Chem Differential Eq HW Solutions Fall 2011 34

# Chem Differential Eq HW Solutions Fall 2011 34 - 34 Chapter...

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Unformatted text preview: 34 Chapter 3 Partial Diﬀerential Equations in Rectangular Coordinates integer and we have n > kL for all n. So only Case III from the solution of π Exercise 12 needs to be considered. Thus ∞ e−.5t sin nx an cos λn t + bn sin λn t, u(x, t) = n=1 where (.5n)2 − 1. λn = Setting t = 0, we obtain ∞ sin x = an sin nx. n=1 Hence a1 = 1 and an = 0 for all n > 1. Now since bn = kan 2 + λn λn L L g (x) sin 0 nπ x dx, L n = 1, 2, . . . , it follows that bn = 0 for all n > 1 and and the solution takes the form u(x, t) = e−.5t sin x cos λ1 t + b1 sin λ1t , where λ1 = (.5)2 − 1 = √ √ .75 = b1 = So 3 2 and ka1 1 =√ . λ1 3 √ u(x, t) = e − . 5t √ 3 3 1 sin x cos( t) + √ sin( t) . 2 2 3 ...
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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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