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Chem Differential Eq HW Solutions Fall 2011 137

Chem Differential Eq HW Solutions Fall 2011 137 - Section...

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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 ! = 1 ( s - a ) n +1 , L ( e at cos bt ) = s - a ( s - a ) 2 + b 2 , we find 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 = 1 So F ( s ) = = 1 s - 2 + 1 s + 1 ; f ( t ) = e 2 t + e - t 37. Partial fractions: 1 s 2 + 3 s + 2) = A s + 2 + B s + 1 ; 1 = A ( s + 1) + B ( s + 2) F ( s ) = - 1 s + 2
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