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An s l type equation with p x e x 2 q x 1 2 xe x 2 σ

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an S-L type equation with p ( x ) = e x/ 2 , q ( x ) = 1 2 xe x/ 2 , σ ( x ) = 1 2 e x/ 2 . 6. The problem is x 2 y ′′ + xy + (2 λ - 1) y = 0 , 0 < x < 1 , y ( x ) , y ( x ) bounded as x 0+ , y (1) - y (1) . Solution. Once again, the only question here is whether this is an S-L problem; if it is, it is singular because the boundary condition at 0 is not of the regular type and, more importantly, while p ( x ) is not x 2 it will still turn out to be 0 at 0. If one proceeds as in the previous problem, one finds that the function μ by which one should multiply the ODE is μ ( x ) = 1 /x , after which the ODE can be rewritten in the form ( xy ) - 1 x y + 2 λy = 0 , an S-L type equation with p ( x ) = x , q ( x ) = - 1 x , σ ( x ) = 2. 7. This was the Quiz 2 problem: Compute the eigenvalues and eigenfunctions of the following regular S-L problem: y ′′ + λy = 0 , 0 < x < π y (0) = 0 , y ( π ) = 0 . 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 , . . .
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3 8. Compute the eigenvalues and eigenfunctions of the following regular S-L problem: y ′′ + λy = 0 , 0 < x < 1 y (0) = 0 , y (1) = 0 .
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