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𝑥0 = 0 𝑦0 = 1 𝑥𝑓 = 3 ℎ = 3
𝒌𝟏 = 𝒇(𝒙𝒏,𝒚𝒏)
 Primea iteración
𝑘1 = (−0,1)(0)(1) = 0
 Segunda iteración
𝑘1 = (−0,1)(0,3)(0,99551) = −0,030
 Tercera iteración
𝑘1 = (−0,1)(0,6)(0,98216) = −0,059
𝒌𝟐 = 𝒇(𝒙𝒏 +
𝟏
𝟐
𝒉, 𝒚𝒏 +
𝟏
𝟐
𝒉𝒌𝟏)
 Primea iteración
𝑘2 = (0 +
(0,3)
2
+ 1 +
1(0)
2
) = −0,015
 Segunda iteración
𝑘2 = (0,3 +
(0,3)
2
(0,99551) +
0,3(−0,030)
2
)(−0,1) = −0,045
 Tercera iteración
𝑘2 = (0,6 +
(0,3)
2
(0,98216) +
0,3(−0,059)
2
) (−0,1) = −0,073
𝒌𝟑 = 𝒇(𝒙𝒏 +
𝟏
𝟐
𝒉, 𝒚𝒏 +
𝟏
𝟐
𝒉𝒌𝟐)
 Primea iteración
𝑘3 = (0 +
(0,3)
2
(1) +
0,3(−0,015)
2
) (−0,1) = −0,015
 Segunda iteración
𝑘3 = (0,3 +
(0,3)
2
)(0,99551)+
0,3(−0,045)
2
(−0,1) = −0,044
 Tercera iteración
𝑘3 = (0,6 +
(0,3)
2
(0,98216) +
0,3(−0,073 )
2
)(−0,1) = −0,073
𝑘4 = 𝑓(𝑥𝑛 + ℎ, 𝑦𝑛 + ℎ,𝑘3)
 Primea iteración
𝑘4 = [0 + 0,3(1)+ 0,3(−0,015)](−0,1) = −0,030
 Segunda iteración
𝑘4 = [0,3 + 0,3(0,99551)+ 0,3(−0,044)](−0,1) = −0,059
 Tercera iteración
𝑘4 = [0,6 + 0,3(0,98216)+ 0,3(−0,044)](−0,073) = −0,086
 Onceava iteración
𝑘4 = [3 + 0,3(0,63763) + 0,3(−0,192)](−0,073) = −0,191
𝑦𝑛+1 = 𝑦𝑛 +
ℎ
6
[𝑘1 + 2𝑘2 + 2𝑘3 + 𝑘4]
 Segunda iteración
𝑦𝑛+1 = 1 +
0,3
6
[0 − 0,015 − 0,015 − 0,030] = 0,99551
 Tercera iteración
𝑦𝑛+1 = 0,99551 +
0,3
6
[−0,030 − 0,045 − 0,044 − 0,059] = 0,98216
 Onceava iteración
𝑦𝑛+1 = 0,69454 +
0,3
6
[−0,188 − 0,190 − 0,190 − 0,191] = 0,63763

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standardisation of garbhpala offhgfffghh
 

Metodo de kutta

  • 1. 𝑥0 = 0 𝑦0 = 1 𝑥𝑓 = 3 ℎ = 3 𝒌𝟏 = 𝒇(𝒙𝒏,𝒚𝒏)  Primea iteración 𝑘1 = (−0,1)(0)(1) = 0  Segunda iteración 𝑘1 = (−0,1)(0,3)(0,99551) = −0,030  Tercera iteración 𝑘1 = (−0,1)(0,6)(0,98216) = −0,059 𝒌𝟐 = 𝒇(𝒙𝒏 + 𝟏 𝟐 𝒉, 𝒚𝒏 + 𝟏 𝟐 𝒉𝒌𝟏)  Primea iteración 𝑘2 = (0 + (0,3) 2 + 1 + 1(0) 2 ) = −0,015  Segunda iteración 𝑘2 = (0,3 + (0,3) 2 (0,99551) + 0,3(−0,030) 2 )(−0,1) = −0,045  Tercera iteración 𝑘2 = (0,6 + (0,3) 2 (0,98216) + 0,3(−0,059) 2 ) (−0,1) = −0,073 𝒌𝟑 = 𝒇(𝒙𝒏 + 𝟏 𝟐 𝒉, 𝒚𝒏 + 𝟏 𝟐 𝒉𝒌𝟐)  Primea iteración 𝑘3 = (0 + (0,3) 2 (1) + 0,3(−0,015) 2 ) (−0,1) = −0,015  Segunda iteración 𝑘3 = (0,3 + (0,3) 2 )(0,99551)+ 0,3(−0,045) 2 (−0,1) = −0,044  Tercera iteración 𝑘3 = (0,6 + (0,3) 2 (0,98216) + 0,3(−0,073 ) 2 )(−0,1) = −0,073 𝑘4 = 𝑓(𝑥𝑛 + ℎ, 𝑦𝑛 + ℎ,𝑘3)  Primea iteración
  • 2. 𝑘4 = [0 + 0,3(1)+ 0,3(−0,015)](−0,1) = −0,030  Segunda iteración 𝑘4 = [0,3 + 0,3(0,99551)+ 0,3(−0,044)](−0,1) = −0,059  Tercera iteración 𝑘4 = [0,6 + 0,3(0,98216)+ 0,3(−0,044)](−0,073) = −0,086  Onceava iteración 𝑘4 = [3 + 0,3(0,63763) + 0,3(−0,192)](−0,073) = −0,191 𝑦𝑛+1 = 𝑦𝑛 + ℎ 6 [𝑘1 + 2𝑘2 + 2𝑘3 + 𝑘4]  Segunda iteración 𝑦𝑛+1 = 1 + 0,3 6 [0 − 0,015 − 0,015 − 0,030] = 0,99551  Tercera iteración 𝑦𝑛+1 = 0,99551 + 0,3 6 [−0,030 − 0,045 − 0,044 − 0,059] = 0,98216  Onceava iteración 𝑦𝑛+1 = 0,69454 + 0,3 6 [−0,188 − 0,190 − 0,190 − 0,191] = 0,63763