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Unformatted text preview: and all the other an ’s and bn ’s are zero, so both of these
solutions are generalized power series solutions.
17 On the other hand, if this method is applied to the diﬀerential equation
x2 d2 y
dy
+ y = 0,
+ 3x
2
dx
dx we obtain
r(r − 1) + 3r + 1 = r2 + 2r + 1,
which has a repeated root. In this case, we obtain only a oneparameter family
of solutions
y = cx−1 .
Fortunately, there is a trick that enables us to handle this situation, the socalled
method of variation of parameters. In this context, we replace the parameter c
by a variable v (x) and write
y = v (x)x−1 .
Then
d2 y
= v (x)x−1 − 2v (x)x−2 + 2v (x)x−3 .
dx2 dy
= v (x)x−1 − v (x)x−2 ,
dx Substitution into the diﬀerential equation yields
x2 (v (x)x−1 − 2v (x)x−2 + 2v (x)x−3 ) + 3x(v (x)x−1 − v (x)x−2 ) + v (x)x−1 = 0,
which quickly simpliﬁes to yield
xv (x) + v (x) = 0, v
1
=− ,
v
x log v  = − log x + a, v= c2
,
x where a and c2 are constants of integration. A further integration yields
v = c2 log x + c1 , y = (c2 log x...
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This document was uploaded on 01/12/2014.
 Winter '14
 Equations

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