Unformatted text preview: AMS 361: Applied Calculus IV by Prof Y. Deng
Homework 9
Assignment Date: Collection Date: Grade: Wednesday (11/29/2009) Monday (12/07/2009) Please note this special date Each problem is worth 10 points Problem 9.1: Perform the following inverse transform. (Hint: Please recall my lecture on periodic function) 1 ? 1 Solution: We are given that ft We also know the formula for the Laplace Transform of a periodic function with period p is: ft ft e f t dt e f t dt Let p=2a here, we obtain So we want to find f(t) such that f(t) is a periodic function of period p=2a, and f(t) satisfies ft That is, e f t dt 1 0 2 0 2 e f t dt (*) , then (*) holds. Note that if we define f(t) as f(t)= In fact, f(t)= 1 0 0 and f(t)=f(t+2a) 1 Problem 9.2: Use Laplace Transform and another method to solve the following DE and compare the results 2 0 Solution: 1. Laplace Transform Method: 2 s 2 Because we know that s0 24 0 2 2, ′ 0 24 0 2 24 2s s 0 ′0 0 0 0 24 s 1 0 2, 24 0 0 3 3, then (1) becomes: 2 2 2 2 6 1 4 (2) 1 24 2 3 0 Now use partial fraction, we set Multiply the common detonator (s6)(s+4) to (2) and simply, we obtain the equations: 2 6 4 1 (3) Then apply the inverse Laplace Transform to (3), we get: 2 9 10 2. Another Method: For I.V.P, we have: ceq: ′′ 2 4, So the general solution is in the form of Use the initial conditions 0 2, ′ 0 0 ′0 4 2′ 11 10 24 0 0 24 6 3, we obtain the equations: 2 6 3 In fact, Problem 9.3: Use Laplace Transform and another method to solve the following DE and compare the results 5 0 0 1. Laplace Transform Method: We apply the Laplace Transform to (1): 5
s s 0 s 0 s ′′ 0 0 36 0 0 2, 0 0 13 1 2 3 36
5s s 0
s0 ′0 36 s 0 3 5 36 5 0 5 ′0 ′′ 0 0 0 4 From (2), we know that 0 5 ′′ 0 36 0 , then (4) becomes: 2 2 5 5 23 36 2 4 3 23 3 4 (5) 3 3 4 to (5) and simply, we obtain the 13 0 2 9 Now use partial fraction, we set Multiply the common detonator equations: 23 9 3 12 0 3 2 12 23 (6) Then apply the inverse Laplace Transform to (6), we get: 31 sin 2 26 2. Another Method: For I.V.P, we have: ceq: 5 5 36 36 0 0 5 78 5 78 4 2 2, So the general solution is in the form of 2 2, 3 3, 3 0 3 sin 2 4 cos 2 Use the initial conditions in (2) (3), we obtain the equations: 0 ′0 ′′ 0 0 2 4 8 3 9 27 0 3 9 27 0 2 0 13 In fact, sin 2 Problem 9.4 (Pob. 32, P. 481) Solve the following equation t ′′ 2t 0 1 ′ 0 2 2 0 1 We apply the Laplace Transform to (1): t 2 2 2 2 ′ 2 2
′ t 1′ 2 2 0 0 2 2 (3) 5 ′ 0 0 2 2 2′ 2 Now we integrate both sides of the equation (3): ln Use partial fractions, we get: ln 2 ln ln ln 2 ln 2 ′ Then we apply the inverse Laplace Transform and get: ′ 2 sinh 2 +C ds ds 2 ln 2 ds Problem 9.5 (Pob. 34, P. 481) Solve the following equation t
′′ 0 4t 0 2 ′ 13t 4 0 1 2 We apply the Laplace Transform to (1):
′′ 4t 4 2′ 2 13 4 13 0 4 13 ′ 4 0 0 2 ′ 4 ′ ′ 4 2 2 (3) We know (3) is separable with solution: 6 Since So our solution to the DE is sin kt sin3 t 2 ktcos kt 3t 2 cos3 t 2 Problem 9.6 (Pob. 36, P. 481, modified) Find the solution for convolution theorem to express the solution as the given form. 4 0 1 2 0 2 0 1 2 by using the We apply the Laplace Transform to (1): 4 Set F(s) = 0 ft (3) f t , then (3) becomes: ′0 4 With the initial condition from (2), we get: 4 (4) Now apply the inverse Laplace Transform to (4) and use the convolution theorem: 1 4 1 sin 2t 2 In fact, ft d 7 sin 2 f t ...
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This note was uploaded on 08/08/2010 for the course AMS 361 taught by Professor Staff during the Fall '08 term at SUNY Stony Brook.
 Fall '08
 Staff

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