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Variation of the Parameters
This is another method we can apply to find a particular solution of a nonhomogeneous
L.D.E.
.
We will develop the method in connection with a second order linear O.D.E. with
variables coefficients,
a
0
(x)y’’ + a
1
(x) y’ + a
2
(x)y = b(x) (1)
Suppose that y
1
(x) and y
2
(x) are linearly independent solutions of the corresponding
homogeneous equation,
a
0
(x)y’’ + a
1
(x) y’ + a
2
(x)y = 0
then the complementary solution of equation (1) is:
y
c
(x) = c
1
y
1
(x) + c
2
y
2
(x)
where c
1
and c
2
are arbitrary constants.
Replace the arbitrary constants c
1
and c
2
in the complementary solution by two functions
v
1
(x) and v
2
(x) to be determined such that
y
p
(x) = v
1
(x)y
1
(x) + v
2
(x)y
2
(x)
is a particular solution of equation (1).
This is the first condition new impose in order to determine v
1
(x) and v
2
(x). Since there
are two functions, we still have a second condition to impose, provided that this second
condition does not violate the first condition.
Let’s differentiate the function y
p
:
y
p
’ = v
1
(x) y
1
’(x) + v
1
’(x)y
1
(x) + v
2
(x) y
2
’(x) + v
2
’(x)y
2
(x).
At this point we impose the second condition. We simplify y
p
’ by demanding
v
1
’(x)y
1
(x) + v
2
’(x)y
2
(x) = 0
So, y
p
’ becomes
y
p
’ = v
1
(x) y
1
’(x) + v
2
(x) y
2
’(x)
And the second derivative is
y
p
’’ = v
1
(x) y
1
’’(x) + v
1
’(x)y
1
’(x) + v
2
(x) y
2
’’(x) + v
2
’(x)y
2
’(x).
Substituting in equation(1)
a
0
(x) [v
1
(x) y
1
’’(x) + v
1
’(x)y
1
’(x) + v
2
(x) y
2
’’(x) + v
2
’(x)y
2
’(x)] + a
1
(x)[ v
1
(x) y
1
’(x) +
v
2
(x) y
2
’(x)] + a
2
(x) [v
1
(x)y
1
(x) + v
2
(x)y
2
(x)] = b(x)
or
v
1
(x)[ a
0
(x)y
1
’’(x) + a
1
(x) y
1
’(x) + a
2
(x)y
1
(x)] + v
2
(x)[ a
0
(x)y
2
’’(x) + a
1
(x) y
2
’(x) +
a
2
(x)y
2
(x)] + a
0
(x)[ v
1
’(x)y
1
’(x) + v
2
’(x)y
2
’(x)] = b(x)
Since y
1
and y
2
are solutions of the corresponding homogeneous equation, the first two
terms on the left side of the equation are zero.
Then, a
0
(x)[ v
1
’(x)y
1
’(x) + v
2
’(x)y
2
’(x)] = b(x)
or
v
1
’(x)y
1
’(x) + v
2
’(x)y
2
’(x) = b(x)/ a
0
(x)
So, the two imposed conditions have created a system of two equations that the
derivatives of the two unknown functions v
1
and v
2
are satisfying.
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1
’(x)y
1
(x) + v
2
’(x)y
2
(x) = 0
v
1
’(x)y
1
’(x) + v
2
’(x)y
2
’(x) =
b(x)
a
0
(x)
!
"
#
$
#
The determinant of the system is the Wronskian of y
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This note was uploaded on 07/23/2011 for the course MAP 2302 taught by Professor Staff during the Fall '08 term at FIU.
 Fall '08
 STAFF

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