e st t u(t 2) dt = lim t dt = T 2 2 e st = T e st lim + e st
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1 Math 46, Profeor David Levermore Anwer to Quetion for Dicuion Friday, 7 October 7 Firt Set of Quetion ( Ue the definition of the Laplace tranform to compute Lf]( for the function f(t = u(t t, where u i the unit tep function. ( What i the exponential order of f(t = u(t t a t? (3 How might we have gueed that Lf]( i defined only for >? (4 If f(t = u(t 5t 4 e t in(3t then for what value of will Lf]( be defined? Anwer. The definition of the Laplace tranform give Lf]( = lim T e t t u(t dt = lim T For we have e t for every t >. Then for every T > t e t dt t dt = T, t e t dt. which clearly diverge to + a T. Hence, Lf]( i undefined for. For > an integration by part how that t e t dt = t e t T e t + dt ( = t e t e t T ( = T e T e T + ( e + e. Hence, for > we have ( Lf]( = lim T e T T = e + e lim T e T + ( e + e (T e T + e T = e ] + e. Anwer. The function f(t = t u(t i piecewie continuou over,. Becaue the exponential order of a product i the um of the exponential order while t i of exponential order a t, u(t i of exponential order a t, we ee that f(t = t u(t i of exponential order + = a t. Alternative Anwer. The function f(t = t u(t i piecewie continuou over,. For every σ > we have lim t e σt f(t = lim e σt t u(t = lim e σt t =. t t Therefore f(t = t u(t i of exponential order a t.
2 Anwer 3. Becaue f(t = t u(t i piecewie continuou over, and i of exponential order a t, it Laplace tranform Lf]( will be defined for every >. Anwer 4. The function f(t = u(t 5t 4 e t in(3t i piecewie continuou over,. Becaue the exponential order of a product i the um of the exponential order while u(t 5 i of exponential order a t, t 4 i of exponential order a t, e t i of exponential order a t, in(3t i of exponential order a t, we ee that f(t = u(t 5t 4 e t in(3t i of exponential order + + ( + = a t. Becaue f(t = u(t 5t 4 e t in(3t i piecewie continuou over, and i of exponential order a t, it Laplace tranform Lf]( will be defined for every >. Conider f(t defined by ( Sketch f(t over t. ( Expre f(t in the form Second Set of Quetion t for t <, 4 for t < 4, f(t = 8 t for 4 t < 8, for 8 t <. f(t = j (t + u(t j (t + u(t 4j (t 4 + u(t 8j 3 (t 8. (3 Compute F ( = Lf](. Anwer. The graph wa hown in dicuion. All the group did it! Anwer. We can expre f(t in term unit tep function a Therefore where f(t = ( u(t t + ( u(t u(t ( u(t 4 u(t 8 (8 t = t + u(t (4 t + u(t 4 ( (8 t 4 + u(t 8 ( (8 t = t + u(t (4 t + u(t 4 (4 t + u(t 8 (t 8. f(t = j (t + u(t j (t + u(t 4j (t 4 + u(t 8j 3 (t 8, j (t = t, j (t = 4 (t + 4 = t, j (t = 4 (t + = t 4t, j 3 (t = (t = t.
3 3 Anwer 3. By applying item 6 in the table we ee that F ( = Lf]( = Lj ]( + e Lj ]( + e 4 Lj ]( + e 8 Lj 3 ](, where item with a = and n = and with a = and n = yield Lj ]( = Lt ]( =, Lj ]( = Lt]( = 3, Lj ]( = Lt + 4t]( = 4 3, Lj 3]( = Lt]( =. Therefore F ( = Lf]( = ( 3 e + 4 e e 8 = ( e 3 + (e 8 e 4 4e. Third Set of Quetion Conider the initial-value problem x + 3x + 3x + x = f(t, x( =, x ( = 3, x ( = 5, where f(t i from the econd et of quetion. ( Solve for X( = Lx](. ( Expand X( into a um of form found in the table below. (3 Take the invere tranform to find x(t = L X](t. Anwer. The Laplace tranform of the initial-value problem i Lx ]( + 3Lx ]( + 3Lx ]( + Lx]( = F (, where F ( wa found in Anwer 3 of the econd et of quetion and Lx]( = X(, Lx ]( = Lx]( x( = X(, Lx ]( = Lx ]( x ( = X( 3, Lx ]( = Lx ]( x ( = 3 X( 3 5. The Laplace tranform of the initial-value problem then become ( 3 X( ( X( ( X( + X( = F (, which become Therefore X( i given by X( = ( X( ( = F (. ( ( e + 3 (e 8 e 4 4e.
4 4 Anwer. We firt oberve that = ( + 3. Therefore in order to anwer the quetion we need partial fraction identitie for ( + 3, ( + 3, 3 ( + 3. Thee have the form = A ( + 3 ( + + A 3 ( + + A +, ( + 3 = B 3 ( + + B 3 ( + + B + + C + C 3 + C, ( + 3 = D ( + + D 3 ( + + D + + E + E. The value of A, A, A, B, B, B, C, C, C, D, D, D, E, and E can be determined by many method. Once they are found the anwer will be X( = A ( + + A 3 ( + + A + ( + ( e B ( + + B 3 ( + + B + + C + C 3 + C ( + (e 8 e 4 4e D ( + + D 3 ( + + D + + E + E All of thee form appear in the table. We will how one way to find the value of A, A, A, B, B, B, C, C, C, D, D, D, E, and E after the table.. Anwer 3. If we define J ( = ( + 3, J ( = then X( ha the form ( + 3 3, J ( = X( = J ( + ( e J ( + (e 8 e 4 4e J ( ( + 3, = J ( + J ( e ( J ( + 4J ( e 4 J ( + e 8 J (. The lat entry in the table then how that ( x(t = L X](t = j (t + j (t u(t ( j (t + 4j (t u(t 4j (t 4 + u(t 8j (t 8,
5 5 where j (t = L J ](t = L (t ( + 3 = L A ( + + A 3 ( + + A ] (t, + j (t = L J ](t = L (t ( = L B ( + + B 3 ( + + B + + C + C 3 + C ] (t, j (t = L J ](t = L (t ( + 3 = L D ( + + D 3 ( + + D + + E + E ] (t. The firt entry in the table with a = and n =,, and repectively how that L ](t = t e t, L (t = t e t, L (t = e t. ( + 3 ( + + The firt entry in the table with a = and n =,, and repectively how that L (t = t, L (t = t, L (t =. 3 Therefore j (t = A t e t + A t e t + A e t, j (t = B t e t + B t e t + B e t + C t + C t + C, j (t = D t e t + D t e t + D e t + E t + E. By placing thee expreion into equation ( we will obtain x(t. A Short Table of Laplace Tranform Lt n e at n! ]( = ( a n+ for > a. Le at a co(bt]( = ( a + b for > a. Le at in(bt]( = b ( a + b for > a. Lt n j(t]( = ( n J (n ( where J( = Lj(t](. Le at j(t]( = J( a where J( = Lj(t](. Lu(t cj(t c]( = e c J( where J( = Lj(t]( and u i the unit tep function.
6 6 We now how one way to determine the value of A, A, A, B, B, B, C, C, C, D, D, D, E, and E. We will ued the reidual method decribed in the write up on partial fraction in the Calculu Review note that are poted on the cla webite. Thi method i eay to apply becaue there are no irreducible quadratic factor in the denominator of the rational function that we are conidering. It i the quicket route to the anwer. Firt, conider the partial fraction identity ( + 3 = A ( A ( + + A +. We ee A =, A = 4, and A = by expanding about = to get Therefore the partial fraction identity i = + 4( + + ( ( + 3 = ( ( Next, conider the partial fraction identity ( + 3 = B 3 ( + + B 3 ( + + B + + C + C 3 + C. We ee B =, B = 6, and B = by Taylor expanding / 3 about = to get = 3 ( 3 = 6( + ( + +. ( + We ee C =, C = 6, and C = by Taylor expanding /( + 3 about = to get ( + = Therefore the partial fraction identity i ( + 3 = 3 ( ( Finally, conider the partial fraction identity ( + 3 = D ( + + D 3 ( + + D + + E + E. We ee D =, D =, and D = 3 by Taylor expanding / about = to get = ( = + ( + + 3( + +. ( + We ee E = and E = 3 by Taylor expanding /( + 3 about = to get ( + = Therefore the partial fraction identity i ( + 3 = ( (
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4e st dt. 0 e st dt. lim. f (t)e st dt. f (t) e st dt + 0. f (t) e. e (2 s)t dt + 0. e (2 s)4 1 ] = 1 = 1. te st dt + = t s e st
![4e st dt. 0 e st dt. lim. f (t)e st dt. f (t) e st dt + 0. f (t) e. e (2 s)t dt + 0. e (2 s)4 1 ] = 1 = 1. te st dt + = t s e st](/thumbs/92/108365672.jpg "4e st dt. 0 e st dt. lim. f (t)e st dt. f (t) e st dt + 0. f (t) e. e (2 s)t dt + 0. e (2 s)4 1 ] = 1 = 1. te st dt + = t s e st")
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