Chem Differential Eq HW Solutions Fall 2011 137

Chem Differential - Section 8.1 The Laplace Transform 137 29 Using L eat tn L = n(s a)n 1 eat tn n = 1(s a)n 1 = sa(s a)2 b2 L eat cos bt we nd

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Section 8.1 The Laplace Transform 137 29. Using L ( e at t n ) = n ! ( s - a ) n +1 , L ± e at t n n ! ² = 1 ( s - a ) n +1 , L ( e at cos bt ) = s - a ( s - a ) 2 + b 2 , we fnd that f ( t )= e 3 t t 4 4! + e 3 t cos t. 33. Partial Fractions: 2 s - 1 s 2 - s - 2 = 2 s - 1 ( s - 2)( s +1) = A s - 2 + B s +1 ; 2 s - 1= A ( s +1)+ B ( s - 2) Take particular values oF s : s = - 1 ⇒- 3= - 3 B B =1 s =2 3=3 A A So F ( s )== 1 s - 2 + 1 s ; f ( t e 2 t + e - t 37. Partial Fractions: 1 s 2 +3 s +2) = A s +2 + B s ; A ( s B ( s F ( s - 1 s + 1 s ; f ( t - e - 2 t + e - t 41. The change oF variables
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This note was uploaded on 12/22/2011 for the course MAP 3305 taught by Professor Stuartchalk during the Fall '11 term at UNF.

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