 ###### 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 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.

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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. 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.