## Info

* In this substitution it should be kept in mind that we have in the present case a2 = 4a2 and b=-at/2.

According to the said theorem, we can write (h ± vi)' = R'(cos Ot ± i sin Ot ) where the value of R (always taken to be positive) is, by (15.10), r^-ï / a} + 4a, — a,2 ,—

(17.8) R = ih1 + v2 = y --^-~ = {ai and 0 is the radian measure of the angle in the interval [0,277), which satisfies the conditions

R R V 4fl2

Therefore, the complementary function can be transformed as follows:

(17.10) yc = AxR'(cos8t + ;sin0f) +A2R'(cos0t - isindt)

= /*'[(/!, + A2)cos6t + (Ax - A2)ism6t] = R'(A<cos6t + A^smOt) where we have adopted the shorthand symbols A5 = Aj + A2 and Ab = (At - A2)i

The complementary function (17.10) differs from its differential-equation counterpart (15.24') in two important respects. First, the expressions cos Ot and sin Ot have replaced the previously used cos vt and sin vt. Second, the multiplicative factor R' (an exponential with base R ) has replaced the natural exponential expression eht. In short, we have switched from the cartesian coordinates (h and u) of the complex roots to their polar coordinates (R and 0). The values of R and 0 can be determined from (17.8) and (17.9) once h and v become known. It is also possible to calculate R and 0 directly from the parameter values a, and a2 via (17.8) and (17.9), provided we first make certain that a2 < 4a2 and that the roots are indeed complex.

Example 5 Find the general solution of y,. 2 + \yt = 5. With coefficients ax = 0 and a2 = \, this constitutes an illustration of the complex-root case of a2 < 4a2. By (17.7), the real and imaginary parts of the roots are h = 0 and v = It follows from (17.8) that

Since the value of 0 is that which can satisfy the two equations h v cos 0 = — = 0 and sin 0 = — = 1

it may be concluded from Table 15.1 that

Consequently, the complementary function is

0 0