Using equation (3) weget
"Y
3
+(1
;
3
"
)
Y
2
+2
"Y
=0
If we compare the p owers in
"
it is obvious that wehavetochoose an asymptotic
expansion in the form
X
=
;
1+
"
1
=
2
X
1
+
:::
Furthermore wehave the relation
X
2
1
;
2=0
This means that wehavetwo solutions
X
(1)
1
=
;
p
2
X
(2)
2
=
p
2
It follows that the rst two terms of the rst two ro ots of equation a) are given by
x
(1)
=
;
1
;
p
2
"
1
=
2
+
O
(
"
)
x
(2)
=
;
1+
p
2
"
1
=
2
+
O
(
"
)
Consider now the scaling
=
1
"
.
We end up with the equation
X
3
+
X
2
+(2+
"
)
X
+
"
2
=0
and
X
=
;
1+
Y
(
"
)
The equation for
Y
(
"
) reads
Y
3
+2
Y
2
+(1+2
"
)
Y
;
2
"
=0
If we compare the powers in
"
we get
X
=
;
1+2
"
+
O
(
"
2
)
Therefore the rst two terms of the third root are given by
x
(3)
=
;
1
"
+2+
O
(
"
2
)
Case b):
Weinvestigate again the scaling
1, where the corresponding equation is given by
"X
3
+
X
2
+(2
;
"
)
X
+1 = 0
9
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