j. The DE
dy
dt
+
y
=
y
3
e
t
is a Bernoulli’s equation, where we make the substitution
u
=
y
1

3
=
y

2
, so
du
dt
=

2
y

3
dy
dt
.
Multiplying the above equation by

2
y

3
, we obtain the linear DE in
u
(
t
)

2
y

3
dy
dt

2
y

2
=

2
e
t
or
du
dt

2
u
=

2
e
t
.
This has the integrating factor
μ
(
t
) =
e

2
t
, so
d
dt
e

2
t
u
(
t
)
=

2
e

t
or
e

2
t
u
(
t
) = 2
e

t
+
C.
It follows that
1
y
2
(
t
)
=
u
(
t
) = 2
e
t
+
Ce
2
t
,
so
1 = 2 +
C
or
C
=

1
.
Thus,
y
(
t
) =
1
√
2
e
t

e
2
t
.
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2. a. The solution to the white lead problem is
P
(
t
) = 10
e

kt
, where
t
= 0 represents 1970. From
the data at 1975, we have 8
.
5 = 10
e

5
k
or
e
5
k
= 10
/
8
.
5 = 1
.
17647. Thus,
k
= 0
.
032504 yr

1
. To
find the halflife, we compute 5 = 10
e

kt
, so
t
= ln(2)
/k
= 21
.
33 yr is the halflife of lead210.
b. The differential equation can be written
P
0
=

k
(
P

r/k
), so we make the substitution
z
(
t
) =
P
(
t
)

r/k
. This leaves the initial value problem
z
0
=

kz,
z
(0) =
P
(0)

r/k
= 10

r/k,
which has the solution
z
(
t
) = (
P
(0)

r/k
)
e

kt
=
P
(
t
)

r/k
. Thus, the solution is
P
(
t
) =
10

r
k
e

kt
+
r
k
= 2
.
3086
e

kt
+ 7
.
6914
,
where
k
= 0
.
032504. In the limit,
lim
t
→∞
P
(
t
) = 7
.
6914 disintegrations per minute of
210
Pb
.
3. a. The differential equation describing the temperature of the tea satisfies
H
0
=

k
(
H

21)
,
H
(0) = 85 and
H
(5) = 81
.
Make the substitution
z
(
t
) =
H
(
t
)

21, which gives the differential equation
z
0
=

kz,
z
(0) =
H
(0)

21 = 64
.
The solution becomes
z
(
t
) = 64
e

kt
=
H
(
t
)

21 or
H
(
t
) = 64
e

kt
+ 21
.
To find
k
, we solve
H
(5) = 81 = 64
e

5
k
+ 21 or
e
5
k
= 64
/
60 = 1
.
0667. Thus,
k
= 0
.
012908 min

1
.
The water was at boiling point when 64
e

kt
+ 21 = 100 or
e

kt
= 79
/
64.
It follows that
t
=

ln(79
/
64)
/k
=

16
.
3 min. This means that the talk went 16.3 min over its scheduled ending.
b. To obtain a temperature of at least 93
◦
C, then we need to find the time that satisfies
H
(
t
) =
93 = 64
e

kt
+ 21, so
e

kt
= 72
/
64 = 1
.
125. Solving for
t
gives
t
=

ln(72
/
64)
/k
=

9
.
125 min. It
follows that you must arrive at the hot water within 16
.
3

9
.
1 = 7
.
2 min of the scheduled end of
the talks.
4. a. Substituting the parameters into the differential equation gives
c
0
=
1
10
6
(22000

2000
c
) =

0
.
002(
c

11)
.
We make the substitution
z
(
t
) =
c
(
t
)

11, which gives the initial value problem
z
0
=

0
.
002
z
with
z
(0) =
c
(0)

11 =

11. The solution of this differential equation is
z
(
t
) =

11
e

0
.
002
t
=
c
(
t
)

11,
so
c
(
t
) = 11

11
e

0
.
002
t
.
b. Solve the equation
c
(
t
) = 11

11
e

0
.
002
t
= 5, so
e
0
.
002
t
= 11
/
6 or
t
= 500 ln(11
/
6) =
303
.
1 days. The limiting concentration
lim
t
→∞
c
(
t
) = 11
.
The graph is below.
0
200
400
600
800
1000
0
2
4
6
8
10
12
t
c
(
t
)
Problem 4
5. The differential equation is separable, so write
Z
T

1
2
dT
=
k
Z
dt
or
2
T
1
2
(
t
) =
kt
+
C.
It follows that
T
(
t
) =
kt
+
C
2
2
.
The initial condition
T
(0) = 1 implies
C
= 2, so
T
(
t
) =
kt
2
+ 1
2
. Since
T
(4) =
4
k
2
+ 1
2
= 25,
2
k
+ 1 = 5 or
k
= 2. Thus, the solution for the spread of the disease in this orchard is
T
(
t
) = (
t
+ 1)
2
.
When
t
= 10,
T
(10) = 121.
6. The differential equation with the information in the problem is given by:
dH
dt
=

k
(
H

25)
,
H
(0) = 35
,
where
t
= 0 is 7 AM. We make the change of variables
z
(
t
) =
H
(
t
)

25, so
z
(0) = 10. The problem
now becomes
dz
dt
=

kz,
z
(0) = 10
,
which has the solution
z
(
t
) = 10
e

kt
or
H
(
t
) = 25 + 10
e

kt
.
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From the information at 9 AM, we see
H
(2) = 33
.
5 = 25 + 10
e

2
k
or
e
2
k
=
10
8
.
5
or
k
=
ln
10
8
.
5
2
= 0
.
081259
.
 Fall '08
 staff