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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 and so e - s ( s - 1)( s 2 + 4) = e - s 5 1 s - 1 - s + 1 s 2 + 4 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 + 4 ; 1 2 sin(2 t ) L -→ 1 4 + s 2 ; cos(2 t ) L -→ s s 2 + 4 ; 1 5 e t - 1 2 sin(2 t ) - cos(2 t ) L -→ 1 5 1 s - 1 - s + 1 s 2 + 4 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)) U 0 ( t - 1) L -→ e - s 5 1 s - 1 - s + 1 s 2 + 4 From this we derive the solution L - 1 e - s 5 1 s - 1 - s + 1 s 2 + 4 = 1 5 e t - 1 - 1 2 sin(2( t - 1)) - cos(2( t - 1)) U 0 ( t - 1) . 41. Taking the Laplace transform of the equations y + 4 y = cos t, y (0) = 0 , y (0) = 0, we obtain s
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