08Ma2aPracSol7

08Ma2aPracSol7 - Math 2a Practical, Fall 2008: Solutions to...

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Unformatted text preview: Math 2a Practical, Fall 2008: Solutions to Homework 7 You are bound by the honor code to not look at this file if you did not take Math 2a in the Fall of 2008 1. Find the Laplace transform of the following functions f ( t ). Specify for which values the integral will converge (6 pts). (a) f ( t ) = te- t Solution. This is of the form t n e at , which is on the table on page 319. Since n = 1 and a =- 1, the answer is 1 ( s +1) 2 . This converges for s >- 1. (b) f ( t ) = t t 2 t > 2 or t < Solution. In this case, the integral for the Laplace transform has finite support and is just Z 2 te- st dt . If s = 0, this is Z 2 t dt = 2. If s 6 = 0, we integrate by parts with u = t,dv = e- st dt and obtain du = dt,v =- 1 s e- st . We thus have Z 2 te- st dt =- t s e- st 2 + Z 2 1 s e- st dt =- 2 s e- 2 s- 1 s 2 e- ts 2 = 1 s 2- e- 2 s 2 s + 1 s 2 . Hence the integral converges for all s R . 2. Find the inverse Laplace transforms of the following functions (9 pts). (a) 2 s ( s +5) Solution. We break this into partial fractions A s + B s +5 , and thus have As + 5 A + Bs = 2. From this we conclude that A = 2 5 and B =- 2 5 . The function can therefore be written as 2 5 1 s- 1 s +5 . The inverse Laplace transforms of the inner terms are on the table: the first is 1, the second is e- 5 t . We thus have 2 5 ( 1- e- 5 t ) . 1 (b) 2 s- 1 s 2 +2 s +1 Solution. We rewrite this as 2 s +1 ( s +1) 2- 3 1 ( s +1) 2 . The first term reduces to 2 1 s +1 , which gives 2 e- t . The second term gives 3....
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This note was uploaded on 01/08/2011 for the course MA 2a taught by Professor Makarov,n during the Fall '08 term at Caltech.

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08Ma2aPracSol7 - Math 2a Practical, Fall 2008: Solutions to...

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