18.03 Problem Set 2: Solutions
4. (W 15 Feb)
[Part I problems: 6 pts.]
(a)
[4]
x
decreases exponentially to zero.
y
grows linearly (with slope 1
/
2) at first, then levels off and finally decays exponentially to
zero.
z
grows quadratically at first, then levels off and converges exponentially to value
1 as
t
→ ∞
.
(b)
[5] The decay constant
l
of an exponential decay is related to half life
t
1
/
2
by
1
2
=
e

lt
1
/
2
, or ln 2 =
lt
1
/
2
.
Thus the decay constant of Lu is
1
2
.
I’ll write
l
for the decay
constant of Lu because the formulas are clearer with this symbol than with
1
2
. So ˙
x
+
lx
=
0, ˙
y
+
ky
=
lx
, ˙
z
=
ky
.
(c)
[5]
x
=
x
(0)
e

lt
. Using
x
(0) = 1 we find
x
=
e

lt
. Thus
y
+
ky
=
le

lt
. Solve this
using the integrating factor
e
kt
:
d
dt
(
e
kt
y
) =
le
(
k

l
)
t
, so
e
kt
y
=
le
(
k

l
)
t
/
(
k

l
)+
c
(assuming
k
6
=
l
, at least), or
y
=
le

lt
/
(
k

l
) +
ce

kt
. To find
c
, evaluate at
t
= 0: 0 =
l
k

l
+
c
, so
y
=
l
e

lt

e

kt
k

l
. Thus
z
=
kl
k

l
e

lt

l

e

kt

k
+
c
. To find the constant of integration here,
evaluate at
t
= 0 again: 0 =
kl
k

l
(
1

l

1

k
)
+
c
, or
c
= 1:
z
=
kl
k

l
e

kt
k

e

lt
l
+ 1.
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 Spring '09
 unknown
 Exponential Function, Radioactive Decay, HalfLife, Cos, Exponential decay

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