Page 330 - Schaum's Outline of Differential Equations

CHAP. 32] SECOND-ORDER BOUNDARY-VALUE PROBLEMS 313

A. = 0: The solution to the differential equation is y = c l + c 2x. Applying the boundary conditions, we obtain the
equations c 1 + c 2 = 0 and c 2 = 0. It follows that c 1 = c 2 = 0, and y = 0. Therefore, X = 0 is not an eigenvalue.
x
A, i= 0: The solution to the differential equation is y = Cj + c 2e^ . Applying the boundary conditions, we obtain

These equations have a nontrivial solution for c 1 and c 2 if and only if

which is an impossibility, since X ^ 0.
Since we obtain only the trivial solution for X = 0 and X ^ 0, can conclude that the problem does not have any
eigenvalues.

32.9. Find the eigenvalues and eigenfunctions of

2
2
As in Problem 32.6, the solution to the differential equation is y = Cie "" + c 2xe "". Applying the boundary
conditions and simplifying, we obtain the equations

Equations (_/) have a nontrivial solution for c 1 and c 2 if and only if the determinant

is zero; that is, if and only if A, = + -j-z. These eigenvalues are complex. In order to keep the differential equation
under consideration real, we require that X be real. Therefore this problem has no (real) eigenvalues and the only
(real) solution is the trivial one: y(x) = 0.

32.10. Find the eigenvalues and eigenfunctions of

2
The characteristic equation is m + X = 0. We consider the cases X = 0, X < 0, and X > 0 separately, since they
lead to different solutions.
A, = 0: The solution is y = Cj + c 2x. Applying the boundary conditions, we obtain Cj = c 2 = 0, which results in the
trivial solution.
A.<0: The solution is y = c\e ^ + C 2e~ ^, where-X and v-^- are positive. Applying the boundary conditions,
we obtain

Here

which is never zero for any value of X < 0. Hence, Cj = c 2 = 0 and y = 0.
A. > 0: The solution is A sin vA x + B cos vA, x. Applying the boundary conditions, we obtain B = 0 and A sin vA = 0
Note that sin 6=0 if and only if 6 = nn, where n = 0, + 1, + 2, ... Furthermore, if 6 > 0, then n must be