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