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Unformatted text preview: Page 568, problem2: Since f ( x ) is even we know that B ( p ) = 0 and a ∞ 2 2 k p A ( p ) = ∫ e − kv cos ( pv ) dv = lim − 2 e − kv cos ( pv ) + 2 e − kv sin ( pv ) 2 2 π0 π a →∞ k + p k +p 0 2k = 2 π ( k + p2 ) ∞ 22 2k cos ( px ) e − c p t dp 2 2 0π(k + p ) ⇒ u ( x, t ) = ∫ Page 568, problem 7: 1 A( p ) = 2 2 cos( p) + p ⋅ sin( p ) − 1 , B( p) = 0 ∫ x cos( px)dx = π π0 p2 ∞ 22 2 cos( p ) + p ⋅ sin( p ) − 1 u ( x, t ) = ∫ cos( px) ⋅ e −c p t dp π0 p2 p. 286, problem 10: −2 6 A = 10 − 3 − 10 5 15 A + B = 14 − 14 − 2 9 4 − 4 1 , B = 4 7 0 − 4 0 11 1 2 − 6 2 − 3 − 6 , A − B = 6 − 10 1 4 1 − 6 5 12 5 − 10 − 140 92 3 − 24 − 119 22 ( A + B)( A − B) = 0 94 − 143 − 140 92 3 − 24 − 1...
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