ODEGuide - A Guide to Solving Simple Ordinary Differential...

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1 A Guide to Solving Simple Ordinary Differential Equations (ODE’s) Brian Frederick Mullins University of Toronto, Department of Physics I. Motivation For many of you, this may be your first encounter with differential equations. The ability to derive and solve differential equations is essential in most scientific and many non- scientific fields. There are many classes of these equations that describe a wide variety of physical phenomena ranging from the vibrating modes on a drumhead to the quantum mechanical description of an electron in a periodic potential. In our discussion, we will concentrate on learning how to solve a very specific and simple differential equation. II. Bridging the Gap At first, the term “ differential equation ” may sound intimidating. However, be assured that differential equations are very similar to equations you have already encountered in your high school math courses such as the quadratic equation. There are two main differences between differential equations and the equations you are already familiar with. The first and most noticeable difference is that differential equations involve taking the derivative of a variable instead of a power . For comparison: Quadratic Equation: ax bx c 2 0 ++ = Differential Equation: a dx dt bx c = 0 The second difference is subtler: In the quadratic case, the variable we are solving for is a number , while in the differential case, we are solving for a function . If you have already taken calculus, you might have guessed this since you can only take the derivative of a function 1 . Let’s look at an example: Suppose we want to solve the following differential equation: dx dt x 2 2 40 += . (1) 1 A constant (number) is considered a function too!
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2 We know this is a differential equation because it involves a derivative (in this case it is the second derivative of x with respect to t ). In order to make this problem easier to read, let’s isolate the derivate term by rewriting equation (1) as: dx dt x 2 2 4 =− (2) Recognize that the variable x is a function of t since we a differentiating with respect to t . Now, by looking at equation (2) we can ask: What function x(t) can be differentiated twice to equal the same function multiplied by constant -4? Let’s randomly guess a solution xt t () = 3 . The second derivative of t 3 is 6 t . Substituting = 3 and dt t 2 2 6 = into our differential equation (1) gives: 64 0 3 tt +≠ (3) Equation (3) shows that our guess t = 3 is not a solution to equation (1) (and hence
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ODEGuide - A Guide to Solving Simple Ordinary Differential...

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