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Boundary Value Problem and Eigenvalues

  1. Oct 27, 2012 #1
    1. The problem statement, all variables and given/known data

    [itex] y'' +λy=0 [/itex]

    [itex]y(1)+y'(1)=0[/itex]

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

    See above.

    3. The attempt at a solution

    So I already showed B=0, what I'm confused about is the α
    part. I got [itex]y(1)+y'(1)=0[/itex] to be [itex]Acos(α)-Asin(α)=0[/itex] which means [itex] tan(α)=1[/itex]. 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?
     
  2. jcsd
  3. Oct 27, 2012 #2
    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
     
  4. Oct 27, 2012 #3
    Well I feel silly now. Thanks.
     
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