means equal to some constant times the number of people in the town which we will call Q. The other thing we see here is what we call that initial condition where 2,000 people were in the town in 1990 and then we have another condition that eight years later there were 2,800 people. So let's write all those particulars down and then solve our differential equation. So we have our equation DQ, the change in the population over change in time is proportional to, so that's why the K is there, the number of people of the population. Again, we were given two conditions, Q of 0 equals 2,000 and Q of 8 is equal to 2,800. Let's solve our differential equation, separate the variables, DQ over Q is equal to KDT, ln of the absolute value of Q equals KT plus C, raise them to the E power and you get Q is equal to E to the C times E to the KT. If we substitute in 0 for T, we get E to the C is equal to 2,000 and you have done problems like this before so this is kind of a natural fit. So we get Q is equal to 2,000 E to the KT. Now we use our other set of information, Q of 8 is equal to 2,800 to solve for K. So we have 2,800 is equal to 2,000 E to the 8K, pulling the 2,000 over to the other side and reducing we get 7 fifths equals E to the 8K taking the ln of both sides, natural log we get ln of 7 fifths equals 8K or K is equal to ln of 7 fifths all over 8. So our final equation becomes Q is equal to 2,000 E and K is ln of 7 fifths all over 8 times T. This is our final equation and our final model for this problem. This was quite simple. Let's go on to one that's more complicated, more like a Newton's law of cooling problem. This one reads body weight, more complicated growth type problem. According to a simple mathematical model for an adult to maintain a certain body weight, an athletic person needs 18 calories per day per pound of body weight. If the person consumes more or less, the weight will change at a rate proportional to the difference between the number of calories consumed and the number needed to maintain the current weight. Suppose that I have a constant caloric intake of 1386 calories per day. When I began my study I weighed 115 pounds and one week later I weighed 117 pounds. What is the equation of my weight W over time T measured in days? Now you notice the sentence that says, if a person consumes more or less, the weight will change at a rate proportional to the difference between the number of calories consumed and the number needed to maintain the current weight. And if you notice also there a sentence says, suppose that I have a constant caloric intake of 1386 calories per day. So let's put all that together with those two conditions. The equation for this model is the change in weight over the change in time is proportional to equals K times the difference between the number of calories and those change in calories. So that's your 18 W. And then remember we had an initial condition of W of 0 is equal to 115 and seven days later gained weight to 117. Now before I solve my differential equation I want to take out the negative in front of the W because when you solve a differential equation and you have a negative there you have to use a UDU form. So let's factor out the negative and in doing so we can also factor out the 18. So then this is equal to negative 18 K times W minus 77. Separate our variables we get DW over W minus 77 is equal to negative 18 K DT. Solve and we get ln of W minus 77 is equal to negative 18 K T plus C. Raise these to E and we get W minus 77 is equal to E to the C times E to the negative 18 K T. Now put in our initial condition of 0 and 115 because remember we still have to find out what E to the C is and that constant K. So the first one we're going to find of course is the E to the C we'll put a 0 in for T all of this piece wipes out. So we get E to the C is equal to 115 minus 77 and that is 38. So that's equal to E to the C substituted in we get W minus 77 is equal to 38 E to the negative 18 K T. Go on and solve for our K try to find that value we will substitute in 7 in 117 and we get 117 minus 77 is equal to 38 E to the negative 18 and this time T is 7 times that K. 117 minus 77 is 40 put that over the 38 and that will equal E to the negative 126 K. Take the natural log of both sides and we get ln of 20 over 19 is equal to negative 126 K or K is equal to ln 20 over 19 all over negative 126. Put that into our original formula of W minus 77 is equal to our E to the C which is 38 times E to the negative 18 our K is ln of 20 over 19 over negative 126 multiply that by T and finalize this by doing W is equal to 77 plus 38 E the negative 18 over negative 126 is 1 7. So we have ln of 20 over 19 all over 7 times T. This is our growth rate model for this problem. This concludes our lesson on modeling using separable differential equations growth and decay.