Unformatted text preview: k > 0. This leaves the case k =μ 2 , where μ > 0. In this case X ±± + μ 2 X = 0 ⇒ X = c 1 cos μx + c 2 sin μx ; X (0) = 0 ⇒ 0 = c 1 ; X = c 2 sin μx ; X ± ( L ) = 0 ⇒ 0 = c 2 μ cos( μL ) ⇒ c 2 = 0 or cos( μL ) = 0 . To avoid the trivial solution, we set cos( μL ) = 0, which implies that μ = (2 k + 1) π 2 L , k = 0 , 1 , . . . . Plugging this value of k in the equation for T , we Fnd T ± + μ 2 c 2 T = 0 ⇒ T ( t ) = B k eμ 2 c 2 t = B k e( (2 k +1) π 2 L ) 2 c 2 t . ±orming the product solutions and superposing them, we Fnd that u ( x, t ) = ∞ ± k =0 B k eμ 2 c 2 t = ∞ ± k =0 B k e( (2 k +1) π 2 L ) 2 c 2 t sin ² (2 k + 1) π 2 L x ³ . To determine the coeﬃcients B k , we use the initial condition and proceed...
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 Fall '11
 StuartChalk
 Constant of integration, Boundary value problem, c2 sin µx

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