# SSM-Ch06 - Chap 6 Laplace Transforms Sec 6 1 Laplace...

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Unformatted text preview: Chap . 6 Laplace Transforms Sec . 6 . 1 Laplace Transform . Inverse Transform . Linearity . s- Shifting This section covers several topics. It begins with the definition (1) of the Laplace transform, discusses what linearity means, and derives (Table 6.1, p. 224) a dozen of the simplest transforms that you will need throughout, and will probably memorize after a while. This includes the damped vibrations e at cos u t , e at sin u t ( a u u , which are obtained from cos u t and sin u t by the so-called s-shifting (Theorem 2, p. 224). The last part of the section on existence and uniqueness of transforms is of lesser practical interest. Nevertheless, you should recognize that on the one hand the transform is very general, so that even discontinuous functions have a Laplace transform; this accounts for the superiority of the method over the classical method of solving ODEs, as you will see in the next sections. On the other hand, not every function has a Laplace transform (see Theorem 3, p. 226), but this is of minor practical interest. Problem Set 6 . 1 . Page 226 7 . Laplace transform . Use the addition formula for the cosine (see p. A61). 15 . Use of the defining integral . In Probs. 13-20 use the defining integral of the Laplace transform. In Prob. 15 integrate by parts: u 2 1 2 te U st dt U 1 2 t e U st U s 2 ¡ 1 2 s u 2 e U st dt U U 1 2 ¡ 2 e U 2 s s ¡ 1 2 s ¡ 1 U s U e U 2 s U 1 u U U 1 s U 1 2 s 2 e U 2 s ¡ 1 2 s 2 . 25 . Nonexistence of the transform . For instance, e t 2 has no Laplace transform because the integrand of the defining integral is e t 2 e U st U e t 2 U st and t 2 U st ¢ 0 for t ¢ s , and the integral from 0 to ¢ of an exponential function with a positive exponent does not exist (is infinite). 29 . Inverse transform . In Probs. 29-40 use Table 6.1 (p. 224) and in some problems reduction by partial fractions. Problem 29 leads to cosine and sine: L U 1 4 s U 3 U s 2 ¡ U 2 U L U 1 4 s s 2 ¡ U 2 U 3 U s 2 ¡ U 2 U 4 cos U t U 3 sin U t . 33 . Inverse transform . Divide both numerator and denominator by L 2 : L U 1 n U L L 2 s 2 ¡ n 2 U 2 U L U 1 n U / L s 2 ¡ U n U / L u 2 U sin n U t L . 37 . Inverse transform . Use partial fractions. 41 . First shifting theorem . Use L U 3.8 t u U 3.8/ s 2 , hence L U 3.8 te 2.4 t u U 3.8/ U s U 2.4 u 2 . Chap. 6 Laplace Transforms 53 49 . First shifting theorem . Use L u 1 2 s 3 u t 2 , hence L u 1 8 s U 2 3 u t 2 e u t 2 8 /2 u 2 t 2 e u t 2 . Sec . 6 . 2 Transforms of Derivatives and Integrals . ODEs The main purpose of the Laplace transform is the solution of differential equations, mainly ODEs. You will see the basics of this in the present section. The idea is to transform the ODE (or a given initial value problem) into an equation (“ subsidiary equation ”) that can be solved by algebra, and then to transform the solution of the subsidiary equation back into the solution of the given ODE (or initial value problem). The great advantage of this “transform method” will become particularly clear in the next two sections, where you will see two...
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SSM-Ch06 - Chap 6 Laplace Transforms Sec 6 1 Laplace...

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