1. Ch 3.4:
Complex Roots of Characteristic Equation
Recall our discussion of the equation
where a, b and c are constants.
Assuming an exponential soln leads to characteristic equation:
Quadratic formula (or factoring) yields two solutions, r1 & r2:
If b2
– 4ac < 0, then complex roots: r1 = λ + iµ, r2 = λ - iµ
Thus
0=+′+′′ cyybya
0)( 2
=++⇒= cbrarety rt
a
acbb
r
2
42
−±−
=
( ) ( )titi
etyety µλµλ −+
== )(,)( 21

2. Euler’s Formula; Complex Valued Solutions
Substituting it into Taylor series for et
, we obtain Euler’s
formula:
Generalizing Euler’s formula, we obtain
Then
Therefore
( ) ( )
tit
n
t
i
n
t
n
it
e
n
nn
n
nn
n
n
it
sincos
!12
)1(
!2
)1(
!
)(
1
121
0
2
0
+=
−
−
+
−
== ∑∑∑
∞
=
−−∞
=
∞
=
tite ti
µµµ
sincos +=
( )
[ ] tietetiteeee ttttitti
µµµµ λλλµλµλ
sincossincos +=+==+
( )
( )
tieteety
tieteety
ttti
ttti
µµ
µµ
λλµλ
λλµλ
sincos)(
sincos)(
2
1
−==
+==
−
+

3. Real Valued Solutions
Our two solutions thus far are complex-valued functions:
We would prefer to have real-valued solutions, since our
differential equation has real coefficients.
To achieve this, recall that linear combinations of solutions
are themselves solutions:
Ignoring constants, we obtain the two solutions
tietety
tietety
tt
tt
µµ
µµ
λλ
λλ
sincos)(
sincos)(
2
1
−=
+=
tietyty
tetyty
t
t
µ
µ
λ
λ
sin2)()(
cos2)()(
21
21
=−
=+
tetytety tt
µµ λλ
sin)(,cos)( 43 ==

4. Real Valued Solutions: The Wronskian
Thus we have the following real-valued functions:
Checking the Wronskian, we obtain
Thus y3 and y4 form a fundamental solution set for our ODE,
and the general solution can be expressed as
tetytety tt
µµ λλ
sin)(,cos)( 43 ==
( ) ( )
0
cossinsincos
sincos
2
≠=
+−
=
t
tt
tt
e
ttette
tete
W
λ
λλ
λλ
µ
µµµλµµµλ
µµ
tectecty tt
µµ λλ
sincos)( 21 +=

5. Example 1
Consider the equation
Then
Therefore
and thus the general solution is
( ) ( )2/3sin2/3cos)( 2/
2
2/
1 tectecty tt −−
+=
0=+′+′′ yyy
i
i
rrrety rt
2
3
2
1
2
31
2
411
01)( 2
±−=
±−
=
−±−
=⇔=++⇒=
2/3,2/1 =−= µλ

6. Example 2
Consider the equation
Then
Therefore
and thus the general solution is
04 =+′′ yy
irrety rt
204)( 2
±=⇔=+⇒=
2,0 == µλ
( ) ( )tctcty 2sin2cos)( 21 +=

7. Example 3
Consider the equation
Then
Therefore the general solution is
023 =+′−′′ yyy
irrrety rt
3
2
3
1
6
1242
0123)( 2
±=
−±
=⇔=+−⇒=
( ) ( )3/2sin3/2cos)( 3/
2
3/
1 tectecty tt
+=

8. Example 4: Part (a) (1 of 2)
For the initial value problem below, find (a) the solution u(t)
and (b) the smallest time T for which |u(t)| ≤ 0.1
We know from Example 1 that the general solution is
Using the initial conditions, we obtain
Thus
( ) ( )2/3sin2/3cos)( 2/
2
2/
1 tectectu tt −−
+=
1)0(,1)0(,0 =′==+′+′′ yyyyy
3
3
3
,1
1
2
3
2
1
1
21
21
1
===⇒





=+−
=
cc
cc
c
( ) ( )2/3sin32/3cos)( 2/2/
tetetu tt −−
+=

9. Example 4: Part (b) (2 of 2)
Find the smallest time T for which |u(t)| ≤ 0.1
Our solution is
With the help of graphing calculator or computer algebra
system, we find that T ≅ 2.79. See graph below.
( ) ( )2/3sin32/3cos)( 2/2/
tetetu tt −−
+=

10. Example 4: Part (b) (2 of 2)
Find the smallest time T for which |u(t)| ≤ 0.1
Our solution is
With the help of graphing calculator or computer algebra
system, we find that T ≅ 2.79. See graph below.
( ) ( )2/3sin32/3cos)( 2/2/
tetetu tt −−
+=