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Unformatted text preview: Differential Equations The subject of ordinary differential equations encompasses such a large field that you can make a profession of it. There are however a small number of techniques in the subject that you have to know. These are the ones that come up so often in physical systems that you not need both the skills to use them and the intuition about what they will do. That small group of methods is what I’ll concentrate on in this chapter. 4.1 Linear Constant-Coefficient A differential equation such as d 2 x dt 2 3 + t 2 x 4 + 1 = 0 relating acceleration to position and time, is not one that I’m especially eager to solve, and one of the things that makes it difficult is that it is non-linear. This means that starting with two solutions x 1 ( t ) and x 2 ( t ) , the sum x 1 + x 2 is not a solution; look at all the cross-terms you get if you try to plug the sum into the equation and have to cube the sum of the second derivatives. Also if you multiply x 1 ( t ) itself by 2 you no longer have a solution. An equation such as e t d 3 x dt 3 + t 2 dx dt- x = 0 may be a mess to solve, but if you have two solutions, x 1 ( t ) and x 2 ( t ) then the sum αx 1 + βx 2 is also a solution. Proof? Plug in: e t d 3 ( αx 1 + βx 2 ) dt 3 + t 2 d ( αx 1 + βx 2 ) dt- ( αx 1 + βx 2 ) = α e t d 3 x 1 dt 3 + t 2 dx 1 dt- x 1 + β e t d 3 x 2 dt 3 + t 2 dx 2 dt- x 2 = 0 This is called a linear, homogeneous equation because of this property. A similar-looking equation, e t d 3 x dt 3 + t 2 dx dt- x = t does not have this property, though it’s close. It is called a linear, inhomogeneous equation. If x 1 ( t ) and x 2 ( t ) are solutions to this, then if I try their sum as a solution I get 2 t = t , and that’s no solution, but it misses working only because of the single term on the right, and that will make it not too far removed from the preceding case. One of the most common sorts of differential equations that you see is an especially simple one to solve. That’s part of the reason it’s so common. This is the linear, constant-coefficient, differential equation. If you have a mass tied to the end of a spring and the other end of the spring is fixed, the force applied to the mass by the spring is to a good approximation proportional to the distance that the mass has moved from its equilibrium position. If the coordinate x is measured from the mass’s equilibrium position, the equation ~ F = m~a says x m d 2 x dt 2 =- kx (4 . 1) James Nearing, University of Miami 1 4—Differential Equations 2 If there’s friction (and there’s always friction), the force has another term. Now how do you describe friction mathematically? The common model for dry friction is that the magnitude of the force is independent of the magnitude of the mass’s velocity and opposite to the direction of the velocity. If you try to write that down in a compact mathematical form you get something like ~ F friction =- μ k F N ~v | ~v | (4 . 2) This is hard to work with. It can be done, but I’m going to do something different. (See problemThis is hard to work with....
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This note was uploaded on 01/30/2012 for the course PHYS 315 taught by Professor Nearing during the Fall '08 term at University of Miami.
- Fall '08