Page 22 - Schaum's Outline of Differential Equations

CHAP. 1] BASIC CONCEPTS 5

y'(0) = 2c 1 cos 0-2c 2 sin 0 = 2c 1. To satisfy the second initial condition, y'(0) = 1, we choose 2c 1=l, or
c l = j. Substituting these values of c 1 and c 2 into y(x), we obtain y(x) = ysin2;t as the solution of the initial-value
problem.

1.10. Find a solution to the boundary-value problem y" + 4y = 0; y(n!8) = 0, y(n!6) = 1, if the general solution
to the differential equation is y(x) = c 1 sin 2x + c 2 cos 2x.
Note that

To satisfy the condition y(n/S) = 0, we require

Furthermore,

To satisfy the second condition, 31(71/6) = 1, we require

Solving (_/) and (2) simultaneously, we find

Substituting these values into y(x), we obtain

as the solution or the boundary-value problem.

1.11. Find a solution to the boundary-value problem if the general solution
to the differential equation is known to be y(x) = c 1 sin 2x + c 2 cos 2x.
Since y(0) = c 1 sin 0 + c 2 cos 0 = c 2, we must choose c 2 = 1 to satisfy the condition y(0) = 1. Since y(7tl2)
= GI sin n+ c 2 cos n= — c 2, we must choose c 2 = -2 to satisfy the second condition, y(7tl2) = 2. Thus, to satisfy both
boundary conditions simultaneously, we must require c 2 to equal both 1 and - 2, which is impossible. Therefore,
there does not exist a solution to this problem.

1.12. Determine c, and c 2 so that y(x) = c 1 sin 2x + c 2 cos 2x + 1 will satisfy the conditions y(7t!8) = 0 and

Note that

To satisfy the condition y(n/8) = 0, we require or equivalently,