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Help with nonhomogeneous linear equation

  1. Nov 27, 2006 #1
    Let L be a nonzero real #
    (a) Show that the boundary-value problem y''+vy=0, y(0)=0, y(L)=0, has only the trivial solution y=0 for the cases v=0 and v<0.

    I get (a), but I dont know how to do (b)

    (b) For the cases v>0, find the values of v for which this problem has a nontrivial solution and give the corresponding solution.

    Basically, I get the y=0 for (b) again. And this is wrong, but I don't know what is wrong.
    r=+-sqrt(v)i because v>0.
    Then y=c1 cos sqrt(v)x+c2 sin sqrt(v)x
    From y(0)=0 & y(L)=0, I get the same answer as (a)....
  2. jcsd
  3. Nov 27, 2006 #2


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    Homework Helper

    You have:
    [tex]y(x)=c_1 \cos (\sqrt{v}x) + c_2 \sin (\sqrt{v}x)[/tex]

    Which very conveniently leads to something.

    From there, you can use
    to figure out something about [itex]\sqrt{v}[/itex]
  4. Nov 27, 2006 #3
    I got c1=0, and c2=0
    But they're not right..
  5. Nov 27, 2006 #4
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