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.
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)
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.