Question

how to solve auxillary equation

Answer 1

Auxiliary Equations with Complex Roots



For homogeneous second-order constant-coefficient differential equations,
ay00 + by0 + cy = 0, (1)
we focused our analysis on the auxiliary equation
ar2 + br + cr = 0. (2)
In particular, we considered the discriminant d = b
2 − 4ac.
In the previous section we considered the cases d. > 0 and d = 0. Now, we turn to the one remaining
case d < 0 where the roots of (2), r− and r+ are complex and distinct. In particular, the roots are
r− = α − iβ and r+α + iβ.
where
α = −
b
2a
and β =
√
4ac − b
2
2a
.
If we follow the same procedure, then we would say that the solutions to (1) are
y1(t) = e
(α+iβ)t
and y2(t) = e
(α−iβ)t
1 Introduction to Exponentials and Complex Numbers
To make sense of e to a complex power, we define
e
λ+iθ = e
λ
(cos θ + isin θ) (3)
To check that (3) has some of the expected properties, let
ζ1 = λ1 + iθ1 and ζ2 = λ2 + iθ2.
We would like to say that
e
ζ1 e
ζ2 = e
ζ1+ζ2
.
Let’s check this e
ζ1 e
ζ2 = e
λ1+iθ1 e
λ2+iθ2 = (e
λ1
(cos θ1 + isin θ1))(e
λ2
(cos θ2 + isin θ2))
= e
λ1 e
λ2
(cos θ1 + isin θ1))(cos θ2 + isin θ2))
(Multplication is commutative.)
= e
λ1+λ2
(cos θ1 cos θ2 + isin θ1 cos θ2 + i cos θ1 sin θ2 + i
2
sin θ1 sin θ2)
(Properties of real exponents and complex multiplication.)
= e
λ1+λ2
((cos θ1 cos θ2 − sin θ1 sin θ2) + i(sin θ1 cos θ2 + cos θ1 sin θ2))
(i
2 = −1)
= e
λ1+λ2
((cos(θ1 + θ2) + i(sin(θ1 + θ2))
(Identities for the sine and cosine of a sum of angles.)
= e
(λ1+iθ1)+(λ+iθ2) = e
ζ1+ζ2
.
(The definition in (3).)
Next, we take a derivative
d
dt e
i(α+iβ)t =
d
dt
The general solution is
y(t) = c1 cost + c2 sin t
If y(0) = 3 and y
0
(0) = −4, then
3 = y(0) = c1
y
0
(t) = −c1 sin t + c2 cost
−4 = y
0
(0) = c2
Thus,
y(t) = 3 cost − 4 sin t
Exercise 3. For the general solution to a frictionless spring
my00 + ky = 0
What do the values c1 and c2 represent.
Example 4. Returning to the damped oscillator with m = 1, b = 2, and k = 2., we have the governing
equation
y
00 + 2y
0 + 2y = 0. (4)
The auxiliary equation is
r
2 + 2r + 2 = 0
with roots
r =
−2 ±
√
2
2 − 4 · 1 · 2
2
=
−2 ± i
√
4
2
= −1 ± i.
Thus, the two roots are
r− = −1 − i and r+ = −1 + i.
At this point, we have two choices. We can write the solutions as
y1(t) = e
(−1−i)t = e
−t
(cost − isin t) and y2(t) = e
(−1+i)t = e
−t
(cost + isin t) (5)
with a general solution
y(t) = c1y1(t) + c2y2(t).
or
y˜1(t) = e
−t
cost and ˜y2(t) = e
−t
sin t (6)
y(t) = ˜c1y˜1(t) + ˜c2y˜2(t).
We know from the exercise above that we can write the solutions in (5) as a linear combination of the
solutions in (6) and vice versa.
For y
0
(0) = and y
0
(0) = −3, then
0 = y(0) = c1 + c2
y
0
(t) = c1(−1 − i)e
(−1−i)t + c2(−1 + i)e
(−1+i)t
−3 = y
0
(0) = c1(−1 − i) + c2(−1 + i)