Ch 3.4: Complex Roots of Characteristic Equation Recall our discussion of the equation where a, b...
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Transcript of Ch 3.4: Complex Roots of Characteristic Equation Recall our discussion of the equation where a, b...
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 = + i, r2 = - iThus
0 cyybya
0)( 2 cbrarety rt
a
acbbr
2
42
titi etyety )(,)( 21
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
titn
ti
n
t
n
ite
n
nn
n
nn
n
nit sincos
!12
)1(
!2
)1(
!
)(
1
121
0
2
0
tite ti sincos
tietetiteeee ttttitti sincossincos
tieteety
tieteetyttti
ttti
sincos)(
sincos)(
2
1
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
tietetytt
tt
sincos)(
sincos)(
2
1
tietyty
tetytyt
t
sin2)()(
cos2)()(
21
21
tetytety tt sin)(,cos)( 43
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 sin)(,cos)( 43
0
cossinsincos
sincos
2
t
tt
tt
e
ttette
teteW
tectecty tt sincos)( 21
Example 1
Consider the equation
Then
Therefore
and thus the general solution is
2/3sin2/3cos)( 2/2
2/1 tectecty tt
0 yyy
ii
rrrety rt
2
3
2
1
2
31
2
41101)( 2
2/3,2/1
Example 2
Consider the equation
Then
Therefore
and thus the general solution is
04 yy
irrety rt 204)( 2
2,0
tctcty 2sin2cos)( 21
Example 3
Consider the equation
Then
Therefore the general solution is
023 yyy
irrrety rt
3
2
3
1
6
12420123)( 2
3/2sin3/2cos)( 3/2
3/1 tectecty tt
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
33
3,1
12
3
2
1
1
21
21
1
cc
cc
c
2/3sin32/3cos)( 2/2/ tetetu tt
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