Help with nonhomogeneous linear equation

[Song]

18.1
#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)...

 

[NateTG]

You have:
y(x)=c_1 \cos (\sqrt{v}x) + c_2 \sin (\sqrt{v}x)
and
y(0)=0

Which very conveniently leads to something.

From there, you can use
y(L)=0
to figure out something about \sqrt{v}

 

[Song]

I got c1=0, and c2=0
But they're not right..

 

[Song]

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