y = ce −x x 2 + y 2 = 2 cx y = c sen x y = 1 lncx N 0 t = 1 3 2 N 0 P t P 2 P 1000 5 98 ◦ C 21 ◦ C 10 88 ◦ C 35 ◦ C 50 ◦ F 100 ◦ F 4 60 ◦ F 75 ◦ F 20 30 ◦ F 10 0 ◦ F 20 15 ◦ F 100 (l) 100 k g 5 l /s
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7/17/2019 Problemario aplicaciones de ecuaciones diferenciales
http://slidepdf.com/reader/full/problemario-aplicaciones-de-ecuaciones-diferenciales 1/2
y = ce−x
x2 + y2 = 2cx
y = c sen x y = 1
lncx
N 0
t = 1
32N 0
P t
P 2
P
1000
5
98C
21C
10
88C
35C 50F 100F
4
60F
75F
20
30F
10
0F
20 15F
100 (l) 100 kg
5 l/s
7/17/2019 Problemario aplicaciones de ecuaciones diferenciales
http://slidepdf.com/reader/full/problemario-aplicaciones-de-ecuaciones-diferenciales 2/2
10 kg
500
300
50
3
4gal/min
60
3 lb
2 gal/min 2.5 gal/min
t 30 g al
1636 km3
0.25%
209 km3
0.05%
0.10%
32 m3
t = 0
4%
0.002 m3/min
x (t)
t > 0
0.00012
500 gal
100 g al
50 lb
t = 0
2 lb
5 gal/s
3 gal/s
100 lt
40 kg
3 kg/l 2 l/min
B
100 lt
0.1 kg/l
B 1 l/min
0.1 H 50 Ω
30 V
i (t)
i (t) = 0
200 Ω
10−4F
100 V
q (0) = 0
q (t)
i (t)
L t L = 0.5 + 0.001t
40 V
10 Ω
i (t) t > 0
Ω
5 H
20 A t = 0 t > 0
t > 0
5x10−3
F
25 Ω
50cos6t
t > 0 t = 0
20 Ω 0.01 F
40e−3t+ 20e−6t
q (0) = 0
0.25
R E L
Lln(2)R
s
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