1550078481-Ordinary_Differential_Equations__Roberts_

372 Ordinary Differential Equations

SOLUTION

Differentiating Yp, we find

Yp / = (-1) 2x.

Computing A(x)yp + b(x), yields

= (-2-1+2) = (-1).

-2x + 4x 2x

Since y; = A(x)yp + b(x), the vector Yp is a particular solution of (12). In
the previous example, we found the general solution of the associated homo-
geneous equation y' = A(x)y to be

Therefore, the general solution of the nonhomogeneous linear system (12) is

y(x) =Ye+ Yp = C1 ( 2 ~) + C2 (:2) + ( ) = (2c: ~~:2-:x2)

where c 1 and c2 are arbitrary constants.

Now let us consider the general, homogeneous linear system with constant
coefficients

(13) y'=Ay

where y and y' are n x 1 column vectors and A is an n x n matrix of real
numbers. In order to solve an n-th order linear homogeneous differential
equation, we assumed there were solutions of the form y = erx where r is an
unknown constant. By analogy, we seek a solution of (13) of the form

(14)

where v is an unknown constant vector and r is an unknown scalar constant.

Differentiating (14), we find y' = rverx. Substituting into (13), we see v and

r must be chosen to satisfy