Boundary Value Problem and Eigenvalues

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SUMMARY

The discussion centers on solving the boundary value problem defined by the differential equation y'' + λy = 0 with the endpoint condition y(1) + y'(1) = 0. It is established that the solution y = Acos(αx) + Bsin(αx) satisfies the conditions if and only if B = 0 and α is a positive root of the equation tan(z) = 1/z. The eigenvalues are given by α²_n, where α_n are the positive roots of the intersection points of y = tan(x) and y = 1/z. The confusion arises from correctly identifying the positive roots and their derivation from the boundary conditions.

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  • Knowledge of trigonometric functions and their properties.
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Homework Statement



y'' +λy=0

y(1)+y'(1)=0

Show that y=Acos(αx)+Bsin(αx) satisfies the endpoint conditions if and only if B=0 and α is a positive root of the equation tan(z)=1/z. These roots <br /> (a_{n})^{∞}_{1} are the abscissas of the points of intersection of the curves y=tan(x) and y=1/z. Thus the eigen values and eigen functions of this problem are the numbers (α^{2}_{n})^{∞}_{1} and the functions {cos(α_{n}x)}^{∞}_{1} respectively.

Homework Equations



See above.

The Attempt at a Solution



So I already showed B=0, what I'm confused about is the α
part. I got y(1)+y&#039;(1)=0 to be Acos(α)-Asin(α)=0 which means tan(α)=1. When I graph the functions of z and find their intersection, I don't get the same values of α as I would expect from what I just solve. I'm kind of just guessing at what the question is really even asking me, because I have not idea where a positive root could even come from. The picture provided makes it seem that α should just be the value of z where they intersect, but that does seem to be right either. Can anyone figure out where a positive root could even come from?
 
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You forgot to multiply by "alpha" when deriving y'(1), that's where you're missing the alpha to get your tan(z)=1/z equation
 
Well I feel silly now. Thanks.
 

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