Section 7.3, Page 527
Z
dz
dz
+ et+z = 0 ⇒
= −et ez ⇒ e−z dz = −
dt
dt
− z = ln(et − C) ⇒ z = − ln(et − C).
• 8.
Z
et dt ⇒ −e−z = −et + C ⇒ e−z = et − C ⇒
• 34.
– (a) Use 1 billion dollars as the x−unit and 1 day as the t−unit. Initially, there is $10 billion
of old currency in circulation, so all of the $50 million returned to the banks is old. At time t,
the amount of new currency is x(t) billion dollars, so 10 − x(t) billion dollars of currency is old.
The fraction of circulating money that is old is (10 − x(t))/10, and the amount of old currency
(10 − x(t))
0.05 billion dollars. This amount of new currency
returned to the banks each day is
10
dx
(10 − x(t))
per day is introduced into circulation, so
=
0.05 = 0.005(10 − x) billion dollars
dt
10
per day.
Z
Z
dx
dx
= 0.005dt ⇒
= 0.005dt ⇒ − ln |10 − x| = 0.005t + c ⇒
– (b)
10 − x
10 − x
ln |10 − x| = −0.005t − c ⇒ 10 − x = Ce−0.005t , where C = e−c ⇒ x = 10 − Ce−0.005t . From
x(0) = 0, we get C = 10, so x = 10 − 10e−0.005t = 10(1 − e−0.005t )
– (c) The new bills make up 90% of the circulating currency when x(t) = 0.9·10 = 9billion dollars.
9 = 10(1 − e−0.005t ) ⇒ 0.9 = 1 − e−0.005t ⇒ e−0.005t = 0.1 ⇒ −0.005t = ln(0.1) ⇒
ln 0.1
≈ 465.517 ≈ 466days.
t=
−0.005
• 36.
– (a) If y(t) is the amount of salt in (kg) after t minutes, then y(0) = 0 and the total amount of
liquid in the tank remains constant at 1000L.
dy
dt
It follows that
kg
L
L
= (0.05 kg
L )(5 min ) + (0.04 M )(10 min ) − (
= 0.25 + 0.40 − 0.015y
= 0.65 − 0.015y
130 − 3y kg
=
.
200 min
Z
dy
=
130 − 3y
and
Z
y(t) kg
L
)(15
)
1000 L
min
dt
200
1
t
− ln |130 − 3y| =
+C.
3
200
1
1
t
1
Since y(0) = 0, we have C = − ln 130, so − ln |130 − 3y| =
− ln 130 ⇒
3
3
200 3
3t
3t
|130 − 3y|
3t
ln |130 − 3y| = −
+ ln 130 ⇒ ln |130 − 3y| − ln 130 = −
⇒ ln(
) = −
⇒
200
200
130
200
3t
3t
|130 − 3y|
= e− 200 ⇒ |130−3y| = 130e− 200 . Since y is continuous, y(0) = 0, and the right-hand
130
side is never zero, we deduce that 130 − 3yand the right-hand side is never zero, we deduce that
3t
3t
130
130− 3y is always positive. thus, 130− 3y > 0. Thus, 130− 3y = 130e− 200 ⇒ y =
(1− e− 200 )
3
kg
1
3·(60)
130
(1 − e− 200 ) ≈ 25.7kg.
3
130
1
= 43 kg.
Note: As t → ∞, y(t) →
3
3
– (b) After one hour, y =
2