Let

be a positive constant. Consider the equation

with boundary conditions

and

.

For certain discrete values of

, this equation can have non-zero solutions. Find the three smallest values of

for which this is the case.

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I've been trying to do this, but I get nowhere. Maybe I'm doing something wrong... Let me know.

So the characteristic equation is k^2 + 8k + a = 0.

The roots are:

k1 = -4 + sqrt(16-a)

k2 = -4 - sqrt(16-a)

General solution:

y = Ae^[(-4+sqrt(16-a))x] + Be^[(-4-sqrt(16-a))x]

From the given conditions:

0 = A+B --> B = -A

0 = Ae^[-8+2sqrt(16-a)] + Be^[-8-2sqrt(16-a)]

Substituting -A for B and taking e^-8 out:

0 = (Ae^(-8))[e^(2sqrt(16-a)) - e^(-2sqrt(16-a)]

So I thought that for A and B to not equal 0, I need to make the term in the second parantheses 0 and solve for a. However I only get one answer: a=0 which seems to be wrong.

Maybe I'm completely missing the point here. I would appreciate it if anyone could help me.