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The Differential Equation Model for Exponential Growth  Problem 3
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.
The solution to a differential equation dy/dx = ky is y = ce^{kx}. 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) = ce^{kx}. Use the initial condition to find the value of c.
We have a couple new general solutions to differential equations, that we can use in some problems. We have dy/dx equals k times y. We know that this differential equation has the general solution equals Ce to the kx. So these are the exponential functions, this is the exponential differential equation. And in a very similar differential equation; dy/dx equals k times the quantity y minus a, gives me the solution y minus a equals c to the kx. Very similar solutions for very similar differential equations. So I have two examples here and we’re going to use these formulas to solve them.
First solve this differential equation subject to the initial condition; y times zero equals 10. First of all, I need to get this differential equation into one of these two forms. I’m going to move this 0.2y to the right hand side, because I want my dy/dx to be isolated on the left. When I do that, I have to subtract this I get 0.2y. You can see that this differential equation is in the form of the first one. K is 0.2, otherwise, it’s exactly the same. So the solution is going to be y equals Ce to the 0.2x. That’s my general solution.
To get the particular solution that satisfies this initial condition, I have to apply the initial condition to the solution so when x equals zero, y has to equals 10. So I’ll plug 10 in for y, Ce to the 0.2, zero in for x and of course e to the zero, I just 1. 1 times C is 10 so C is 10. That gives me my particular solution y equals 10e to the 0.2x. That’s the solution to this differential equation with this initial condition.
Let’s take a look at another example. Part b; dp/dt equals 0.1p plus 200 with the initial condition p(0) equals 1000. What’s going to make this one tricky is that, we’ve changed the variables here. We don’t have y and x, we have p and t. But I’ve rewritten this rule up here just to remind me that I have a formula. How to solve this differential equation, I can just use this general solution. If I can get this, into the form of this, I can use my formula.
Well the key is factoring out a 0.1 from both of these terms. You factor 0.1 out of 0.1p you’re left with just p. If you factor a 0.1 out of 200, you’re left with 2000, because 2000 times 0.1 is 200. So this is very much in this form. Where 0.1 is k, my variable y is now p. This is the name of my function, P. The independent variable x is now t, so I have to keep that in mind. Instead of having y minus A, I have p plus 200. And that’s okay because if you look at the solution, the solution starts with the y minus A. So here my solution will start with a p plus 2000. So p plus 2000 equals Ce to the k times, this should be t. Ce to the k, and k is 0.1,t.
The only thing that’s left to do is to find out what the value of C is that goes with my initial condition. Let me pull this over here. P(0) equals 1000. So when t is 0, p is 1000, so I’ll –plug 1000 in here, plus 2000 from the left side, equals Ce to the 0.1 times 0. Again e to the 0 is 1 and I get C equals 3000. And so my final solution is P, let me write it in the same form. P plus 2000 equals C which was 3000e to the 0.1t. Of course you can write your final solution as P equals 2000 plus 3000e to the 0.1t. That’s how to use some simple formulas to solve differential equations that are related to exponential growth.
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Norm Prokup
PhD. in Mathematics, University of Rhode Island
B.S. in Mechanical Engineering, Cornell University
He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.
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