Chem Differential Eq HW Solutions Fall 2011 45

Chem Differential Eq HW Solutions Fall 2011 45 - Section...

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Section 3.6 Heat Conduction in Bars: Varying the Boundary Conditions 45 Also, ± 1 / 2 0 sin μ n xdx = 1 μ n ( 1 - cos μ n 2 ) . Applying (11), we fnd c n = ± 1 / 2 0 100 sin μ n ² ± 1 0 sin 2 μ n = 100 μ n ( 1 - cos μ n 2 ) ² 1 2 ( 1 + cos 2 μ n ) = 200 ( 1 - cos μ n 2 ) μ n (1 + cos 2 μ n ) . Thus the solution is u ( x, t )= ³ n =1 200 ( 1 - cos μ n 2 ) μ n (1 + cos 2 μ n ) e - μ 2 n t sin μ n x. 17. Part (a) is straightForward as in Example 2. We omit the details that lead to the separated equations: T ± - kT =0 , X ±± - kX ,X ± (0) = - X (0) ± (1) = - X (1) , where k is a separation constant. (b) IF k = 0 then X ±± X = ax + b, X ± (0) = - X (0) a = - b X ± (1) = - X (1) a = - ( a + b ) 2 a = - b ; a = b . So k = 0 leads to trivial solutions. (c) IF k = α 2 > 0, then X ±± - μ 2 X X = c 1 cosh μx + c 2 sinh μx ; X ± (0) = - X (0) μc 2 = - c 1 X ±
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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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