1 Solutions
27
and simplifying gives
y
= (5+
t
)+
c
(5+
t
)
1
. The initial condition
y
(0) = 0
implies
c
=
25
so
y
= (5+
t
)
25(5+
t
)
1
;
for
0
t
45
. At
t
=
T
= 45
we get
y
(45) = 50
25
50
= 49
:
5
lbs salt. Once the tank is full, the inflow
and outflow rates will be equal and the brine in the tank will (in the
limit as
t
! 1
) stabilize to the concentration of the incoming brine, i.e.,
0
:
5
lb/gal. Since the tank holds 100 gal, the total amount present will
approach
0
:
5
100 = 50
lbs. Thus
lim
t
!1
y
(
t
) = 50
.
31. input rate:
input rate
=
rc
output rate:
output rate
=
r
P
(
t
)
V
Let
P
0
denote the amount of pollutant at time
t
= 0
. Since
P
0
=
input rate
output rate it follows that
P
(
t
)
is a solution of the initial
value problem
P
0
=
rc
rP
(
t
)
V
;
P
(0) =
P
0
:
Rewriting this equation in standard form gives the differential equation
P
0
+
r
V
P
=
rc
. The coefficient function is
p
(
t
) =
r=V
and the integrating
factor is
(
t
) =
e
rt=V
. Thus
(
e
rt
V
P
)
0
=
rce
rt
V
. Integrating and simplifying
gives
P
(
t
) =
cV
+
ke
rt
V
;
where
k
is the constant of integration. The initial
condition
P
(0) =
P
0
implies
c
=
P
0
cV
so
P
(
t
) =
cV
+ (
P
0
cV
)
e
rt
V
:
(a)
lim
t
!1
P
(
t
) =
cV:
(b) When the river is cleaned up at
t
= 0
we assume the input con
centration is
c
=
0
. The amount of pollutant is therefore given by
P
(
t
) =
P
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 Fall '08
 BELL,D
 Constant of integration, Lake Erie, Lake Ontario, output rate, 100 gal

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