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WrittenHW10-Solutions

# WrittenHW10-Solutions - Math 216(Section 50 1(7.114 Written...

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Math 216 (Section 50) Written Homework #10 – Solutions Fall 2007 1 ( 7.1–14 ) L{ f ( t ) } = L{ t 3 / 2 - e - 10 t } = L{ t 3 / 2 } - L{ e - 10 t } = Γ(5 / 2) s 5 / 2 - 1 s + 10 ( s > 0) . However, Γ(5 / 2) = 3 2 Γ(3 / 2) = 3 4 Γ(1 / 2) = 3 4 π, and so L{ f ( t ) } = 3 π 4 s 5 / 2 - 1 s + 10 ( s > 0) . 2 ( 7.1–16 ) L{ f ( t ) } = L{ sin 2 t + cos 2 t } = L{ sin 2 t } + L{ cos 2 t } = 2 s 2 + 4 + s s 2 + 4 ( s > 0) . 3 ( 7.1–21 ) Let us consider the Laplace transform t e 2 it = t cos 2 t + it sin 2 t : L{ t e 2 it } = L{ t cos 2 t } + i L{ t sin 2 t } , and evaluate the left-hand side. By definition, L{ t e 2 it } := 0 e - st t e 2 it dt = 0 t e (2 i - s ) t dt Integration by parts yields, assuming s > 0, 0 t e (2 i - s ) t dt = t e (2 i - s ) t 2 i - s 0 - 0 e (2 i - s ) t 2 i - s dt = - 0 e (2 i - s ) t 2 i - s dt = - e (2 i - s ) t (2 i - s ) 2 0 = 1 (2 i - s ) 2 = 1 ( s 2 - 4) - 4 s i We can rewrite the result as follows: 1 ( s 2 - 4) - 4 s i = ( s 2 - 4) - 4 s i ( s 2 - 4) 2 + (4 s ) 2 = ( s 2 - 4) - 4 s i s 4 + 8 s 2 + 16 = s 2 - 4 ( s 2 + 4) 2 + i - 4 s ( s 2 + 4) 2 . As a result, we have L{ t cos 2 t } + i L{ t sin 2 t } = ( s 2 - 4) ( s 2 - 4) 2 + (4 s ) 2 + i - 4 s ( s 2 - 4) 2 + (4 s ) 2 1

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and thus taking the real and imaginary parts, L{ t cos 2 t } = ( s 2 - 4) ( s 2 - 4) 2 + (4 s ) 2 , L{ t sin 2 t } = - 4 s ( s 2 - 4) 2 + (4 s ) 2 ( s > 0) , 4 ( 7.2–2
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