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The Differential Equation Model for Exponential Growth  Problem 2
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. Recall that various methods of integration, such as substitution, can be used when solving differential equations.
I want to solve a differential equation that’s related to exponential growth. Recall our resolve about exponential growth; dy/ dx equals k times y. This is the exponential growth differential equation, implies y equals Ce to the kx. This is the exponential growth function. And if k is negative, these will both be exponential to k. Use change of variables to solve this differential equation which is very similar. Dy/dx equals k times the quantity y minus A. It’s not exactly the same as this but a change of variables will make it very much the same. Let’s change to the variable u.
Let u equals y minus A. This is just renaming y minus, u. If I do that, and by the way A is a constant in this problem, then the derivative with respect to xdu/dx will just be equal to dy/dx. The derivative of the constant will be zero. I can make a replacement in this equation, this equation becomes, on the right side, k times y minus A which is just u, and in the left side, dy/dx is the same as du/dx. This is just the same as this differential equation, only the variable’s name has changed to u. The solution of this is going to be u equals C times e to the kx.
Now recall that u was just y minus A. this means that y minus A is equals to Ce to the kx. So this would be the general solution of this differential equation. Let’s use this fact in part b of this problem. Use the result from part a to solve the differential equation dy/dx equals 0.5y minus 50.
Let me write the result back down again here. Dy/dx, the differential equation, in part a, was k times y minus a. And we found that the solution was y minus a equals Ce to the kx. All I have to do is write this differential equation in this form and I can use this rule to solve it. So dy/dx equals, I can factor out 0.5 out of this. 0.5 times y minus, 0.5 times what would equals 50? That would have to be 100. 0.5 times 100 equals 50. That tells me that k is 0.5 and A is 100, but otherwise this is in exactly the same form as my differential equation from part a. That tells me that the solution is y minus 100 equals Ce to the kx, and k is 0.5. C e to the 0.5x.
In other words, y equals 100 plus Ce to the 0.5x. There’s still a parameter. This is still a general solution to this differential equation. For every different value of c, you’ll get a different solution including for example C equals zero. Y equals 100 is also a solution to this differential equation. But for every real number C, this function is a solution to my differential equation.
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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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