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Exponential models & differential equations (Part 2)

Given the general solution P=Ceᵏᵗ and the conditions P(0)=100 and P(50)=200, we find the solution to an exponential modeling problem.

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Video transcript

- [Voiceover] In the last video, we established that if we say the rate of change of a population with respect to time is going to be proportional to the population, we were able to solve that differential equation, find a general solution, which involves an exponential. That the population is going to be equal to some constant times e to some other constant, times time. And in the last video we assumed time was days. So let's just apply this, just to feel good that we can truly model population this way. So let's use it with some concrete numbers. Once again, you've probably done that before. You've probably started with the assumption that you can model with an exponential function, and then you used some information, some conditions, to figure out what the constants are. You probably did this earlier in pre-calculus or algebra class. Let's just do it again, just so that we can feel that this thing right over here is useful. So let's give you some information. Let's say, that at time equals zero, the population is equal to 100 insects or whatever we're measuring the population of. And let's say that at time equals 50, the population, so after 50 days, the population is at 200. So notice it doubled after 50 days. So given this information, can we solve for C and k? I encourage you to pause the video and try to work through it on your own. So this first initial condition's pretty straighforward to you, because when t is equal to zero, P is 100. So we could say, based on this first piece of information, we could say that 100 must be equal to C times e to the k times zero. Well that's just going to be e to the zero. Well e to the zero is just one. So this is just the same thing as C times one, and just like that we have figured out what C is. We can now write that the population is going to be equal to 100e to the kt. So you can see, expressed this way, our C is always going to be our initial population. So, e to the kt. And now, we can use this second piece of information. So our population is 200, let's write that down. So our population is 200, when time is equal to 50. After 50 days. So 200 is equal to 100e to the k times 50, right? t is now 50. Let me just write that. k times 50. Now we can divide both sides by 100, and we will get, two is equal to e to the 50k. Then we can take the natural log of both sides. Natural log on the left hand side, we get the natural log of two. And on the right hand side, the natural log of e to the 50k, well that's just going to be, that's the power that you need to raise e to, to get e to the 50k. Well that's just going to be 50k. All you did was took the natural log of both sides. Notice, that this equation that I've just written expresses the same thing. Natural log of two is equal to 50k, that means e to the 50k is equal to two, which is exactly what we had written there. And now we can solve for k. Divide both sides by 50, and we are left with, k is equal to the natural log of 2, over 50. And we're done. We can now write the particular solution that meets these conditions. So we can now write that our population, and I can even write our population as a function of time, is going to be equal to 100 times e to the, now k is natural log of two over 50, so I'll write that. So, natural log of two over 50, and then that times t. If you assume rate of change of population is going to be proportional to population, you assume this information right over here, this function is what is going to describe your growth of population.