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