boyce/diprima 9 th ed, ch 4.2: homogeneous equations with constant coefficients elementary...
TRANSCRIPT
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