fall 2008 course description
fall 2008 syllabus
Differential Equations is both the course which applies calculus and the motivation for inventing it. You may have already had some experience solving simple differential equations in a calculus course:
|dy/dx = 2x|
|dy/dx = y|
|dy/dx = sin(x2)|
All of these are "simple" because there is an algorithm for solving them, even though the algorithm does not always have the most satisfying conclusion. In the last example, the algorithm is to integrate, but then we get "stuck." An important thing to remember about this field, however, is that most differential equations do not even have an algorithm:
One could say that Isaac Newton, in studying the motions of the planets, invented the field of differential equations by writing down the first one. It is still famous today as an insolvable problem: the "n-body problem." This is a system of differential equations which describes the changing positions of n bodies with mass interacting with each other under the influence of gravity.
A couple of examples may help to give the flavor. The ingredients of a differential equation are
variables - There is at least one each of independent and dependent variables. If there is only one of each, what you have is an ordinary differential equation. If there is more than one independent variable, it's a partial differential equation. And if there is more than one dependent variable, it's a system.
rules - There is one rule for an ordinary differential equation, more for PDE's and systems. The rule is an assumption about how the dependent variables change, stated in terms of derivatives.
Describe the velocity of an object falling near the earth's surface; neglect air resistance.
Variables: Velocity changes as a function of time. Let t represent the independent variable time, and let v represent the dependent variable velocity.
Rule: Without air resistance and near the earth's surface, the acceleration due to gravity is a constant 9.8 meters per second per second directed downward.
Differential Equation: This translates the rule into mathematical notation. Recall from calculus that acceleration is the derivative of velocity with respect to time. Thus,
v'(t) = -9.8
Solution: This is found easily by integrating with respect to t: v(t) = -9.8t + c. Plugging in t = 0, the constant c is found to be v(0), the initial velocity. The complete solution is
|v(t) = -9.8t + v(0)|
Some observations: a differential equation is an equation involving a derivative. The unknown in this equation is a function, and to solve the DE means to find a rule for this function. The general solution is not just one function, but a whole family of functions. The word "family" indicates that all the solutions are related to each other. The particular solution is just one function, and makes use of a known value of the function, for example the initial value.
The next example is more interesting, and shows why the field of differential equations is not just calculus.
Describe the velocity of an object falling near the earth's surface; incorporate air resistance.
Variables: As before, let t represent the independent variable time, and let v represent the dependent variable velocity.
Rule: Newton's Second Law of Motion helps us here. It says that the mass of an object times its acceleration is equal to the sum of the forces acting on it. There are now two forces, gravity and air resistance. The force of gravity is the weight of the object: 9.8 * mass, directed downward. Air resistance can be handled in several ways; we choose the simplest. Assume the force of air resistance is proportional to velocity and in the opposite direction: k * velocity, where k is a negative constant.
Differential Equation: Remember, mass * acceleration = gravity + air resistance.
m*v'(t) = -9.8*m + k*v(t)
In practice, k is determined by the size and shape of the object. (Feathers have more air resistance than rocks, for example.) In shorthand, this equation can be written as
mv' = -9.8m + kv
Solution: A moment's thought will reveal that this is a much different situation than in Example 1. The unknown function v(t) appears on both sides of the equation. Roughly, the equation says that the derivative of v involves the original function. This is an example of a first order linear differential equation, and I don't intend to give away the solution method right here. But I will tell you the solution, and you can check it by plugging it into the original equation. Remember that m, k, and v(0) are constants.
|v(t) = 9.8m/k + (v(0)+9.8m/k)ekt/m|
Another observation. Compare the solutions in the two examples. The first solution is a line with negative slope, indicating that the velocity decreases (larger negative numbers) with time. This is approximately right: objects do gain speed as they fall, and the direction is negative. But it's not exactly right: there is an upper limit to how fast any object can travel in free fall.
The second solution involves an exponential. Good: exponential functions do look like their derivatives. Further, as t approaches infinity, the exponential term approaches zero, and the velocity approaches 9.8m/k, a fixed negative number. This is what we call terminal velocity. In the graphs below, m/k = -1 and v(0) = 0.
|Velocity function without air resistance||Velocity function with air resistance|
fall 2008 course description
fall 2008 syllabus
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