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Unformatted text preview: Chapter 2 Laplace Transforms 2.1 Preliminary Remarks Laplace transforms provide a very efficient method for finding the solution of linear differential equations, particularly those with constant coefficients. 2.1.1 Linear Differential Equations The following are examples of linear differential equations: 1. dx dt + ax = 0 a const. 2. dx dt + ax = e t 3. d 2 x dt 2 + a dx dt + bx = 0 a, b const. 4. d 2 x dt 2 + 3 dx dt 12 x = e t cos 2 t The first equation can be solved by separation of variables: The equation can be rewritten 1 x dx dt = a and integrating with respect to t gives ln  x  = at + c so  x  = e at + c = e c e at where c is arbitrary. Hence the solution is x = Ae at where  A  = e c . We can determine the constant A if we know the value of x at some particular point, usually the initial value x (0) of x. 2.1.2 Some Useful Limits We will sometimes need to consider the behaviour of functions such as f ( t ) = e t t, as t → ∞ . More generally we consider limits such as lim t →∞ e at t n . The general result that we need is the following. Suppose a > 0 and n is an integer, then lim t →∞ t n e a t = lim t →∞ e a t t n = 0 . 53 54 CHAPTER 2. LAPLACE TRANSFORMS Essentially, this is saying that the exponential e a t function eventually grows more rapidly than any power function t n . For example lim t →∞ e 5 t t 2 = 0 , lim t →∞ e 2 t t 7 = 0 , and so on. 2.1.3 Improper Integrals Integrals such as ∞ a f ( t ) dt where the region of integration is infinite, or 3 1 1 t 2 dt where the integrand is infinite in the interval of integration are called improper integrals. Integrals of the form ∞ a f ( t ) dt are evaluated by defining ∞ a f ( t ) dt = lim R →∞ R a f ( t ) dt Example 1 Evaluate ∞ 1 1 t 2 dt Solution ∞ 1 1 t 2 dt = lim R →∞ R 1 1 t 2 dt = lim R →∞ R 1 t 2 dt = lim R →∞ t 1 R 1 = lim R →∞ R 1 ( (1) 1 ) = lim R →∞ 1 1 R = 1 since lim R →∞ 1 R = 0 . Example 2 Evaluate ∞ e 3 t dt Solution We have 2.1. PRELIMINARY REMARKS 55 ∞ e 3 t dt = lim R →∞ R e 3 t dt = lim R →∞ e 3 t 3 R = lim R →∞ e 3 R 3 e 3 = lim R →∞ 1 3 1 3 e 3 R = 1 3 since lim R →∞ e 3 R = 0 . Example 3 Evaluate ∞ 1 1 + t 2 dt Solution We have ∞ 1 1 + t 2 dt = lim R →∞ R 1 1 + t 2 dt = lim R →∞ arctan( t ) R = lim R →∞ arctan( R ) arctan(0) = lim R →∞ arctan( R ) = π 2 2.1.4 Laplace Transforms If f ( t ) is a function then its Laplace transform is defined to be the function F ( s ) , s > 0, where F ( s ) = ∞ e st f ( t ) dt We write L{ f ( t ) } = F ( s ) to indicate F ( s ) is the Laplace transform of f ( t ) . The transform operation transforms a function f ( t ) of the independent variable t to a function F ( s ) of the variable s . The domain of f ( t ) is often called the time domain and the domain of F ( s ) is called the frequency domain....
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This note was uploaded on 05/27/2008 for the course ENG 121 taught by Professor Smith during the Spring '08 term at The School of the Art Institute of Chicago.
 Spring '08
 smith
 Laplace

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