Exercises 2.2
Therefore,
dA
dt
= 3
−
A
40
=
120
−
A
40
.
Separating this differential equation and integrating yield
40
120
−
A
dA
=
dt
⇒
−
40 ln

120
−
A

=
t
+
C
⇒
ln

120
−
A

=
−
t
40
+
C,
where
−
C
40
is replaced by
C
⇒
120
−
A
=
Ce
−
t/
40
,
where
C
can now be positive or negative
⇒
A
= 120
−
Ce
−
t/
40
.
There are 2 kg of salt in the tank initially, thus
A
(0) = 2. Using this initial condition, we find
2 = 120
−
C
⇒
C
= 118
.
Substituting this value of
C
into the solution, we have
A
= 120
−
118
e
−
t/
40
.
Thus
A
(10) = 120
−
118
e
−
10
/
40
≈
28
.
1 kg
.
Note: There is a detailed discussion of mixture problems in Section 3.2.
35.
In Problem 34 we saw that the differential equation
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations, Harshad number, Nontotient, differential equation dt

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