Page 108 - Schaum's Outline of Differential Equations

CHAP. 10] wth-ORDER LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS 91

which can be rewritten, using Euler's relations (see Problem 9.16) as

10.5. Solve

The characteristic equation, has roots and
. The solution is

If, using Euler's relations, we combine the first two terms and then similarly combine the last two terms,
we can rewrite the solution as

4)
10.6. Solve / + 8/" + 24y" + 32/ + I6y = 0.
4
2
4
3
The characteristic equation, X + 8X + 24X + 32X +16 = 0, can be factored into (k + 2) = 0. Here Xj = -2 is
a root of multiplicity four; hence the solution is

10.7. Solve

3
2
The characteristic equation can be factored into (X - 1) (X + I) = 0; hence, A,j = 1 is a root of multiplicity three
and X 2 = -1 is a root of multiplicity two. The solution is

10.8. Solve

The characteristic equation has roots 2 + z'2 and 2 + z'2; hence A,j = 2 + z'2 and X 2 = 2 - z'2 are both roots of
multiplicity two. The solution is

10.9. Find the general solution to a fourth-order linear homogeneous differential equation for y(x) with real
4x
numbers as coefficients if one solution is known to be J?e .
2 4x
4x
3 4x
4x
If x e is a solution, then so too are X e , xe , and e . We now have four linearly independent solutions to a
fourth-order linear, homogeneous differential equation, so we can write the general solution as

10.10. Determine the differential equation described in Problem 10.9.
The characteristic equation of a fourth-order differential equation is a fourth-degree polynomial having
3 4x
exactly four roots. Because X e is a solution, we know that X = 4 is a root of multiplicity four of the corresponding