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4 the problem is y y λy 0 x 2 y 0 0 y 2 y 2 0 this

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4. The problem is y ′′ - y + λy = 0 , 0 < x < 2 , y (0) = 0 , y (2) - y (2) = 0 . This problem does nt look like an S-L problem at first glance, but it might be rewritten so as to be an S-L problem. Is there a function p ( x ) such that p ( x ) > 0 in [0 , 2] and multiplying the differential equation by p ( x ) it is of the form ( py ) + qy + λσy = 0 for some q, σ ?. We want these two equations to be the same, for some function p : 0 = p ( y ′′ - y + λy ) = py ′′ - py + λpy, 0 = ( py ) + qy + λσy = py ′′ + p y + qy + λσy. Clearly (I hope) we need (and it suffices) p such that p = - p . Solving this differential equation, we get p ( x ) = Ce x , for constant C . We can take C = 1. We conclude: The differential equation is equivalent to d dx ( e x y ) + λe x y = 0 , so that we have a regular S-L problem, with p ( x ) = e x , q ( x ) = 0 , σ ( x ) = e x , κ 1 = 0 , κ 2 = 1 , κ 3 = 1 , κ 4 = - 1 .
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2 5. The problem is 2 y ′′ + y + ( λ + x ) y = 0 , 0 < x < , y (0) = 0 , y ( x ) , y ( x ) bounded as x → ∞ . Solution. The only question here is whether this is an S-L problem; if it is, it is singular because the interval is infinite among other reasons. To see if it is S-L, we must see if there is a function μ ( x ) such that after multiplying the differential equation by μ ( x ), the equation has the form ( py ) + qy + λσy = 0. In ur case, multiplication by a function μ ( x ) gives 0 = 2 μy ′′ + μy + μxy + λμy. We want this to be 0 = py ′′ + p y + qy + λσy = 0 . Thus 2 μ = p , μ = p , q = μx , σ = μ . From the first two equations, p = 1 2 p ; a non zero solution of this ODE is p ( x ) = e x/ 2 . Then μ = 1 2 e x/ 2 . The equation can be rewritten in the form d dx ( e x/ 2 y ) + 1 2 xe x/ 2 y + 1 2 xe x/ 2 λy = 0 ,
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