Page 329 - Schaum's Outline of Differential Equations

312 SECOND-ORDER BOUNDARY-VALUE PROBLEMS [CHAP. 32

y
32.4. Solve /' = 2; y(-l) = 5, (l) - 2/(l) = 1.
This is a nonhomogeneous boundary-value problem of forms (32.1) and (32.2), where (f>(x) = 2, Yi = 5, and
Ji=\. Since the associated homogeneous problem has nontrivial solutions (Problem 32.2), this problem does not
have a unique solution. There are, therefore, either no solutions or more than one solution. Solving the differential
2
equation, we find that y = c l + c 2x + x . Then, applying the boundary conditions, we obtain the equations c l - c 2 = 4
2
and c l - c 2 = 4; thus, c l = 4 + c 2, c 2 arbitrary. Finally, y = c 2(1 + x) + 4 + x ; and this problem has infinitely many
solutions, one for each value of the arbitrary constant c 2.
y
32.5. Solve /' = 2; y(-l) = 0, (l) - 2/(l) = 0.
This is a nonhomogeneous boundary-value problem of forms (32.1) and (32.2), where (f>(x) = 2 and = y 2 = 0.
y 1
As in Problem 32.4, there are either no solutions or more than one solution. The solution to the differential equa-
2
tion is y = c l + c 2x + x . Applying the boundary conditions, we obtain the equations c 1-c 2 = -l and c 1-c 2 = 3.
Since these equations have no solution, the boundary-value problem has no solution.

32.6. Find the eigenvalues and eigenfunctions of

The coefficients of the given differential equation are constants (with respect to x)', hence, the general solution
can be found by use of the characteristic equation. We write the characteristic equation in terms of the variable m,
2
2
since X now has another meaning. Thus we have m - 4km + 4X = 0, which has the double root m = 2X; the solution
2
r :
to the differential equation is y = c le " + c 2xe ^ Applying the boundary conditions and simplifying, we obtain
It now follows that c l = 0 and either c 2 = 0 or X = -1. The choice c 2 = 0 results in the trivial solution y = 0; the
2x
choice X = -1 results in the nontrivial solution y = c 2xe" , c 2 arbitrary. Thus, the boundary-value problem has the
2x
eigenvalue X = -1 and the eigenfunction y = c 2xe" .
32.7. Find the eigenvalues and eigenfunctions of

As in Problem 32.6 the solution to the differential equation is y = c 1e 2Xjr +c 2xe 2Xjr Applying the boundary
conditions and simplifying, we obtain the equations

This system of equations has a nontrivial solution for Cj and c 2 if and only if the determinant

j
is zero; that is, if and only if either A, = — or^, = ^. When A, = — , (1) has the solution c l = 0, c 2 arbitrary; when
^
A, = -j, (1) has the solution c l = -3c 2 , c 2 arbitrary. It follows that the eigenvalues are 'k l = — j and A, 2 = ^ and the
x
xl2
corresponding eigenfunctions are y^ = c 2xe~ and y 2= (—3 + x)e .
c 2
32.8. Find the eigenvalues and eigenfunctions of

2
In terms of the variable m, the characteristic equation is m + 'km = 0. We consider the cases X = 0 and X + 0
separately, since they result in different solutions.