1. a. X0 + λX =0 X (0) = 0 X ( π )=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 = A 3 cosh √ − λx + B 3 Or X = A 4 sinh √ − λx + B 4
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