Chem Differential Eq HW Solutions Fall 2011 138

Chem Differential Eq HW Solutions Fall 2011 138 - 138...

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138 Chapter 8 The Laplace and Hankel Transforms with Applications Laplace transformt the equation and use the given initial conditions: L ( y ±± )+ L ( y )= L ( - cos τ ) s 2 Y - sy (0) - y ± (0) + Y = - s s 2 +1 Y ( s 2 +1) = = - s s 2 Y = - s ( s 2 +1) 2 = 1 2 d ds 1 s 2 y = - 1 2 τ sin τ = - 1 2 ( t - π ) sin( t - π ) = 1 2 ( t - π ) sin t 45. Laplace transform the equation y ±± - y ± - 6 y = e t cos t and use the initial conditions y (0) = 0 ,y ± (0) = 1: s 2 Y - sy (0) - y ± (0) - sY + y (0) - 6 Y = s - 1 ( s - 1) 2 ; Y ( s 2 - s - 6) = 1 + s - 1 ( s - 1) 2 ; So Y = 1 ( s - 3)( s +2) + 1 ( s - 3)( s s - 1 ( s - 1) 2 ; Y = - 1 5( s + 1 5( s - 3) + 1 ( s - 3)( s s - 1 s 2 - 2 s +2 = - 1 5( s + 1 5( s - 3) + 1 ( s + 2)( s 2 - 2 s + 2 ( s - 3)( s + 2)( s 2 - 2 s We now Fnd the partial frctions decomposition of the last term on the right. Write
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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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