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you know that:
ln
ln
ycxcx
x
xxxx
2
12
22
=+



^^
hh
In fact, you can absorb the final –
x
term into
c
1
x
, giving you the general
solution:
ln
ln
x
xxx
2


What a pair! The variation of parameters
method meets the Wronskian
As noted in the previous sections, the method of variation of parameters
allows you to tackle linear differential equations, such as this second order
differential equation:
y"
+
p
(
x
)
y'
+
q
(
x
)
y
=
g
(
x
)
The method of variation of parameters relies on the solution to the homoge
neous equation:
y"
+
p
(
x
)
y'
+
q
(
x
)
y
= 0
The solution to the homogeneous equation is:
y
h
=
c
1
y
1
(
x
) +
c
2
y
2
(
x
)
The method of variation of parameters says that you then try to find a partic
ular solution of the following form:
y
p
=
u
1
(
x
)
y
1
(
x
) +
u
2
(
x
)
y
2
(
x
)
Substituting
y
p
into the differential equation gives you these two equations:
u'
1
(
x
)
y
1
(
x
) +
u'
2
(
x
)
y
2
(
x
) = 0
u'
1
(
x
)
y
1
'
(
x
) +
u'
2
(
x
)
y
2
'
(
x
) =
g
(
x
)
You can formally solve these equations for
u'
1
(
x
) and
u'
2
(
x
) as follows:
ux
yxy x y xyx
yxgx
1
1
2
2
=


l
ll
^
^
^
h
h
h
and:
2
1
2
1
=

l
^
^
^
h
h
h
142
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 Spring '11
 DELALLAVE
 Differential Equations, Equations

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