Boyce/DiPrima 9th ed, Ch 4.2: Homogeneous Equations with Constant CoefficientsElementary Differential Equations and Boundary Value Problems, 9th edition, by William E. Boyce and Richard C. DiPrima, ©2009 by John Wiley & Sons, Inc.

Consider the nth order linear homogeneous differential equation with constant, real coefficients:

As with second order linear equations with constant coefficients, y = ert is a solution for values of r that make characteristic polynomial Z(r) zero:

By the fundamental theorem of algebra, a polynomial of degree n has n roots r1, r2, …, rn, and hence

01)1(

1)(

0 yayayayayL nn

nn

0)( polynomial sticcharacteri

11

10

rZ

nnnnrtrt arararaeeL

)())(()( 210 nrrrrrrarZ

Real and Unequal Roots

If roots of characteristic polynomial Z(r) are real and unequal, then there are n distinct solutions of the differential equation:

If these functions are linearly independent, then general solution of differential equation is

The Wronskian can be used to determine linear independence of solutions.

trtrtr neee ,,, 21

trn

trtr necececty 2121)(

Example 1: Distinct Real Roots (1 of 3)

Consider the initial value problem

Assuming exponential soln leads to characteristic equation:

Thus the general solution is

03211

067)( 234

rrrr

rrrrety rt

tttt ececececty 34

2321)(

1)0(,2)0(,0)0(,1)0(

067)4(

yyyy

yyyyy

Example 1: Solution (2 of 3)

The initial conditions

yield

Solving,

Hence

1278

294

032

1

4321

4321

4321

4321

cccc

cccc

cccc

cccc

1)0(,2)0(,0)0(,1)0( yyyy

tttt ececececty 33

2321)(

8

1,

3

2,

12

5,

8

114321 cccc

tttt eeeety 32

8

1

3

2

12

5

8

11)(

Example 1: Graph of Solution (3 of 3)

The graph of the solution is given below. Note the effect of the largest root of the characteristic equation.

tttt eeeety 32

8

1

3

2

12

5

8

11)(

0 .2 0 .4 0 .6 0 .8 1 .0 1 .2 1 .4t

5

4

3

2

1

1

yt

Complex Roots

If the characteristic polynomial Z(r) has complex roots, then they must occur in conjugate pairs, i.

Note that not all the roots need be complex.

Solutions corresponding to complex roots have the form

As in Chapter 3.4, we use the real-valued solutions

tietee

tieteettti

ttti

sincos

sincos

tete tt sin,cos

Example 2: Complex Roots (1 of 2)

Consider the initial value problem

Then

The roots are 1, -1, i, -i. Thus the general solution is

Using the initial conditions, we obtain

The graph of solution is given on right.

01101)( 224 rrrety rt

2)0(,2/5)0(,4)0(,2/7)0(,0)4( yyyyyy

tctcececty tt sincos)( 4321

tteety tt sincos2

130)(

Example 2: Small Change in an Initial Condition (2 of 2)

Note that if one initial condition is slightly modified, then the solution can change significantly. For example, replace

with

then

The graph of this soln and original soln are given below.

tteety tt sincos2

130)(

2)0(,2/5)0(,4)0(,2/7)0( yyyy

8/15)0(,2/5)0(,4)0(,2/7)0( yyyy

tteety tt sin16

17cos

2

1

32

95

32

1)(

Repeated Roots

Suppose a root rk of characteristic polynomial Z(r) is a repeated root with multiplicty s. Then linearly independent solutions corresponding to this repeated root have the form

If a complex root + i is repeated s times, then so is its conjugate - i. There are 2s corresponding linearly independent solns, derived from real and imaginary parts of

or

trstrtrtr kkkk etettee 12 ,,,,

,sin,cos

,,sin,cos,sin,cos11 teettet

ttettetetettrstrs

tttt

kk

tiustiutiutiu etettee 12 ,,,,

Example 4: Repeated Roots

Consider the equation

Then

The roots are i, i, -i, -i. Thus the general solution is

011012)( 224 rrrrety rt

02)4( yyy

ttcttctctcty sincossincos)( 4321

2 4 6 8 1 0t

1 5

1 0

5

5

1 0

ytSample Solution: y= (1 + t) Cos t + (1 + t) Sin t

Example 4: Complex Roots of -1 (1 of 2)

For the general solution of , the characteristic equation is .

To solve this equation, we need to use Euler’s equation to find the four 4th roots of -1:

Letting m = 0, 1, 2, and 3, we get the roots:

014 r0)4( yy

)2/4/sin()2/4/cos()1(

integer any for )2sin()2cos(1

orsincos1

4/)2(4/1

)2(

mime

memim

ei

mi

mi

i

ly.respective,2

1,

2

1,

2

1,

2

1 iiii

Example 4: Complex Roots of -1 (2 of 2)

Given the four complex roots, extending the ideas from Chapter 4, we can form four linearly independent real solutions.

For the complex conjugate pair , we get the solutions

For the complex conjugate pair , we get the solutions

So the general solution can be written as

2

1,

2

1,

2

1,

2

1 iiiir

2

1 i

)/sin(ey),/cos(ey 22 2t/2

2t/1 tt

2

1 i

)/sin(ey),/cos(ey 22 2-t/4

2-t/3 tt

44332211 yyc cycyc