boyce/diprima 9 th ed, ch 3.4: repeated roots; reduction of order elementary differential equations...

Boyce/DiPrima 9th ed, Ch 3.4: Repeated Roots; Reduction of OrderElementary Differential Equations and Boundary Value Problems, 9th edition, by William E. Boyce and Richard C. DiPrima, ©2009 by John Wiley & Sons, Inc.

Recall our 2nd order linear homogeneous ODE

where a, b and c are constants.

Assuming an exponential soln leads to characteristic equation:

Quadratic formula (or factoring) yields two solutions, r1 & r2:

When b2 – 4ac = 0, r1 = r2 = -b/2a, since method only gives one solution:

0 cyybya

0)( 2 cbrarety rt

a

acbbr

2

42

atbcety 2/1 )(

Second Solution: Multiplying Factor v(t)

We know that

Since y1 and y2 are linearly dependent, we generalize this approach and multiply by a function v, and determine conditions for which y2 is a solution:

Then

solution a )()(solution a )( 121 tcytyty

atbatb etvtyety 2/2

2/1 )()( try solution a )(

atbatbatbatb

atbatb

atb

etva

betv

a

betv

a

betvty

etva

betvty

etvty

2/2

22/2/2/

2

2/2/2

2/2

)(4

)(2

)(2

)()(

)(2

)()(

)()(

Finding Multiplying Factor v(t)

Substituting derivatives into ODE, we seek a formula for v:

0 cyybya

43

2

222

22

22

2

22/

)(0)(

0)(4

4)(

0)(4

4

4)(0)(

4

4

4

2

4)(

0)(24

)(

0)()(2

)()(4

)()(

0)()(2

)()(4

)()(

ktktvtv

tva

acbtva

tva

ac

a

btvatv

a

ac

a

b

a

btva

tvca

b

a

btva

tcvtva

btvbtv

a

btvbtva

tcvtva

btvbtv

a

btv

a

btvae atb

General Solution

To find our general solution, we have:

Thus the general solution for repeated roots is

abtabt

abtabt

abtabt

tecec

ektkek

etvkekty

2/2

2/1

2/43

2/1

2/2

2/1 )()(

abtabt tececty 2/2

2/1)(

Wronskian

The general solution is

Thus every solution is a linear combination of

The Wronskian of the two solutions is

Thus y1 and y2 form a fundamental solution set for equation.

abtabt tececty 2/2

2/1)(

abtabt tetyety 2/2

2/1 )(,)(

te

a

bte

a

bte

ea

btea

btee

tyyW

abt

abtabt

abtabt

abtabt

allfor 0

221

21

2))(,(

/

//

2/2/

2/2/

21

Example 1 (1 of 2)

Consider the initial value problem

Assuming exponential soln leads to characteristic equation:

So one solution is and a second solution is found:

Substituting these into the differential equation and simplifying yieldswhere are arbitrary constants.

044 yyy

20)2(044)( 22 rrrrety rt

ttt

tt

t

etvetvetvty

etvetvty

etvty

2222

222

22

)(4)(4)()(

)(2)()(

)()(

tety 21 )(

211 )(,)(',0)(" ktktvktvtv 21 and cc

Example 1 (2 of 2)

Letting

So the general solution is

Note that both tend to 0 as regardless of thevalues ofUsing initial conditions

Therefore the solution to theIVP is

tt tececty 22

21)(

5,132

13)0('and1)0(

2121

1

cccc

cyy

tt teety 22 5)(

ttetyttvkk 2221 )(and)(,0 and1

21 and yy t21 and cc

0 . 0 0 . 5 1 . 0 1 . 5 2 . 0 2 . 5 3 . 0t

0 . 5

1 . 0

1 . 5

2 . 0y t

tetty 2)51()(

Example 2 (1 of 2)

Consider the initial value problem

Assuming exponential soln leads to characteristic equation:

Thus the general solution is

Using the initial conditions:

Thus

3/10,20,025.0 yyyyy

2/10)2/1(025.0)( 22 rrrrety rt

2/2

2/1)( tt tececty

3

2,2

3

1

2

12

2121

1

cc

cc

c

2/2/

3

22)( tt teety

1 2 3 4t

2

1

1

2

3

4yt

)3/22()( 2/ tety t

Example 2 (2 of 2)

Suppose that the initial slope in the previous problem was increased

The solution of this modified problem is

Notice that the coefficient of the second term is now positive. This makes a bigdifference in the graph, since theexponential function is raised to apositive power:

1 2 3 4t

2

2

4

6yt

20,20 yy

02/1

2/2/2)( tt teety

)3/22()(:blue

)2()(:red2/

2/

tety

tetyt

t

Reduction of Order

The method used so far in this section also works for equations with nonconstant coefficients:

That is, given that y1 is solution, try y2 = v(t)y1:

Substituting these into ODE and collecting terms,

Since y1 is a solution to the differential equation, this last equation reduces to a first order equation in v :

0)()( ytqytpy

)()()()(2)()()(

)()()()()(

)()()(

1112

112

12

tytvtytvtytvty

tytvtytvty

tytvty

02 111111 vqyypyvpyyvy

02 111 vpyyvy

Example 3: Reduction of Order (1 of 3)

Given the variable coefficient equation and solution y1,

use reduction of order method to find a second solution:

Substituting these into the ODE and collecting terms,

,)(;0,032 11

2 ttytyytyt

3212

212

12

)(2 )(2 )()(

)( )()(

)()(

ttvttvttvty

ttvttvty

ttvty

)()( where,02

02

033442

03222111

1213212

tvtuuut

vvt

vtvtvvtvtv

vtvttvtvttvtvt

Example 3: Finding v(t) (2 of 3)

To solve

for u, we can use the separation of variables method:

Thus

and hence

.0 since,

ln2/1ln2

102

2/12/1

tctuetu

Ctudttu

duu

dt

dut

C

)()(,02 tvtuuut

2/1ctv

ktctv 2/33/2)(

Example 3: General Solution (3 of 3)

Since

Recall

So we can neglect the second term of y2 to obtain

The Wronskian of can be computed

Hence the general solution to the differential equation is

12/112/32 3/23/2)( tktctktcty

2/12 )( tty

11 )( tty

2/12

11)( tctcty

ktctv 2/33/2)(

)(and)( 21 tyty

0,02/3))(,( 2/321 tttyyW