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 2nd part of the solution can be
obtained by choosing:
c = [c32 + 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
m2 + 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: yiv + 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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