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# solving differential equation by variation of parameters

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5577-6.4-9E AID: 2058 | 28/09/2012
If
A
gas a repeated eigenvalue
1 2
λ λ
=
with only one corresponding (linearly independent)
eigenvector
1
v
, two linearly independent solutions of
'
X AX=
are
1
1 1
t
X v e
λ
=
and
( )
1
2 1 2
t
X v t w e
λ
= +
, where
2
w
satisfies equation
( )
1 2 1
A I w v
λ
=
The eigenvalues of
are
1,2
2
λ
=
,and
3
3
λ
=
with corresponding eigenvectors
1
0
1
0
v
÷
=
÷
÷
,
2
1
0
0
v
÷
=
÷
÷
Using these above values we have the solution of the system
2
1
0
( ) 1
0
t
X t e
÷
=
÷
÷
2
2 2
0
( ) 1
0
t
X t t w e
÷
÷
= +
÷
÷
÷
÷
……
( )
1
To find
2
2 2
2
x
w y
z
÷
=
÷
÷
in a second linearly independent solution of the form given by Eq.
( )
1
,
we solve
( )
1 2 1
A I w v
λ
=
2
2
2
3 0 0 1 0 0 0
0 2 1 2 0 1 0 1
0 0 2 0 0 1 0
x
y
z
÷
÷ ÷ ÷ ÷
=
÷
÷ ÷ ÷ ÷
÷ ÷ ÷ ÷
÷
2
2
2
1 0 0 0
0 0 1 1
0 0 0 0
x
y
z
÷ ÷ ÷
=
÷ ÷ ÷
÷ ÷ ÷
Which indicates that
2
0x =
,
2
1z =
and
2
0y =
Therefore the second linearly independent solution of the system is
2
2
0 0
( ) 1 0
0 1
t
X t t e
÷
÷ ÷
= +
÷
÷ ÷
÷ ÷
÷
The third linearly independent solution of the system corresponding to eigenvector
2
v
is
3
3
1
( ) 0
0
t
X t e
÷
=
÷
÷
Hence, the general solution of the system in matrix form is
3 2 2
3
2 2
2
1 0 0 0
( ) 0 1 1 0
0 0 0 1
0 0
0
0 0
t t t
t
t t
t
X t e e t e
e
e te C
e
÷
÷ ÷ ÷ ÷
= + + +
÷
÷ ÷ ÷ ÷
÷ ÷ ÷ ÷
÷
÷
=
÷
÷
Therefore the general solution is
3
2 2
2
0 0
0
0 0
t
t t
t
e
X e te C
e
÷
=
÷
÷

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5577-6.4-9E AID: 2058 | 28/09/2012 If gas a repeated eigenvaluewith only one corresponding (linearly independent) eigenvector , two linearly independent solutions of are and , where satisfies equationThe eigenvalues ofare,andwith corresponding eigenvectors, Usi ...
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