We will use k to denote the square root of the absolute value of .
If then
:
solves the ODE.
Substituted boundary conditions give that both "A" and "B" are equal to zero.
For positive we obtain that
:
solves the ODE.
Substitution of boundary conditions again yields "A" = "B" = 0.
For negative it is easy to show that
:
solves the ODE.
From the first boundary condition,
:.
Now, after the cosine is gone, we will substitute the second boundary condition:
:.
So either "A" = 0 or "k" is an integer.
Thus we get that the eigenfunctions which solve the "boundary value problem" are:.
One may easily check that they satisfy the boundary conditions.
Example (partial)
Consider the elliptic eigenvalue problem (boundary value problem):