Exponential growth and decay / Applications of ODE, equilibria, and stability

10.4 and 10.5

1) Main points

Chapter 10.4 begins to apply our knowledge of differential equations to exponential functions. We know that the solution to the differential equation: dy/dx = y must be a function whose derivative is the same as the function because the derivative = y. We know that the derivative of e^t = e^t so y = e^t must be a possible solution. But we must also account for the constants involved so the family of solutions is y = Ce^kt. It is also important to remember that in the differential equation y can be multiplied by a constant k so: { dy/dx = ky }. This is k is equal to the k in y = Ce^kt because when you take the derivative of y = Ce^kt the constant C is ignored but the constant k matters.

Example:

This means that a differential equation such as dy/dx = .05y has the general solution of y = Ce^kt and k = .05. To find the particular solution you plug in the given constraints, in this example they are y=50 and t=0. So The particular solution of this differential equation would be 50 = Ce^(.05)(0) which gives us C=50 therefore the solution is y=50e^(.05)(t).

The main idea of this chapter is approaching an exponential growth/decay problem from the top down by using what is given about the rate of change to find the actual function.

Chapter 10.5 works off of 10.4 but instead of starting with the differential equation dy/dx = ky it works from the equation dy/dx = k(y-A). We find that the general solution for dy/dt = k(y-A0 is (y = A + Ce^kt). The first example of the chapter introduces a function in which there is an equilibrium solution meaning the function increases or decrease to approach a certain value that creates an equilibrium. This value is the value of the independent variable that makes the differential function equal to zero. If a small change in the initial conditions of a function makes it veer towards the equilibrium solution then it is called a stable equilibrium solution. If it veers away it is unstable.

2) Challenges

When using the dy/dx (Libnitz) notation it is really important to notice which variable (in this case y or x) is represented in the differential equation. For example dy/dx = 2y is different from dy/dx = 2x. The book's examples often jump to the conclusion that from a differential equation such as dP/dt = .02p, C = P0. This comes from the assumption that when t = 0 That the population will be at the initial amount, this information plugged back into the equation gives C = P0. This is an important thing for me to remember. In general it is hard for me to think of these examples backwards. This chapter is centered around using differential equations to find the functions so it is important to distinguish what the different constants mean.


3) Reflections

These concepts allow us to tackle more complicated real life examples. Equilibriums are seen very often in medicine and blood concentration. I think we will find differential equations really useful in analyzing real world things especially because it is often easier to calculate a rate of change than to find an actual function.