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Parameter sensitivity equation
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Past exam of the mathematics course of the University of Cambridge
/
2023
/
ia
/
Paper 2
/
6A
/
Solution
Created
2026-09-23
Updated
2026-10-03
View more
The
parameter sensitivity equation
follows by differentiating with
respect
to
μ
and
writing
u
=
y
μ
:
u
x
=
u
+
x
+
y
2
+
2
μ
y
u
,
u
(
0
,
μ
)
=
0.
(1)
At
μ
=
0
,
y
x
=
y
,
y
(
0
)
=
1
, so
y
(
x
,
0
)
=
e
x
. Then
u
x
−
u
=
x
+
e
2
x
,
u
(
0
,
0
)
=
0
,
(2)
and the
integrating factor
e
−
x
yields
u
(
x
,
0
)
=
e
2
x
−
x
−
1.
(3)
Equating
powers
in
y
=
y
0
+
μ
y
1
+
μ
2
y
2
+
⋯
gives
y
0
=
e
x
,
y
1
=
e
2
x
−
x
−
1
,
(4)
y
2
′
−
y
2
=
2
y
0
y
1
,
y
2
(
0
)
=
0.
(5)
Consequently
y
2
=
e
x
(
e
2
x
−
1
−
x
2
−
2
x
)
.
(6)
Total
articles
:
1