Chem Differential Eq HW Solutions Fall 2011 147

# Chem Differential Eq HW Solutions Fall 2011 147 - Section...

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Section 8.3 The Laplace Transform Method 147 9. Transforming the problem, we Fnd (see Exercise 7 for similar details) s 2 U ( x, s ) - su ( x, 0) - u t ( x, 0) = U xx ( x, 0); s 2 U ( x, s ) - 1= U xx ( x, 0); U xx ( x, 0) - s 2 U ( x, s )= - 1; U ( x, s A ( s ) e - sx + 1 s 2 ; U (0 ,s L (sin t 1 1+ s 2 ; A ( s )+ 1 s 2 = 1 s 2 A ( s 1 s 2 - 1 s 2 U ( x, s ± 1 s 2 - 1 s 2 ² e - sx + 1 s 2 u ( x, t t - ( t - x ) U 0 ( t - x ) + sin( t - x ) U 0 ( t - x ) . 13. Verify that u ( x, t u 1 ( x, t u 2 ( x, t ) , where u 1 is a solution of u t = u xx ; u (0 ,t )=7 0 ; u ( x, 0) = 70; and u 2 is a solution of
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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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