Chem Differential Eq HW Solutions Fall 2011 143

# Chem Differential Eq HW Solutions Fall 2011 143 - Section...

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Section 8.2 Further Properties of The Laplace Transform 143 Hence 1 ( s - 1)( s 2 +4) = 1 5 ± 1 s - 1 - s +1 s 2 +4 ² and so e - s ( s - 1)( s 2 = e - s 5 ± 1 s - 1 - s s 2 ² You can use the Table of Laplace transforms at the end of the book to verify the following computations: e t L -→ 1 s - 1 sin(2 t ) L 2 s 2 ; 1 2 sin(2 t ) L 1 4+ s 2 ; cos(2 t ) L s s 2 ; 1 5 ± e t - 1 2 sin(2 t ) - cos(2 t ) ² L 1 5 ± 1 s - 1 - s s 2 ² U 0 ( t - a ) f ( t - a ) L e - as F ( s ); 1 5 ± e t - 1 - 1 2 sin(2( t - 1)) - cos(2( t - 1)) ² U 0 ( t - 1) L e - s 5 ± 1 s - 1 - s s 2 ² From this we derive the solution L - 1 ³ e - s 5 ± 1 s - 1 - s s 2 ²´ = 1 5 ± e t - 1 - 1 2 sin(2( t - 1)) - cos(2( t - 1)) ² U 0 ( t - 1) . 41. Taking the Laplace transform of the equations y ±± y = cos t, y (0) = 0 ,y ± (0) = 0, we obtain s 2 Y Y = s s 2 Y = 1 2 s s 2 · 2 s 2 +2 2 . So y = 1 2 cos t * sin(2 t ) . 45. T = 2, for 0 <t< 1 ,f ( t )= t and for 1 2, f ( t )=2 - t ; so, by the
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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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