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Section 3-3 : Complex Roots
In this section we will be looking at solutions to the differential equation

ay′′ + by′ + cy =
0

in which roots of the characteristic equation,

ar 2 + br + c =
0
are complex roots in the form r1,2= λ ± µ i
...

(λ + µ i ) t
(λ −µ i ) t

=
y1 ( t ) e=
y2 ( t ) e
and

Now, these two functions are “nice enough” (there’s those words again… we’ll get around to defining
them eventually) to form the general solution
...
Since we started with
only real numbers in our differential equation we would like our solution to only involve real numbers
...

To do this we’ll need Euler’s Formula
...


e − iθ= cos ( −θ ) + i sin ( −θ )= cos θ − i sin θ

Now, split up our two solutions into exponentials that only have real exponents and exponentials that
only have imaginary exponents
...

λ t iµ t
=
y1 ( t ) e=
e
eλ t ( cos ( µ t ) + i sin ( µ t ) )

λ t −iµ t
=
y2 ( t ) e=
e
eλ t ( cos ( µ t ) − i sin ( µ t ) )

This doesn’t eliminate the complex nature of the solutions, but it does put the two solutions into a form
that we can eliminate the complex parts
...
In other words,

=
y ( t ) c1 y1 ( t ) + c2 y2 ( t )

will also be a solution
...


y1 ( t ) + y2 ( t ) =
2eλ t cos ( µ t )

This is a real solution and just to eliminate the extraneous 2 let’s divide everything by a 2
...


1
1
u ( t ) = y1 ( t ) + y2 ( t ) =eλ t cos ( µ t )
2
2
Note that this is just equivalent to taking

c=
c=
1
2

1
2

Now, we can arrive at a second solution in a similar manner
...


y1 ( t ) − y2 ( t ) =
2i eλ t sin ( µ t )

On the surface this doesn’t appear to fix the problem as the solution is still complex
...
This is equivalent to taking

c1 =

1
2i

and

c2 = −

1
2i

Our second solution will then be

v (t ) =

1
1
y1 ( t ) − y2 ( t ) =eλt sin ( µ t )
2i
2i

We now have two solutions (we’ll leave it to you to check that they are in fact solutions) to the
differential equation
...

So, if the roots of the characteristic equation happen to be r1,2= λ ± µ i the general solution to the
differential equation is
...


Example 1 Solve the following IVP
...


r 2 − 4r + 9 =
0
The roots of this equation are r1,2= 2 ± 5 i
...


=
y ( t ) c1e 2t cos

( 5t ) + c e

2t

2

sin

( 5t )

Now, you’ll note that we didn’t differentiate this right away as we did in the last section
...
While the differentiation is not terribly difficult, it can get a little messy
...


=
0 y=
( 0 ) c1

In other words, the first term will drop out in order to meet the first condition
...
Now, apply the second initial
condition to the derivative to get
...


y (t ) = −

8 2t
e sin
5

Example 2 Solve the following IVP
...


r 2 − 8r + 17 =
0
The roots of this are r1,2= 4 ± i
...
Applying the initial conditions gives the following system
...
The actual solution to the IVP is then
...

4 y′′ + 24 y′ +=
37 y 0

=
y (π ) 1

′ (π ) 0
y=

Solution
The characteristic equation this time is
...
The general solution as well as its derivative is

t
t
y ( t ) c1e −3t cos   + c2e −3t sin  
=
2
2
t c
t
t c
t
y′ ( t ) =
−3c1e −3t cos   − 1 e −3t sin   − 3c2e −3t sin   + 2 e −3t cos  
2 2
2
2 2
2
Applying the initial conditions gives the following system
...
Also, make sure that you evaluate the trig functions as much as
possible in these cases
...
Solving this system gives

c1 =
c2 =
−6e3π
e3π
The actual solution to the IVP is then
...


Example 4 Solve the following IVP
...


r 2 + 16 =
0

⇒

r=
±4 i

Be careful with this characteristic polynomial
...
Students however, tend to just start at r 2 and write
times down until they run out of terms in the differential equation
...

Okay, back to the problem
...


=
y ( t ) c1 cos ( 4t ) + c2 sin ( 4t )

y′ ( t ) =
−4c1 sin ( 4t ) + 4c2 cos ( 4t )
Plugging in the initial conditions gives the following system
...


3
y (t ) =
−10 cos ( 4t ) + sin ( 4t )
4



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