Differential Equations Lecture Work Solutions 23

# Differential Equations Lecture Work Solutions 23 - 3 cosh...

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1. e. X 0 + λX =0 X (0) = 0 X 0 ( L )+ X ( L )=0 Try e rx . As we know from ODEs, this leads to the characteristic equation for r r 2 + λ =0 Or r = ± λ We now consider three cases depending on the sign of λ Case 1: λ< 0 In this case r is the square root of a positive number and thus we have two real roots. In this case the solution is a linear combination of two real exponentials X = A 1 e λx + B 1 e λx It is well known that the solution can also be written as a combination of hyperbolic sine and cosine, i.e. X = A 2 cosh λx + B 2 sinh λx The other two forms are may be less known, but easily proven. The solution can be written as a shifted hyperbolic cosine (sine). The proof is straight forward by using the formula for cosh( a + b )(s inh( a + b )) X =
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Unformatted text preview: 3 cosh √ − λx + B 3 Or X = A 4 sinh √ − λx + B 4 Which form to use, depends on the boundary conditions. Recall that the hypebolic sine vanishes ONLY at x = 0 and the hyperbolic cosine is always positive. If we use the following form of the general solution X = A 4 sinh √ − λx + B 4 then the derivative X wil be X = √ − λA 4 cosh √ − λx + B 4 The Frst boundary condition X (0) = 0 yields B 4 = 0 and clearly to satisfy the second boundary condition √ − λA 4 cosh √ − λL = 0 we must have A 4 = 0 (recall cosh x is never zero thus the coeﬃcient A 4 must vanish). 23...
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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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