1. Complex Roots of the Characteristic Equation
We established previously that if we had a solution of the form ert to the second order
equation
ay ′′ + by ′ + cy = 0
then r must 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 (r1 and r2 ), we get
two linearly independent solutions (er1 t and er2 t ) and therefore the general solution.
In the case with one repeated real root r, we used reduction of order to find two
linearly independent solutions: ert and tert .
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 equation y ′′ + y = 0, using our usual method
of looking for solutions of the form ert .
Here, the characteristic equation is r 2 + 1 = 0, which has solutions r = ±i, where i is
√
the imaginary number i = −1. Hence, we could get two “solutions”
y1 = eit , y2 = e−it
1

2. and a “general” solution
y(t) = c1 eit + c2 e−it .
However, this is extremely unsastifactory. We started with a real differential equation,
and ended with a complex function as a solution! We want to have a real solution.
It is worth noting that we can use Euler’s identity
eit = cos(t) + i sin(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 equation y ′′ + y = 0 we have found two possible “solutions”:
y1 = eit , and y2 = e−it .
or
y1 (t) = cos(t) + i sin(t)
y2 (t) = cos(−t) + i sin(−t) = cos(t) − i sin(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 it 1
e + e−it = [(cos(t) + i sin(t)) + (cos(t) − i sin(t))]
2 2
1
= [2 cos(t)]
2
= cos(t)
Eureka! The function cos(t) is a real function, and is a solution to y ′′ + y = 0! (In
fact, it’s easy enough to check this directly.)
To get our second solution, we need to multiply both equations by i/2 and then
subtract in the right order:
i −it i i2 i i2
(e − eit ) = cos(t) − sin(t) − cos(t) − sin(t)
2 2 2 2 2
= −i2 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: If y = a(t) + ib(t) and y = a(t) − ib(t) are both solutions to a
homogeneous equation, the principle of superposition guarantees that both a(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)
= cos2 (t) + sin2 (t) = 1
− sin(t) cos(t)
So these are linearly independent, and the general solution to y ′′ + y = 0 must be
y(t) = c1 cos(t) + c2 sin(t).
Example:
Find two real fundamental solutions to y ′′ + 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( 3 t) ± i sin( 3 t).
The real and imaginary parts give two real solutions
√ √
y1 (t) = cos( 3 t) and y2 = sin( 3 t).
We can see that these are linearly independent as the Wronskian is
√ √
cos( 3 t) sin( √t)
3 √ √ √ √ √
√ √ √ = 3 cos2 ( 3 t) + 3 sin2 ( 3 t) = 3
− 3 sin( 3 t) 3 cos( 3 t)
which is not zero.
It is worth noting that complex roots to polynomials always come in conjugate pairs:
if a + bi is a solution, then so is a − 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, with b real and not zero.
Then our real and imaginary parts of e±bt i are cos(b t) and sin(b t), and the Wronskian
for these is non-zero:
cos(b t) sin(b t)
= b cos2 (b t) + b sin2 (b t) = b
−b sin(b t) b cos(b t)
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 form a + bi where a and b are real and non-zero. Of
course, since complex roots come in conjugate pairs, a − bi is also a root. Thus, we
have the following two solutions to the original differential equation:
e(a+bi)t = eat ebt i
= eat cos(bt) + ieat sin(bt) and
e(a−bi)t = eat e−bt i
= eat cos(bt) − ieat sin(bt)
Using exactly the same trick as in the pure imaginary case, we extract the real and
imaginary parts of these solutions:
1 (a+bi)t
e + e(a−bi)t = eat cos(bt) and
2
i (a−bi)t
e − e(a+bi)t = eat sin(bt)
2
Will these two solutions be linearly independent? We check the Wronskian:
eat cos(bt) eat sin(bt)
eat (a cos(bt) − b sin(bt)) eat (b cos(bt) + a sin(bt))
= e2at (b cos2 (bt) + a cos(bt) sin(bt)) − e2at (a sin(bt) cos(bt) − b sin2 (bt))
= be2at
which is nonzero, since we are assuming b = 0.
Example:
Solve the initial value problem y ′′ − 2y ′ + 2y = 0, y(π) = eπ , 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) = eπ means that
Then y ′(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) = c1 cos(bt) + c2 sin(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 = R cos(ωt − δ)
Then our amplitude will be R, the period will be 2π/ω, and δ will be a phase shift.
(It simply shifts the graph right or left of a typical cosine. The graph of cos(ωt − δ)
is shifted right of the graph of cos(ωt) by the amount δ/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
R cos(ωt − δ) = R cos(δ) cos(ωt) + R sin(δ) sin(ωt)
= c1 cos(bt) + c2 sin(bt)
In other words, we want ω = b, R cos(δ) = c1 , and R sin(δ) = c2 . So we know ω and
we get
R 2 = c2 + c2
1 2
c2
tan(δ) =
c1

6. If all we want is the amplitude and period, we do not need to actually solve for the
phase shift δ. (Note that solving for R and δ like 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 of c1 and c2 to make sure to choose δ in the correct quadrant.)
Example:
Solve the initial value problem y ′′ + 2y = 0, y(0) = 1, and y ′ (0) = 3. Then find the
amplitude and period of the solution. √
The characteristic equation is r 2 + 2 = 0, so r = ± 2i. Thus the general solution is
√ √
y(t) = c1 cos( 2t) + c2 sin( 2t)
The initial condition y(0) = 1 gives us c1 = 1, and since
√ √ √ √
y ′(t) = − 2 sin( 2t) + c2 2 cos( 2t)
√
we see that y ′(0) = 3 means that c2 = 3/ 2, so we have the solution
√ 3 √
y(t) = cos( 2t) + √ sin( 2t).
2
To find the amplitude, we simply need
2
3 9 11
R= 12 + √ = 1+ = ≈ 2.345.
2 2 2
√ √
The period is 2π/ω, where here ω = 2, so the period is 2 π.
We can also write down the frequency, which is the reciprocal of the period:
1 1
frequency = =√
period 2π
Finally, we sometimes refer to the angular frequency or radian frequency, which is the
frequency multiplied by 2π. (In other words, how often we repeat within a circle of
2π radians. If you follow all the twists and turns here, angular frequency is just ω.)
√
Here of course the angular frequency is ω = 2.
We can rewrite our solution in the form (approximately)
√
y(t) = 2.345 cos( 2t − δ)
3
where we can determine that since tan(δ) = √2 and we have (c1 , c2 ) in the first
√
quadrant, δ = tan−1 (3/ 2) ≈ 1.13. (If c1 and c2 had both been negative, we would
√
have required a δ in 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