We now consider the situation where the auxiliary equation
with real coefficients contains a complex root, say a + bi, which means it must
also contain its conjugate, a – bi. The complementary function for the given
differential equation would then contain the terms:

c1 exp(a +bi)x + c2(a- bi)x

which may be written, alternatively,

c1exp (ax)[ cos bx + i sin bx] +
c2exp (ax)[ cos bx - sin bx]

The terms can then be re-arranged to get:

exp (ax) [(c1 + c2) cos bx + i(c1 – c2) sin bx]

If we let c3 = c1 + c2 and i(c1 – c2) = c4, we have:

y= exp(ax) [c3 cos bx + c4 sin bx]

so that part of the solution corresponding to the two
complex roots, m1 = a +bi and m2 = a – bi, has real numbers c3 and c4. (if we choose c1 and c2 as complex conjugate
numbers).

Note that a 2

^{nd}part of the solution can be obtained by choosing:
c = [c3

^{2}+ c4^{2}]^{1/2 }and tan a = c3/ c4
Then we have:

sin a =
c3/ c and cos a = c4/ c

Multiply the y solution in c3, c4 by c/c = 1 and one gets:

y= exp(ax) c [c3/ c
cos bx + c4/c sin bx]

or:

y= c exp(ax)
[sin a cos bx + cos a sin bx]

= c exp(ax) sin ( bx +
a)

In this last expression c and a are two arbitrary
constants.

Example:

Solve the differential equation:

*y” + y’ + y = 0***We write out the auxiliary equation:**

*m*^{2}+ m + 1 = 0
Solve using the quadratic formula to obtain:

*m1 = - ½ +*

*Ö3 i / 2*
and:

*m2 = - ½ -*

*Ö3 i/ 2*

*y= exp (-x/2)[ c1 cos*

*Ö3 x/ 2 + c2 sin*

*Ö3 x/ 2 ]*

*Problems for the Math Maven:*

1) Find
the general solution of: y”’ – 3y” + 7y’ – 5y = 0

2) Find
the general solution of: y

^{iv}+ 18y” + 81 = 0
In each case put the solution in a form analogous to:

y= exp(ax) [c3 cos bx + c4 sin bx]

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