Euler y Runge Kutta

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Euler y Runge Kutta 
Realiza de manera ordenada tu proceso. Analiza si con los valores dados de n es posible aplicar el método 
indicado. De no ser posible sólo indica que no es posible y explica tu conclusión. 
1. Apliqué el método de Euler para aproximar las soluciones de los siguientes problemas de valor 
inicial y comparé después los resultados con los valores reales. 
a) 𝒚′ = 𝒙𝒆𝟑𝒙 − 𝟐𝒚 𝟎 ≤ 𝒙 ≤ 𝟏 𝒚(𝟎) = 𝟎 𝒄𝒐𝒏 𝒉 = 𝟎. 𝟐 
La solución real es 𝑦 =
1
5
𝑥𝑒3𝑥 −
1
25
𝑒3𝑥 +
1
25
𝑒−2𝑥 → 𝑽𝒂𝒍𝒐𝒓 𝑹𝒆𝒂𝒍 = 𝟑. 𝟐𝟏𝟗𝟏𝟎 𝒄𝒐𝒏 𝑿 𝒆𝒗𝒂𝒍𝒖𝒂𝒅𝒂 𝒆𝒏 𝟏 
Formula 
𝒚𝒏+𝟏 = 𝒚𝒏 + 𝒉𝒇(𝒙𝒏, 𝒚𝒏) 
1er Iteración: 
𝑦𝑛+1 = 0 + (0.2)𝑓(0, 0) = 0 + (0.2)(0) = 0 
2da Iteración: 
𝑦𝑛+1 = 0 + (0.2)𝑓(0.2, 0) = 0 + (0.2)(0.36442) = 0.07288 
3er Iteración: 
𝑦𝑛+1 = 0.07288 + (0.2)𝑓(0.4, 0.07288) = 0.07288 + (0.2)(1.18228) = 0.30934 
4ta Iteración: 
𝑦𝑛+1 = 0.30934 + (0.2)𝑓(0.6, 0.30934) = 0.30934 + (0.2)(3.01111) = 0.91156 
5ta Iteración: 
𝑦𝑛+1 = 0.91156 + (0.2)𝑓(0.8, 0.91156) = 0.91156 + (0.2)(6.99542) = 2.31065 
 
 
 
𝒙𝒏 0.2 0.4 0.6 0.8 1 
𝒚𝒏 0 0.07288 0.30934 0.91156 2.31065 
𝒇(𝒙𝒏, 𝒚𝒏) 0.36442 1.18228 3.01111 6.99542 15.46425 
 
𝑬𝒓𝒓𝒐𝒓 𝑷𝒐𝒓𝒄𝒆𝒏𝒕𝒖𝒂𝒍 = |
𝑉. 𝑉𝑒𝑟𝑑𝑎𝑑𝑒𝑟𝑜 + 𝑉. 𝐴𝑝𝑟𝑜𝑥
𝑉. 𝑉𝑒𝑟𝑑𝑎𝑑𝑒𝑟𝑜
| × 100 = |
3.21910 + 2.31065
3.21910
| × 100 = 𝟐𝟖. 𝟐𝟐𝟎𝟕𝟓% 
 
b) 𝒚′ = 𝒙𝒆𝟑𝒙 − 𝟐𝒚 𝟎 ≤ 𝒙 ≤ 𝟏 𝒚(𝟎) = 𝟎 𝒄𝒐𝒏 𝒉 = 𝟎. 𝟏 
La solución real es 𝑦 =
1
5
𝑥𝑒3𝑥 −
1
25
𝑒3𝑥 +
1
25
𝑒−2𝑥 → 𝑽𝒂𝒍𝒐𝒓 𝑹𝒆𝒂𝒍 = 𝟑. 𝟐𝟏𝟗𝟏𝟎 𝒄𝒐𝒏 𝑿 𝒆𝒗𝒂𝒍𝒖𝒂𝒅𝒂 𝒆𝒏 𝟏 
Formula 
𝒚𝒏+𝟏 = 𝒚𝒏 + 𝒉𝒇(𝒙𝒏, 𝒚𝒏) 
1er Iteración: 
𝑦𝑛+1 = 0 + (0.1)𝑓(0,0) = 0 + (0.1)(0) = 0 
2da Iteración: 
𝑦𝑛+1 = 0 + (0.1)𝑓(0.1,0) = 0 + (0.1)(0.13499) = 0.01350 
3er Iteración: 
𝑦𝑛+1 = 0.01350 + (0.1)𝑓(0.2, 0.01350) = 0.01350 + (0.1)(0.33743) = 0.04724 
4ta Iteración: 
𝑦𝑛+1 = 0.04724 + (0.1)𝑓(0.3, 0.04724) = 0.04724 + (0.1)(0.64340) = 0.11158 
5ta Iteración: 
𝑦𝑛+1 = 0.11158 + (0.1)𝑓(0.4, 0.11158) = 0.11158 + (0.1)(1.10488) = 0.22207 
6ta Iteración: 
𝑦𝑛+1 = 0.22207 + (0.1)𝑓(0.5, 0.22207) = 0.22207 + (0.1)(1.79671) = 0.40174 
7ma Iteración: 
𝑦𝑛+1 = 0.40174 + (0.1)𝑓(0.6, 0.40174) = 0.40174 + (0.1)(2.82631) = 0.68437 
8va Iteración: 
𝑦𝑛+1 = 0.68437 + (0.1)𝑓(0.7, 0.68437) = 0.68437 + (0.1)(4.34758) = 0.30934 
 
9na Iteración: 
𝑦𝑛+1 = 0.30934 + (0.1)𝑓(0.8,0.30934) = 0.30934 + (0.1)(3.01111) = 1.77716 
10ma Iteración: 
𝑦𝑛+1 = 1.77716 + (0.1)𝑓(0.9, 1.77716) = 1.77716 + (0.1)(9.83744) = 2.76090 
 
 
𝒙𝒏 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 
𝒚𝒏 0 0.01350 0.04724 0.11158 0.22207 0.40174 0.68437 1.11913 1.77716 2.76090 
𝒇(𝒙𝒏, 𝒚𝒏) 0.13499 0.33743 0.64340 1.10488 1.79671 2.82631 4.34758 6.58028 9.83744 14.56373 
 
𝑬𝒓𝒓𝒐𝒓 𝑷𝒐𝒓𝒄𝒆𝒏𝒕𝒖𝒂𝒍 = |
𝑉. 𝑉𝑒𝑟𝑑𝑎𝑑𝑒𝑟𝑜 + 𝑉. 𝐴𝑝𝑟𝑜𝑥
𝑉. 𝑉𝑒𝑟𝑑𝑎𝑑𝑒𝑟𝑜
| × 100 = |
3.21910 + 14.56373
3.21910
| × 100 = 𝟏𝟒. 𝟐𝟑𝟑𝟕𝟑% 
 
2. Apliqué el método de Runge Kutta de cuarto orden para aproximar las soluciones de los 
siguientes problemas de valor inicial y comparé después los resultados con los valores reales. 
c) 𝑦′ = 𝑥𝑒3𝑥 − 2𝑦 0 ≤ 𝑥 ≤ 1 𝑦(0) = 0 𝑐𝑜𝑛 ℎ = 0.5 
La solución real es 𝑦 =
1
5
𝑥𝑒3𝑥 −
1
25
𝑒3𝑥 +
1
25
𝑒−2𝑥 → 𝑉𝑎𝑙𝑜𝑟 𝑅𝑒𝑎𝑙 = 3.21910 𝑐𝑜𝑛 𝑋 𝑒𝑣𝑎𝑙𝑢𝑎𝑑𝑎 𝑒𝑛 1 
Formulas 
𝒌𝟏 = 𝒉𝒇(𝒙𝒏, 𝒚𝒏) 
𝒌𝟐 = 𝒉𝒇(𝒙𝒏 +
𝒉
𝟐
, 𝒚𝒏 +
𝒌𝟏
𝟐
) 
𝒌𝟑 = 𝒉𝒇(𝒙𝒏 +
𝒉
𝟐
, 𝒚𝒏 +
𝒌𝟐
𝟐
) 
𝒌𝟒 = 𝒉𝒇(𝒙𝒏 + 𝒉, 𝒚𝒏 + 𝒌𝟑) 
𝒚𝒏+𝟏 = 𝒚𝒏 +
𝟏
𝟔
(𝒌𝟏 + 𝟐𝒌𝟐 + 𝟐𝒌𝟑 + 𝒌𝟒) 
 
1er Iteración: 
𝑘1 = (0.5)𝑓(0, 0) = (0.5)(0) = 0 
𝑘2 = (0.5)𝑓 (0 +
0.5
2
, 0 +
0
2
) = (0.5)(0.52925) = 0.26463 
𝑘3 = (0.5)𝑓 (0 +
0.5
2
, 0 +
0.26463
2
) = (0.5)(0.26462) = 0.13231 
𝑘4 = (0.5)𝑓(0 + 0.5, 0 + 0.13231) = (0.5)(1.97622) = 0.98811 
𝑦𝑛+1 = 0 +
1
6
(0 + 2(0.26463) + 2(0.13231) + 0.98811) = 0.29700 
 
2da Iteración: 
𝑘1 = (0.5)𝑓(0.5, 0.29700) = (0.5)(1.64684) = 0.82342 
𝑘2 = (0.5)𝑓 (0.5 +
0.5
2
, 0.29700 +
0.82342
2
) = (0.5)(5.69838) = 2.84919 
𝑘3 = (0.5)𝑓 (0.5 +
0.5
2
, 0.29700 +
2.84919
2
) = (0.5)(3.67261) = 1.83631 
𝑘4 = (0.5)𝑓(0.5 + 0.5, 0.29700 + 1.83631) = (0.5)(15.81892) = 7.90946 
𝑦𝑛+1 = 0.29700 +
1
6
(0.82342 + 2(2.84919) + 2(1.83631) + 7.90946) = 3.31431 
 
 
 
 
 
 
 
𝑬𝒓𝒓𝒐𝒓 𝑷𝒐𝒓𝒄𝒆𝒏𝒕𝒖𝒂𝒍 = |
𝑉. 𝑉𝑒𝑟𝑑𝑎𝑑𝑒𝑟𝑜 + 𝑉. 𝐴𝑝𝑟𝑜𝑥
𝑉. 𝑉𝑒𝑟𝑑𝑎𝑑𝑒𝑟𝑜
| × 100 = |
3.21910 + 3.31431
3.21910
| × 100 = 𝟐. 𝟗𝟓𝟕𝟕𝟒% 
 
d) 𝑦′ = 𝑥𝑒3𝑥 − 2𝑦 0 ≤ 𝑥 ≤ 1 𝑦(0) = 0 𝑐𝑜𝑛 ℎ = 0.2 
La solución real es 𝑦 =
1
5
𝑥𝑒3𝑥 −
1
25
𝑒3𝑥 +
1
25
𝑒−2𝑥 → 𝑉𝑎𝑙𝑜𝑟 𝑅𝑒𝑎𝑙 = 3.21910 𝑐𝑜𝑛 𝑋 𝑒𝑣𝑎𝑙𝑢𝑎𝑑𝑎 𝑒𝑛 1 
Formulas 
𝒌𝟏 = 𝒉𝒇(𝒙𝒏, 𝒚𝒏) 
𝒌𝟐 = 𝒉𝒇(𝒙𝒏 +
𝒉
𝟐
, 𝒚𝒏 +
𝒌𝟏
𝟐
) 
𝒌𝟑 = 𝒉𝒇(𝒙𝒏 +
𝒉
𝟐
, 𝒚𝒏 +
𝒌𝟐
𝟐
) 
𝒌𝟒 = 𝒉𝒇(𝒙𝒏 + 𝒉, 𝒚𝒏 + 𝒌𝟑) 
𝒚𝒏+𝟏 = 𝒚𝒏 +
𝟏
𝟔
(𝒌𝟏 + 𝟐𝒌𝟐 + 𝟐𝒌𝟑 + 𝒌𝟒) 
 
1er Iteración: 
𝒙𝒏 0 0.5 1 
𝒌𝟏 − 0 0.82342 
𝒌𝟐 − 0.26463 2.84919 
𝒌𝟑 − 0.13231 1.83631 
𝒌𝟒 − 0.98811 7.90946 
𝒚𝒏 0 0.29700 3.31431 
𝑘1 = (0.2)𝑓(0,0) = (0.2)(0) = 0 
𝑘2 = (0.2)𝑓 (0 +
0.2
2
, 0 +
0
2
) = (0.2)(0.13499) = 0.02700 
𝑘3 = (0.2)𝑓 (0 +
0.2
2
, 0 +
0.02700
2
) = (0.2)(0.10799) = 0.02160 
𝑘4 = (0.2)𝑓(0 + 0.2,0 + 0.02160) = (0.2)(0.32122) = 0.06425 
𝑦𝑛+1 = 0 +
1
6
(0 + 2(0.02700) + 2(0.02160) + 0.06425) = 0.02691 
 
2da Iteración: 
𝑘1 = (0.2)𝑓(0.2, 0.02691) = (0.2)(0.31060) = 0.06212 
𝑘2 = (0.2)𝑓 (0.2 +
0.2
2
, 0.02691 +
0.06212
2
) = (0.2)(0.62286) = 0.12439 
𝑘3 = (0.2)𝑓 (0.2 +
0.2
2
, 0.02691 +
0.12439
2
) = (0.2)(0.55967) = 0.11194 
𝑘4 = (0.2)𝑓(0.2 + 0.2, 0.02691 + 0.11194) = (0.2)(1.05035) = 0.21007 
𝑦𝑛+1 = 0.02691 +
1
6
(0.06212 + 2(0.12439) + 2(0.11194) + 0.21007) = 0.15105 
 
3er Iteración: 
𝑘1 = (0.2)𝑓(0.4, 0.15105) = (0.2)(1.02595) = 0.20519 
𝑘2 = (0.2)𝑓 (0.4 +
0.2
2
, 0.15105 +
0.20519
2
) = (0.2)(1.73355) = 0.34671 
𝑘3 = (0.2)𝑓 (0.4 +
0.2
2
, 0.15105 +
0.34671
2
) = (0.2)(1.59203) = 0.31841 
𝑘4 = (0.2)𝑓(0.4 + 0.2, 0.15105 + 0.31841) = (0.2)(2.69087) = 0.53818 
𝑦𝑛+1 = 0.15105 +
1
6
(0.20519 + 2(0.34671) + 2(0.31841) + 0.53818) = 0.49665 
 
4ta Iteración: 
𝑘1 = (0.2)𝑓(0.6, 0.49665) = (0.2)(2.63649) = 0.52730 
𝑘2 = (0.2)𝑓 (0.6 +
0.2
2
, 0.49665 +
0.52730
2
) = (0.2)(4.19572) = 0.83914 
𝑘3 = (0.2)𝑓 (0.6 +
0.2
2
, 0.49665 +
0.83914
2
) = (0.2)(3.88388) = 0.77678 
𝑘4 = (0.2)𝑓(0.6 + 0.2,0.49665 + 0.77678) = (0.2)(6.27168) = 1.25434 
𝑦𝑛+1 = 0.49665 +
1
6
(0.52730 + 2(0.83914) + 2(0.77678) + 1.25434) = 1.33223 
 
5ta Iteración: 
𝑘1 = (0.2)𝑓(0.8, 1.33223) = (0.2)(6.15408) = 1.23082 
𝑘2 = (0.2)𝑓 (0.8 +
0.2
2
, 1.33223 +
1.23082
2
) = (0.2)(9.49648) = 1.89930 
𝑘3 = (0.2)𝑓 (0.8 +
0.2
2
, 1.33223 +
1.89930
2
) = (0.2)(8.82800) = 1.76560 
𝑘4 = (0.2)𝑓(0.8 + 0.2, 1.33223 + 1.76560) = (0.2)(13.88988) = 2.77798 
𝑦𝑛+1 = 1.33223 +
1
6
(1.23082 + 2(1.89930) + 2(1.76560) + 2.77798) = 3.22199 
 
 
 
 
 
 
𝑬𝒓𝒓𝒐𝒓 𝑷𝒐𝒓𝒄𝒆𝒏𝒕𝒖𝒂𝒍 = |
𝑉. 𝑉𝑒𝑟𝑑𝑎𝑑𝑒𝑟𝑜 + 𝑉. 𝐴𝑝𝑟𝑜𝑥
𝑉. 𝑉𝑒𝑟𝑑𝑎𝑑𝑒𝑟𝑜
| × 100 = |
3.21910 + 3.22199
3.21910
| × 100 = 𝟎. 𝟎𝟖𝟗𝟖𝟖% 
 
 
3. Organiza tus resultados en la tabla: 
𝑬𝒖𝒍𝒆𝒓 𝑽𝒂𝒍𝒐𝒓 𝑨𝒑𝒓𝒐𝒙𝒊𝒎𝒂𝒅𝒐 𝑬𝒓𝒓𝒐𝒓 𝑹𝒖𝒏𝒈𝒆 𝑲𝒖𝒕𝒕𝒂 𝑽𝒂𝒍𝒐𝒓 𝑨𝒑𝒓𝒐𝒙𝒊𝒎𝒂𝒅𝒐 𝑬𝒓𝒓𝒐𝒓 
ℎ = 0.2 2.31065 28.22075% ℎ = 0.5 3.31431 2.95774% 
ℎ = 0.1 2.76090 14.23373% ℎ = 0.2 3.22199 0.08988% 
 
 
𝒙𝒏 0 0.2 0.4 0.6 0.8 1 
𝒌𝟏 − 0 0.06212 0.20519 0.52730 1.23082 
𝒌𝟐 − 0.02700 0.12439 0.34671 0.83914 1.89930 
𝒌𝟑 − 0.02160 0.11194 0.31841 0.77678 1.76560 
𝒌𝟒 − 0.06425 0.21007 0.53818 1.25434 2.77798 
𝒚𝒏 0 0.02691 0.15105 0.49665 1.33223 3.22199