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Exercises 3.2
An integrating factor for this equation has the form
µ
(
t
)=exp
±Z
4
100
−
t
dt
²
=
e
−
4 ln(100
−
t
)
= (100
−
t
)
−
4
.
Multiplying the linear equation by the integrating factor yields
(100
−
t
)
−
4
dx
dt
+4
x
(100
−
t
)
−
5
=(1
.
2)(100
−
t
)
−
4
⇒
D
t
³
(100
−
t
)
−
4
x
´
=(1
.
2)(100
−
t
)
−
4
⇒
(100
−
t
)
−
4
x
=1
.
2
Z
(100
−
t
)
−
4
dt
=
1
.
2
3
(100
−
t
)
−
3
+
C
⇒
x
(
t
)=(0
.
4)(100
−
t
)+
C
(100
−
t
)
4
.
To Fnd the value of
C
, we use the initial condition
x
(0) = 1. Therefore,
x
(0) = (0
.
4)(100) +
C
(100)
4
=1
⇒
C
=
−
39
100
4
=
−
3
.
9
×
10
−
7
.
This means that at time
t
there is
x
(
t
)=(0
.
4)(100
−
t
)
−
(3
.
9
×
10
−
7
)(100
−
t
)
4
liters of nitric acid in the tank. When the percentage of nitric acid in the tank is 10%, the
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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