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Unformatted text preview: MA341 Homework 11 Solutions 7.6 5. g ( t ) = , < t < 1; 2 , 1 < t < 2; 1 , 2 < t < 3; 3 , 3 < t . g ( t ) = 2 u ( t 1) u ( t 2) + 2 u ( t 3). The Laplace transform of u ( t a ) is L{ u ( t a ) } ( s ) = e as s . Thus the Laplace transform of g ( t ) is L{ g ( t ) } ( s ) = 2 e s s e 2 s s + 2 e 3 s s = 2 e s e 2 s + 2 e 3 s s . 9. From the graph of g ( t ) we find that the function of g ( t ) can be written as g ( t ) = , < t ≤ 1; t 1 , 1 < t ≤ 2; t + 3 , 2 < t ≤ 3; , 3 < t . Hence, the function g ( t ) in terms of the unit step function is g ( t ) = ( t 1) u ( t 1) + ( 2 t + 4) u ( t 2) + ( t 3) u ( t 3) = ( t 1) u ( t 1) 2( t 2) u ( t 2) + ( t 3) u ( t 3) . Then, by formula (5) of section 7.6, we get: L{ g ( t ) } ( s ) = e s 1 s 2 2 e 2 s 1 s 2 + e 3 s 1 s 2 = e s 2 e 2 s + e 3 s s 2 . 13. L 1 { e 2 s 3 e 4 s s + 2 } = L 1 { e 2 s s + 2 }  3 L 1 { e 4 s s + 2 } To use the translation in t property given by formula (6), we first express e 2 s s +2 as e as F ( s ). For that purpose, we put e as = e 2 s and F ( s ) = 1 s +2 . Thus, a = 2 and f ( t ) = L 1 { 1 s +2 } ( t ) = e 2 t . It now follows from the translation property that L 1 { e 2 s s + 2 } ( t ) = f ( t 2) u ( t 2) = e 2( t 2) u ( t 2) . Similarly, L 1 { e 4 s s + 2 } ( t ) = f ( t 4) u ( t 4) = e 2( t 4) u ( t 4) . 1 MA341 Homework 11 Solutions Consequently, L 1 { e 2 s 3 e 4 s s + 2 } = e 2( t 2) u ( t 2) 3 e 2( t 4) u ( t 4) . 15. To use the translation in t property given by formula (6), we first express se 3 s s 2 +4 s +5 as e as F ( s ). For that purpose, we put e as = e 3 s and F ( s ) = s s 2 +4 s +5 . Thus, a = 3 and f ( t ) = L 1 { s s 2 + 4 s + 5 } ( t ) = L 1 { s + 2 ( s + 2) 2 + 1 } ( t ) 2 L 1 { 1 ( s + 2) 2 + 1...
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 Spring '08
 Schecter
 Differential Equations, Equations, Laplace, L1

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