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Final 2007 reference solution
1. (a) This is a separable equation:
e

2
y
dy
=

e

2
x
dx,

1
2
e

2
y
ﬂ
ﬂ
ﬂ
ﬂ
y
0
=
1
2
e

2
x
ﬂ
ﬂ
ﬂ
ﬂ
x
0
,
y
=

1
2
log(2

e

2
x
)
.
(b) From (a) we see that
y
exists if and only if 2

e

2
x
>
0
,
i.e.
x >

(log 2)
/
2
.
2. (a) The ﬁrst order equation can be rewritten as
y
0
+ (1 +
1
x
)
y
=
e

x
,
thus we get
μ
(
x
) = exp
Z
x
1
(1 +
1
x
)
dx
=
xe
x

1
,
y
(
x
) =
x

1
e
1

x
Z
x
1
ξe
ξ

1
e

ξ
dξ
=
1
2
e

x
(
x

x

1
)
.
(b) Clearly
y
(
x
)
→
0 as
x
→ ∞
, thus the maximum will be obtained at some point
x
where
y
0
(
x
) = 0
,
i.e.
y
0
(
x
) =
1
2
e

x
(1 +
x

2

x
+
x

1
) = 0
,
from which we derive a cubic equation
x
3

x
2

x

1 = 0
.
3. By the hint one has
F
=
mv
dv
dx
=

kv,
mdv
=

kdx,
Z
0
v
0
mdv
=

k
Z
x
0
dx,
x
=
mv
0
k
.
4. (a) The characteristic equation is
r
2
+ 4
r
+ 5 = 0
,
which has two roots
r
=

2
±
i,
thus the
general solution is
x
=
e

2
t
(
A
cos
t
+
B
sin
t
)
.
Apply the conditions
x
(0) = 2, ˙
x
(0) =

5, we get
A
= 2,
B
=

1
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 Spring '09
 T.Qian

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