Differential Equations Lecture Work Solutions 24

# Differential Equations Lecture Work Solutions 24 - λL = 0...

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Any other form will yields the same trivial solution, may be with more work!!! Case 2: λ =0 In this case we have a double root r =0andasweknowfromODEstheso lut ionis X = Ax + B The boundary condition at zero yields X (0) = B =0 and the second condition X 0 ( L )+ X ( L )= A + AL =0 Or A (1 + L )=0 This implies that B = A = 0 and therefore we have again a trivial solution. Case 3: λ> 0 In this case the two roots are imaginary r = ± i λ Thus the solution is a combination of sine and cosine X = A 1 cos λx + B 1 sin λx DiFerentiate X 0 = λA 1 sin λx + λB 1 cos λx Substitute the boundary condition at zero X (0) = A 1 Thus A 1 = 0 and the solution is X = B 1 sin λx Now use the condition at L X 0 ( L )+ X ( L )= λB 1 cos λL + B 1
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Unformatted text preview: λL = 0 If we take B 1 = 0, we get a trivial solution, but we have another choice, namely √ λ cos √ λ L + sin √ λL = 0 If cos √ λL = 0 then we are left with sin √ λL = 0 which is not possible (the cosine and sine functions do not vanish at the same points). Thus cos √ λL 6 = 0 and upon dividing by it we get − √ λ = tan √ λL This can be solved graphically or numerically (see ±gure). The points of intersection are values of √ λ n . The solution is then depending on n , and obtained by substituting λ n X n ( x ) = sin ± λ n x 24...
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## This note was uploaded on 12/22/2011 for the course MAP 2302 taught by Professor Bell,d during the Fall '08 term at UNF.

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