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Solution one sees first there are no non-negative

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Unformatted text preview: Solution. One sees first there are no non-negative eigenvalues. Once this is established, we search for eigenvalues of the form l = μ 2 , μ > 0. As done already many times in this course, the general solution of the ODE is y ( x ) = C 1 cos μx + C 2 sin μx. The boundary condition y (0) = 0 implies C 1 = 0 so that y ( x ) = C 2 sin μx . Then y ′ ( x ) = C 2 μ cos μx and y ′ ( π ) = 0 implies C 2 μ cos μπ = 0. There will be a non zero solution for μ such that cos μπ = 0, thus μπ = (2 n- 1) π/ 2, n = 1 , 2 , . . . . The final answer is: The eigenvalues are the numbers λ n = ( 2 n- 1 2 ) 2 , for n = 1 , 2 , 3 , 4 , . . . ; the corresponding eigenfunctions are the functions y n ( x ) = C sin (2 n- 1) x 2 , n = 1 , 2 , 3 , . . . 3 8. Compute the eigenvalues and eigenfunctions of the following regular S-L problem: y ′′ + λy = 0 , < x < 1 y ′ (0) = 0 , y (1) = 0 . 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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Solution One sees first there are no non-negative...

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