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#33

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 don't 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^2+v=0

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)...