Page 107 - Schaum's Outline of Differential Equations

90 wth-ORDER LINEAR HOMOGENEOUS DIFFERENTIAL EQUATIONS [CHAP. 10

THE GENERAL SOLUTION
The roots of the characteristic equation determine the solution of (10.1). If the roots A are all
real and distinct, the solution is

If the roots are distinct, but some are complex, then the solution is again given by (10.3). As in Chapter 9, those
terms involving complex exponentials can be combined to yield terms involving sines and cosines. If A^ is a root
of multiplicity p [that is, if (A - 'k k) p is a factor of the characteristic equation, but (A - 'k k) p +1 is not] then there
p
k
h
h
k
will be p linearly independent solutions associated with A^ given by e " ,xe ",x e *,...,x ~ e ". These
solutions are combined in the usual way with the solutions associated with the other roots to obtain the complete
solution.
In theory it is always possible to factor the characteristic equation, but in practice this can be extremely
difficult, especially for differential equations of high order. In such cases, one must often use numerical techniques
to approximate the solutions. See Chapters 18, 19 and 20.

Solved Problems

10.1. Solve /" - 6y" + Uy'-6y = 0.
3
2
The characteristic equation is X - 6A + 11A - 6 = 0, which can be factored into
The roots are A : = 1, A 2 = 2, and ^3 = 3; hence the solution is

4)
10.2. Solve / - 9y" + 20y = 0.
4
2
The characteristic equation is A - 9A + 20 = 0, which can be factored into

The roots are hence the solution is

10.3. Solve y' -5y = 0.
5x
The characteristic equation is A, — 5 = 0, which has the single root Aj = 5. The solution is then y = c le .
(Compare this result with Problem 6.9.)

10.4. Solve /" - 6y" + 2y' + 36y = 0.
3
2
The characteristic equation, A - 6A +2A + 36 = 0, has roots and
The solution is