Chapter 5
“output
A
” consists of two ﬂows: one is going out of the system and the other one is going to
the tank B. So,
output
A
=
x
(
t
)
100
kg
/
L
·
(4 + 3) L
/
min =
7
x
(
t
)
100
kg
/
min
,
and the Frst equation in (5.17) becomes
dx
dt
=1
.
2+
y
100
−
7
x
100
.
Similarly, the second equation in (5.17) can be written as
dy
dt
=
3
x
100
−
3
y
100
.
Rewriting this system in the operator form, we obtain
(
D
+0
.
07)[
x
]
−
0
.
01
y
=1
.
2
,
−
0
.
03
x
+(
D
+0
.
03)[
y
]=0
.
(5.18)
Eliminating
y
yields
{
(
D
+0
.
07)(
D
+0
.
03)
−
(
−
0
.
01)(
−
0
.
03)
}
[
x
]=(
D
+0
.
03)[1
.
2] = 0
.
036
,
which simpliFes to
(
D
2
+0
.
1
D
+0
.
0018
)
[
x
]=0
.
036
.
(5.19)
The auxiliary equation,
r
2
+0
.
1
r
+0
.
0018 = 0, has roots
r
1
=
−
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations, Trigraph, corresponding homogeneous equation, ﬁrst equation

Click to edit the document details