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Unformatted text preview: 1 MA360 Handout 1b (week of Jan 9) Sp2006 Kallfelz Assignment: 1,4,6,7,10 (pg5-Sheng) /2.(a),(c),(f),(h) (pg 17) [Read pp1-16, Sheng, and Review techniques and examples in Handouts1a, Handouts 1b] I.) The Laplace Transform – Underlying Motivation(s) Consider an n-th order non-homogenerous differential equations, with constant coefficients, i.e., a differential equation of the form: () () t f t y D n = (Eqn. I.1) where: a.) + + + + = =--- = ∑ 1 1 1 1 ... a dt d a dt d a dt d a dt d a D n n n n n n k k n k k n (i.e., D n is the n-th order differential operator, with constant coefficients: a 0, a 1 ,… a n ) b.) y ( t ) is the unknown function, with initial conditions 1 : ( ) ( ) 1 1 ) 1 ( ) ( , ,..., c y c y c y n n = = ′ =-- c.) f ( t ) is a differentiable function to n-th order You may be used to solving (Eqn I.1) using methods like Undetermined Coefficients (UC) or the Variation of Paramaters (VP), etc. But what you may also recall is the usefulness of such methods have limited applicability—the algebra can get awfully messy in cases when the order n > 2 of (Eqn I.1). As an example, consider the particular case when n = 2, in the case of the following ODE: () ( ) 1 ) ( ) ( 1 2 2 2 = + ′ ′ = + ≡ t y t y t y dt d t y D with initial conditions: y’ (0) = 0, y (0) =1 In other words, the above example is rather simple. Translating in terms of the general notation of (Eqn. I.1) above, the coefficients are: a 0 = 1, a 1 = 0, a 2 = 1. Moreover, the initial constants in this example are: c 0 = 1, c 1 = 0, and f ( t ) = 1 Anyway, you realize using either UC or VP it’s a straightforward to solve. In other words, the degree of n = 2 bounds the (algebraic) complexity of the UC or VP procedures to acceptable limits. But what about something only slightly more complicated? Consider, for example the following 3 rd order equation: 1 The notation y ( k ) (0) is shorthand for the k-th derivative of y ( t ) at t = 0 (for 0 ≤ k ≤ n-1, in the above case of initial conditions.) 2 ( ) ( ) ( ) ( ) t e t y t y t y t y dt d dt d t y D- = + ′- = +- ≡ 2 2 ) ( 3 ) ( 2 3 3 3 3 3 with initial conditions: y (0) = 1, y’ (0) = -2, y” (0) = 1. I.e, the coefficients are: a 0 = 2, a 1 = -3, a 2 = 0, a 3 = 1. Moreover, the initial constants in this example are: c 0 = 1, c 1 = -2, c 2 = 1 and f ( t ) = 2 e- t . It’s only slightly more complicated in form than that of the first example, but trying to solve it using methods like UC or VP proves much messier. Just the homogeneous part of the solution (i.e., the solution of ( ) () 2 ) ( 3 ) ( 3 = + ′- t y t y t y ) involves solving the auxillary equation: r 3- 3 r +2 = 0, which involves some non-trivial algebraic maneuvers....
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This note was uploaded on 11/18/2010 for the course MA 360 taught by Professor Williamm.kallfelz during the Spring '06 term at Capitol College.
- Spring '06