 ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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# The Differential Equation Model for Exponential Growth - Problem 1

Norm Prokup ###### Norm Prokup

Cornell University
PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

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The solution to a differential equation dy/dx = ky is y = cekx. This can be used to solve problems involving rates of exponential growth.

The population of a group of animals is given by a function of time, p(t). So, the rate of growth of the population is p'(t). If the rate of growth is proportional to the population, p'(t) = kp(t), where k is a constant. Now, this is of the form dy/dx = ky, so this differential equation can be solved to find that p(t) = cekx. Use the initial condition to find the value of c.

I want to solve the differential equation problem that involves exponential growth. Recall that the differential equation dy/dx equals ky, implies y equals Ce to the kx. These would be the general solutions of this differential equation. Let’s get started.

Problem says 150 Tribbles are let loose on a space station. Ten days later there are 3000 Tribbles. If the rate of growth of their population is proportional to the population itself, find a formula for the population p(t).

I just want to explain here that y is proportional to x means that y equals a constant times x. So that’s what that phrase means. Again, if the rate of growth of their population is proportional to the population itself, find the population.

Let’s just observe that the population is p(t). The rate of growth of their population would be the derivative of p(t) with respect to time. That’s p'(t). And so for the rate of growth, to be proportional to the population, means I have this differential equation; p'(t) equals k times p(t).

Now this is exactly of the form of the differential equation that I have to the left here, if we take a look. Dy/dx equals k times y. Y is the function, dy/dx is the rate of change of that function. Let’s go back here. That means that the general solution of this differential equation is p(t) equals a constant times e to the kt. So this is my general solution to this problem.

I need to find some of these values, in order to get a very specific population function. First of all I’m told at t equals zero there are 150 tribbles. That initial condition would be p(0) equals 150. I plug 150 in for the population, zero in for time here, and I get 150 equals Ce to the k times zero. Of course e to the zero is just 1. So this means that C equals 150, and my population function is p(t) equals 150e to the kt.

I also like to know what k is. And I can use the other condition that they give me; the fact that after 10 days there are 3000 tribbles. P(10) is 3000. I take my same population equation, plug 3000 in for population and 10 in for t. So I get 3000 equals 150e to the k times 10. I could divide both sides by 150, 3000 divide by 150 is 20. So I have that 20 equals e to the 10 to the k. to solve this equation I need to take a log of both sides; the natural log and I’ll get natural log 20 equals 10k.

That means that k is 1/10 of the natural log of 20. I’m going to need a calculator for this. 0.1 times the natural log of 20. I get 0.29957, I’m going to round that to approximately 0.3. K is approximately 0.3. My final population function from here is p(t) equals 150 e to the 0.3t. This is my population function for the tribbles.