ch 4.4: variation of parameters
DESCRIPTION
Ch 4.4: Variation of Parameters. The variation of parameters method can be used to find a particular solution of the nonhomogeneous n th order linear differential equation provided g is continuous. - PowerPoint PPT PresentationTRANSCRIPT
Ch 4.4: Variation of Parameters
The variation of parameters method can be used to find a particular solution of the nonhomogeneous nth order linear differential equation
provided g is continuous.
As with 2nd order equations, begin by assuming y1, y2 …, yn are fundamental solutions to homogeneous equation.
Next, assume the particular solution Y has the form
where u1, u2,… un are functions to be solved for.
In order to find these n functions, we need n equations.
),()()()( 1)1(
1)( tgytpytpytpyyL nn
nn
)()()()()()()( 2211 tytutytutytutY nn
Variation of Parameters Derivation (2 of 5)
First, consider the derivatives of Y:
If we require
then
Thus we next require
Continuing in this way, we require
and hence
nnnn yuyuyuyuyuyuY 22112211
02211 nn yuyuyu
nnnn yuyuyuyuyuyuY 22112211
02211 nn yuyuyu
1,,1,0)1()1(2
)1(11 2
nkyuyuyu kn
kk
n
1,,1,0,)()(11
)( nkyuyuY knn
kk
Variation of Parameters Derivation (3 of 5)
From the previous slide,
Finally,
Next, substitute these derivatives into our equation
Recalling that y1, y2 …, yn are solutions to homogeneous equation, and after rearranging terms, we obtain
)()(11
)1()1(11
)( nnn
nnnn
nn yuyuyuyuY
1,,1,0,)()(11
)( nkyuyuY knn
kk
gyuyu nnn
n )1()1(11
)()()()( 1)1(
1)( tgytpytpytpy nn
nn
Variation of Parameters Derivation (4 of 5)
The n equations needed in order to find the n functions u1, u2,… un are
Using Cramer’s Rule, for each k = 1, …, n,
and Wk is determinant obtained by replacing k th column of W with (0, 0, …, 1).
gyuyu
yuyu
yuyu
nnn
n
nn
n
)1()1(11
11
111
0
0
))(,,()( where,)(
)()()( 1 tyyWtW
tW
tWtgtu n
kk
Variation of Parameters Derivation (5 of 5)
From the previous slide,
Integrate to obtain u1, u2,… un:
Thus, a particular solution Y is given by
where t0 is arbitrary.
nktW
tWtgtu k
k ,,1,)(
)()()(
nkdssW
sWsgtu
t
t
kk ,,1,
)(
)()()(
0
n
kk
t
t
k tydssW
sWsgtY
1
)()(
)()()(
0
Example (1 of 3)
Consider the equation below, along with the given solutions of corresponding homogeneous solutions y1, y2, y3:
Then a particular solution of this ODE is given by
It can be shown that
tttt etytetyetyeyyyy )(,)(,)(, 3212
3
1
2
)()(
)()(
0kk
t
t
ks
tydssW
sWetY
t
ttt
ttt
ttt
e
eete
eete
etee
tW 4
2
1)(
Example (2 of 3)
Also,
t
tt
tt
tt
tt
tt
tt
tt
tt
tt
e
ete
ete
tee
tW
ee
ee
ee
tW
t
eet
eet
ete
tW
12
01
0
)(
2
1
0
0
)(
12
21
10
0
)(
3
2
1
Example (3 of 3)
Thus a particular solution is
Choosing t0 = 0, we obtain
More simply,
t
t
st
t
t
st
t
t
st
t
t s
sstt
t s
stt
t s
st
kk
t
t
ks
dsee
dsete
dssee
dse
eeeds
e
eteds
e
see
tydssW
sWetY
000
000
0
4
2222
3
1
2
4212
4
44
2
4
12
)()(
)()(
tttt eeteetY 2
3
1
12
1
2
1
4
1)(
tetY 2
3
1)(