Solution this exercise is almost identical to the

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Solution. This exercise is almost identical to the previous one. Once again one establishes that there are no nnon-negative eigenvalues; I’ll omit the details. After that, one searches for eigenvalues of the form l = μ 2 , μ > 0. Once again, the general solution of the ODE is y ( x ) = C 1 cos μx + C 2 sin μx, so y ( x ) = - C 1 μ sin μx + C 2 μ cos μx . The boundary condition y (0) = 0 implies C 2 = 0 so that y ( x ) = C 1 cos μx . Then y (1) = 0 implies C 1 cos μ = 0, so (as mentioned in the previous problem) there will be a non zero solution for μ = (2 n - 1) π/ 2, n = 1 , 2 , . . . . The final answer is: The eigenvalues are the numbers λ n = ( (2 n - 1) π 2 ) 2 , for n = 1 , 2 , 3 , 4 , . . . ; the corresponding eigenfunctions are the functions y n ( x ) = C sin (2 n - 1) πx 2 , n = 1 , 2 , 3 , . . .
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