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Exercises 1.4
for
n
=1,
t
2
=
t
1
+0
.
1=0
.
2
,T
2
=
T
1
.
1(70
−
T
1
)=97+0
.
1(70
−
97) = 94
.
3;
for
n
=2,
t
3
=
t
2
.
.
3
3
=
T
2
.
1(70
−
T
2
)=94
.
3+0
.
1(70
−
94
.
3) = 91
.
87
.
Table 1–C
: Euler’s method approximations for the solutions of
T
0
=
K
(
M
−
T
),
T
(0) = 100, with
K
M
= 70, and
h
=0
.
1.
n
t
n
T
n
n
t
n
T
n
0
0.0
100.00
11
1.1
79.414
1
0.1
97.000
12
1.2
78.473
2
0.2
94.300
13
1.3
77.626
3
0.3
91.870
14
1.4
76.863
4
0.4
89.683
15
1.5
76.177
5
0.5
87.715
16
1.6
75.559
6
0.6
85.943
17
1.7
75.003
7
0.7
84.349
18
1.8
74.503
8
0.8
82.914
19
1.9
74.053
9
0.9
81.623
20
2.0
73.647
10
1.0
80.460
By continuing this way and rounding results to three decimal places, we Fll in Table 1-C.

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