Complex roots of the characteristic equation

1. Complex Roots of the Characteristic Equation
We established previously that if we had a solution of the forme
rt
to the second order
equation
ay
′′
+by
′
+cy= 0
thenrmust satisfy
ar
2
+br+c= 0
which we called the characteristic equation. We also established that for a second-
degree linear homogeneous equation, we needed 2 linearly independent solutions to
write down the general solution, and that a non-zero Wronskian would tell us when
we had linearly independent solutions.
In the case where the characteristic equation has two real roots (r1andr2), we get
two linearly independent solutions (e
r1t
ande
r2t
) and therefore the general solution.
In the case with one repeated real rootr, we used reduction of order to find two
linearly independent solutions:e
rt
andte
rt
.
Today we discuss how to deal with complex roots of the characteristic equation:
1. An Unsatisfactory Result
2. Pure Imaginary Roots
3. Complex Roots
4. Amplitude and Period
5. Practice
1 An Unsatisfactory Result
Let us attempt to find solutions to the equationy
′′
+y= 0, using our usual method
of looking for solutions of the forme
rt
.
Here, the characteristic equation isr
2
+ 1 = 0, which has solutionsr=±i, whereiis
the imaginary numberi=
√
−1. Hence, we could get two “solutions”
y1=e
it
, y2=e
−it
1

2. and a “general” solution
y(t) =c1e
it
+c2e
−it
.
However, this is extremely unsastifactory. We started witha real differential equation,
and ended with acomplexfunction as a solution! We want to have a real solution.
It is worth noting that we can use Euler’s identity
e
it
= cos(t) +isin(t)
to rewrite these solutions into terms of sines and cosines, and thus make the imaginary
part of the solution explicit.
We will now attempt to find some real valued solutions to differential equations with
complex roots to the characteristic equation.
2 Pure Imaginary Roots
For the equationy
′′
+y= 0 we have found two possible “solutions”:
y1=e
it
,andy2=e
−it
.
or
y1(t) = cos(t) +isin(t)
y2(t) = cos(−t) +isin(−t) = cos(t)−isin(t)
Now let us apply the principal of superposition: given that these two functions are
solutions, so is any linear combination of the two. Particularly, we have one solution
of the form
1
2

e
it
+e
−it

=
1
2
[(cos(t) +isin(t)) + (cos(t)−isin(t))]
=
1
2
[2 cos(t)]
= cos(t)
Eureka! The function cos(t) is a real function, and is a solution toy
′′
+y= 0! (In
fact, it’s easy enough to check this directly.)
To get our second solution, we need to multiply both equations byi/2 and then
subtract in the right order:
i
2
(e
−it
−e
it
) =
i
2
cos(t)−
i
2
2
sin(t)−
i
2
cos(t)−
i
2
2
sin(t)
=−i
2
sin(t) = sin(t)
So (perhaps not surprisingly) the other solution is sin(t). (Note: The approach we
have taken here will work any time we have a solution and its complex conjugate.

3. In other words: Ify=a(t) +ib(t) andy=a(t)−ib(t) are both solutions to a
homogeneous equation, the principle of superposition guarantees that botha(t) and
b(t) are solutions to the differential equation also.)
Are these two solutions linearly independent? We check the Wronskian:





cos(t) sin(t)
−sin(t) cos(t)





= cos
2
(t) + sin
2
(t) = 1
So these are linearly independent, and the general solutiontoy
′′
+y= 0 must be
y(t) =c1cos(t) +c2sin(t).
Example:
Find two real fundamental solutions toy
′′
+ 3y= 0, and show that they are linearly
independent.
The characteristic equation is
r
2
+ 3r= 0
which has roots±
√
3i. Thus we get two solutions,e
±
√
3t i
= cos(
√
3t)±isin(
√
3t).
The real and imaginary parts give two real solutions
y1(t) = cos(
√
3t) andy2= sin(
√
3t).
We can see that these are linearly independent as the Wronskian is





cos(
√
3t) sin(
√
3t)
−
√
3 sin(
√
3t)
√
3 cos(
√
3t)





=
√
3 cos
2
(
√
3t) +
√
3 sin
2
(
√
3t) =
√
3
which is not zero.
It is worth noting that complex roots to polynomials always come in conjugate pairs:
ifa+biis a solution, then so isa−bi. (This is a fact that you encountered briefly
in Vector Geometry.) So if we have two purely imaginary roots, they must be of the
form±bi, withbreal and not zero.
Then our real and imaginary parts ofe
±bt i
are cos(b t) and sin(b t), and the Wronskian
for these is non-zero:





cos(b t) sin(b t)
−bsin(b t)bcos(b t)





=bcos
2
(b t) +bsin
2
(b t) =b
Thus, we have completely solved the case in which we have purely imaginary roots.
There will always be two linearly independent real solutions of the form cos(bt) and
sin(bt) which we can use to write down the general solution.

4. 3 Complex Roots
Let us now suppose that our characteristic equation has a complex root. In other
words, we will have a root of the forma+biwhereaandbare real and non-zero. Of
course, since complex roots come in conjugate pairs,a−biis also a root. Thus, we
have the following two solutions to the original differential equation:
e
(a+bi)t
=e
at
e
bt i
=e
at
cos(bt) +ie
at
sin(bt) and
e
(a−bi)t
=e
at
e
−bt i
=e
at
cos(bt)−ie
at
sin(bt)
Using exactly the same trick as in the pure imaginary case, weextract the real and
imaginary parts of these solutions:
1
2

e
(a+bi)t
+e
(a−bi)t

=e
at
cos(bt) and
i
2

e
(a−bi)t
−e
(a+bi)t

=e
at
sin(bt)
Will these two solutions be linearly independent? We check the Wronskian:





e
at
cos(bt) e
at
sin(bt)
e
at
(acos(bt)−bsin(bt))e
at
(bcos(bt) +asin(bt))





=e
2at
(bcos
2
(bt) +acos(bt) sin(bt))−e
2at
(asin(bt) cos(bt)−bsin
2
(bt))
=be
2at
which is nonzero, since we are assumingb6= 0.
Example:
Solve the initial value problemy
′′
−2y
′
+ 2y= 0,y(π) =e
π
,y
′
(π) = 0.
The characteristic equation is
which has roots
So our fundamental solutions are
y1(t) =
and
y2(t) =
for a general solution

5. y(t) =
Then we use the initial conditions.y(π) =e
π
means that
Theny
′
(π) = 0 means that
So the unique solution to this initial value problem is
4 Amplitude and Period
Suppose we have a differential equation which has a solution of the form
y(t) =c1cos(bt) +c2sin(bt)
Then we have a periodic solution, and we sometimes want to know what the amplitude
and period of our solution is. We will answer this by rewriting our solution above in
the form
y=Rcos(ωt−δ)
Then our amplitude will beR, the period will be 2π/ω, andδwill be a phase shift.
(It simply shifts the graph right or left of a typical cosine.The graph of cos(ωt−δ)
is shiftedrightof the graph of cos(ωt) by the amountδ/ω.)
We can rewrite the solution using the fact that
cos(A−B) = cos(A) cos(B) + sin(A) sin(B)
In our case, we want
Rcos(ωt−δ) =Rcos(δ) cos(ωt) +Rsin(δ) sin(ωt)
=c1cos(bt) +c2sin(bt)
In other words, we wantω=b,Rcos(δ) =c1, andRsin(δ) =c2. So we knowωand
we get
R
2
=c
2
1+c
2
2
tan(δ) =
c2
c1

6. If all we want is the amplitude and period, we do not need to actually solve for the
phase shiftδ. (Note that solving forRandδlike this is exactly like finding the radius
and angle to put the point (c1, c2) into polar coordinates. Caution: You will need to
check the signs ofc1andc2to make sure to chooseδin the correct quadrant.)
Example:
Solve the initial value problemy
′′
+ 2y= 0,y(0) = 1, andy
′
(0) = 3. Then find the
amplitude and period of the solution.
The characteristic equation isr
2
+ 2 = 0, sor=±
√
2i. Thus the general solution is
y(t) =c1cos(
√
2t) +c2sin(
√
2t)
The initial conditiony(0) = 1 gives usc1= 1, and since
y
′
(t) =−
√
2 sin(
√
2t) +c2
√
2 cos(
√
2t)
we see thaty
′
(0) = 3 means thatc2= 3/
√
2, so we have the solution
y(t) = cos(
√
2t) +
3
√
2
sin(
√
2t).
To find the amplitude, we simply need
R=
v
u
u
t
1
2
+

3
√
2
!
2
=
s
1 +
9
2
=
s
11
2
≈2.345.
The period is 2π/ω, where hereω=
√
2, so the period is
√
2π.
We can also write down the frequency, which is the reciprocalof the period:
frequency =
1
period
=
1
√
2π
Finally, we sometimes refer to theangular frequencyorradian frequency, which is the
frequency multiplied by 2π. (In other words, how often we repeat within a circle of
2πradians. If you follow all the twists and turns here, angularfrequency is justω.)
Here of course the angular frequency isω=
√
2.
We can rewrite our solution in the form (approximately)
y(t) = 2.345 cos(
√
2t−δ)
where we can determine that since tan(δ) =
3
√
2
and we have (c1, c2) in the first
quadrant,δ= tan
−1
(3/
√
2)≈1.13. (Ifc1andc2had both been negative, we would
have required aδin the third quadrant, so we would have had tan
−1
(3/
√
2) +π≈
4.27.)

7. 5 Practice
Now find general solutions to each of the following:
•y
′′
+ 6y
′
+ 13y= 0
•y
′′
+ 2y
′
−3y= 0
•4y
′′
−4y
′
+y= 0