Calcular a solução do PVI \left\{\begin{array}{l}\frac{d y}{d x}=-x y \\\ y(0)=1\end{array}\right. com h=0,1, no interior do intervalo [ 0,1 ], pelo método de Runge-Kutta de quarta ordem.
Resolução: \quad y_{j+1}=y_{j}+\frac{h}{6}\left(k_{1}+2 k_{2}+2 k_{3}+k_{4}\right), para j=0,1,2, \ldots, 9.
\begin{aligned} & k_{1}= \\\ & k_{2}= \\\ & k_{3}= \\\ & k_{4}= \end{aligned}
\_\_\_\_
\_\_\_\_
\_\_\_\_
\_\_\_\_
\begin{tabular}{|l|l|l|l|l|l|l|}
\hline j & x_{j} & y_{j} & k_{1} & k_{2} & k_{3} & k_{4} \
\hline 0 & 0 & 1 & & & & \
\hline 1 & & & & & & \
\hline 2 & & & & & & \
\hline 3 & & & & & & \
\hline 4 & & & & & & \
\hline 5 & & & & & & \
\hline 6 & & & & & & \
\hline 7 & & & & & & \
\hline 8 & & & & & & \
\hline 9 & & & & & & \
\hline 10 & & & & & & \
\hline

Passo 1

O problema dá o PVI
\begin{cases} \dfrac{dy}{dx} = -x\,y,\\[6pt] y(0)=1, \end{cases}
o passo h=0{,}1 e pede aplicar o método de Runge–Kutta de quarta ordem no interior de [0,1] (ou seja, calcular y_j para j=0,1,\dots,10 onde x_j= jh).
As fórmulas dadas são
y_{j+1}=y_j+\frac{h}{6}\bigl(k_1+2k_2+2k_3+k_4\bigr),
para j=0,1,\dots,9. O enunciado deixou em branco as definições de k_1,k_2,k_3,k_4 e a tabela a ser preenchida.]

Passo 2

Primeiro vamos identificar a função do PVI e definir os k's usados no método (a notação aqui segue o esquema habitual do RK4, em que os k's são as avaliações da derivada f(x,y), não multiplicadas por h).
  • A função do campo é \;f(x,y)=-x\,y\;.
  • Para cada passo com passo h temos as definições
\begin{aligned} k_1 &= f\bigl(x_j,y_j\bigr),\\[4pt] k_2 &= f\Bigl(x_j+\frac{h}{2},\,y_j+\frac{h}{2}k_1\Bigr),\\[4pt] k_3 &= f\Bigl(x_j+\frac{h}{2},\,y_j+\frac{h}{2}k_2\Bigr),\\[4pt] k_4 &= f\bigl(x_j+h,\,y_j+h k_3\bigr). \end{aligned}
Explicando cada símbolo (pra você não ficar perdido): x_j é o ponto atual, y_j a aproximação de y(x_j). O k_1 é a inclinação em (x_j,y_j). O k_2 e k_3 são inclinações em pontos intermediários (meio passo), avaliando a função f numa estimativa previsível do y. O k_4 é a inclinação no próximo ponto x_j+h usando a previsão com k_3. Depois juntamos essas inclinações ponderadas para obter y_{j+1}.
Vamos aplicar isso passo a passo com h=0{,}1. Em cada passo vou mostrar o cálculo dos k's e o incremento \dfrac{h}{6}(k_1+2k_2+2k_3+k_4).]

Passo 3

Dados iniciais: \;x_0=0,\;y_0=1,\;h=0{,}1.
Calculemos k's:
\begin{aligned} k_1 &= f(0,1) = -0\cdot 1 = 0,\\[4pt] k_2 &= f\!\Bigl(0+\tfrac{h}{2},\,1+\tfrac{h}{2}k_1\Bigr) = f(0{,}05,\,1)= -0{,}05\cdot 1 = -0{,}05,\\[4pt] k_3 &= f\!\Bigl(0{,}05,\,1+\tfrac{h}{2}k_2\Bigr) = f\bigl(0{,}05,\,1-0{,}0025\bigr) = -0{,}05\cdot 0{,}9975 = -0{,}049875,\\[4pt] k_4 &= f\!\Bigl(0{,}1,\,1+h k_3\Bigr) = f\bigl(0{,}1,\,1-0{,}0049875\bigr) = -0{,}1\cdot 0{,}9950125 = -0{,}09950125. \end{aligned}
Agora o incremento:
\Delta y = \frac{h}{6}\bigl(k_1+2k_2+2k_3+k_4\bigr) = \frac{0{,}1}{6}\bigl(0+2(-0{,}05)+2(-0{,}049875)+(-0{,}09950125)\bigr) = -0{,}00498752083333\ldots
Logo
y_1 = 1 - 0{,}00498752083333\ldots = 0{,}99501247916667\ldots]

Passo 4

Dados: \;x_1=0{,}1,\;y_1\approx 0{,}99501247916667.
Calculemos k's:
\begin{aligned} k_1 &= f(0{,}1,\,0{,}99501247916667) = -0{,}1\cdot 0{,}99501247916667 = -0{,}0995012479166667,\\[4pt] k_2 &= f\!\Bigl(0{,}15,\,0{,}99501247916667+\tfrac{h}{2}k_1\Bigr) = f\bigl(0{,}15,\,0{,}9900374167708334\bigr) = -0{,}15\cdot 0{,}9900374167708334 = -0{,}148505612515625,\\[4pt] k_3 &= f\!\Bigl(0{,}15,\,0{,}99501247916667+\tfrac{h}{2}k_2\Bigr) = f\bigl(0{,}15,\,0{,}9875871985408854\bigr) = -0{,}15\cdot 0{,}9875871985408854 = -0{,}1481380797811328,\\[4pt] k_4 &= f\!\Bigl(0{,}2,\,0{,}99501247916667+h k_3\Bigr) = f\bigl(0{,}2,\,0{,}9801986711885535\bigr) = -0{,}2\cdot 0{,}9801986711885535 = -0{,}1960397342377107. \end{aligned}
Incremento e y_2:
\Delta y = \frac{0{,}1}{6}\bigl(k_1+2k_2+2k_3+k_4\bigr)\approx -0{,}0148138061124649,
y_2 \approx 0{,}99501247916667 - 0{,}0148138061124649 = 0{,}9801986730542019.]

Passo 5

Dados: \;x_2=0{,}2,\;y_2\approx 0{,}9801986730542019.
Cálculos:
\begin{aligned} k_1 &= -0{,}2\cdot 0{,}9801986730542019 = -0{,}1960397346108404,\\[4pt] k_2 &= -0{,}25\cdot\bigl(0{,}9801986730542019+\tfrac{0{,}1}{2}k_1\bigr) = -0{,}25\cdot 0{,}9703966863236598 = -0{,}24259917158091495,\\[4pt] k_3 &= -0{,}25\cdot\bigl(0{,}9801986730542019+\tfrac{0{,}1}{2}k_2\bigr) = -0{,}25\cdot 0{,}9680687144751562 = -0{,}24201717861878905,\\[4pt] k_4 &= -0{,}3\cdot\bigl(0{,}9801986730542019+0{,}1k_3\bigr) = -0{,}3\cdot 0{,}955996955192323 = -0{,}2867990865576969. \end{aligned}
Incremento:
\Delta y\approx -0{,}0242011920261324. Assim
y_3 \approx 0{,}9801986730542019 - 0{,}0242011920261324 = 0{,}9559974810280695.]

Passo 6

Dados: \;x_3=0{,}3,\;y_3\approx 0{,}9559974810280695.
Cálculos:
\begin{aligned} k_1 &= -0{,}3\cdot 0{,}9559974810280695 = -0{,}28679924430842085,\\[4pt] k_2 &= -0{,}35\cdot\bigl(0{,}9559974810280695+0{,}05k_1\bigr) = -0{,}35\cdot 0{,}9416575188126485 = -0{,}32958013158442697,\\[4pt] k_3 &= -0{,}35\cdot\bigl(0{,}9559974810280695+0{,}05k_2\bigr) = -0{,}35\cdot 0{,}9395184744488481 = -0{,}32883146605709685,\\[4pt] k_4 &= -0{,}4\cdot\bigl(0{,}9559974810280695+0{,}1k_3\bigr) = -0{,}4\cdot 0{,}9231143344223598 = -0{,}3692457337689439. \end{aligned}
\Delta y\approx -0{,}0328811362226736, então
y_4 \approx 0{,}9559974810280695 - 0{,}0328811362226736 = 0{,}9231163448053959.]

Passo 7

Dados: \;x_4=0{,}4,\;y_4\approx 0{,}9231163448053959.
Cálculos:
\begin{aligned} k_1 &= -0{,}4\cdot 0{,}9231163448053959 = -0{,}36924653792215836,\\[4pt] k_2 &= -0{,}45\cdot\bigl(0{,}9231163448053959+0{,}05k_1\bigr) = -0{,}45\cdot 0{,}904654017909288 = -0{,}4070943070591796,\\[4pt] k_3 &= -0{,}45\cdot\bigl(0{,}9231163448053959+0{,}05k_2\bigr) = -0{,}45\cdot 0{,}9027616294524369 = -0{,}4062427332535966,\\[4pt] k_4 &= -0{,}5\cdot\bigl(0{,}9231163448053959+0{,}1k_3\bigr) = -0{,}5\cdot 0{,}8824920714800363 = -0{,}44124603574001815. \end{aligned}
\Delta y\approx -0{,}0406194442381288, logo
y_5 \approx 0{,}9231163448053959 - 0{,}0406194442381288 = 0{,}8824969005672671.]

Passo 8

Dados: \;x_5=0{,}5,\;y_5\approx 0{,}8824969005672671.
Cálculos:
\begin{aligned} k_1 &= -0{,}5\cdot 0{,}8824969005672671 = -0{,}44124845028363353,\\[4pt] k_2 &= -0{,}55\cdot\bigl(0{,}8824969005672671+0{,}05k_1\bigr) = -0{,}55\cdot 0{,}8604344780530854 = -0{,}473238962929197,\\[4pt] k_3 &= -0{,}55\cdot\bigl(0{,}8824969005672671+0{,}05k_2\bigr) = -0{,}55\cdot 0{,}8588349524208073 = -0{,}472359223831443,\\[4pt] k_4 &= -0{,}6\cdot\bigl(0{,}8824969005672671+0{,}1k_3\bigr) = -0{,}6\cdot 0{,}8352609781841228 = -0{,}5011565869104737. \end{aligned}
\Delta y\approx -0{,}0472266901785898, então
y_6 \approx 0{,}8824969005672671 - 0{,}0472266901785898 = 0{,}8352702103886773.]

Passo 9

Dados: \;x_6=0{,}6,\;y_6\approx 0{,}8352702103886773.
Cálculos:
\begin{aligned} k_1 &= -0{,}6\cdot 0{,}8352702103886773 = -0{,}5011621262332064,\\[4pt] k_2 &= -0{,}65\cdot\bigl(0{,}8352702103886773+0{,}05k_1\bigr) = -0{,}65\cdot 0{,}810212104077017 = -0{,}5266378686505609,\\[4pt] k_3 &= -0{,}65\cdot\bigl(0{,}8352702103886773+0{,}05k_2\bigr) = -0{,}65\cdot 0{,}8089383169561493 = -0{,}526809905021496,\\[4pt] k_4 &= -0{,}7\cdot\bigl(0{,}8352702103886773+0{,}1k_3\bigr) = -0{,}7\cdot 0{,}7825892198865277 = -0{,}5478124539205694. \end{aligned}
\Delta y\approx -0{,}0525978354582982, assim
y_7 \approx 0{,}8352702103886773 - 0{,}0525978354582982 = 0{,}7826723749303792.]

Passo 10

Dados: \;x_7=0{,}7,\;y_7\approx 0{,}7826723749303792.
Cálculos:
\begin{aligned} k_1 &= -0{,}7\cdot 0{,}7826723749303792 = -0{,}5478706624512655,\\[4pt] k_2 &= -0{,}75\cdot\bigl(0{,}7826723749303792+0{,}05k_1\bigr) = -0{,}75\cdot 0{,}7552788418078159 = -0{,}5664591313558619,\\[4pt] k_3 &= -0{,}75\cdot\bigl(0{,}7826723749303792+0{,}05k_2\bigr) = -0{,}75\cdot 0{,}7543494183625861 = -0{,}5657620637719396,\\[4pt] k_4 &= -0{,}8\cdot\bigl(0{,}7826723749303792+0{,}1k_3\bigr) = -0{,}8\cdot 0{,}7260961685531852 = -0{,}5808769348425481. \end{aligned}
\Delta y\approx -0{,}0565531664591570, então
y_8 \approx 0{,}7826723749303792 - 0{,}0565531664591570 = 0{,}7261192084712222.]

Passo 11

Dados: \;x_8=0{,}8,\;y_8\approx 0{,}7261192084712222.
Cálculos:
\begin{aligned} k_1 &= -0{,}8\cdot 0{,}7261192084712222 = -0{,}5808953667769778,\\[4pt] k_2 &= -0{,}85\cdot\bigl(0{,}7261192084712222+0{,}05k_1\bigr) = -0{,}85\cdot 0{,}6970744401323733 = -0{,}5925132741125173,\\[4pt] k_3 &= -0{,}85\cdot\bigl(0{,}7261192084712222+0{,}05k_2\bigr) = -0{,}85\cdot 0{,}6964935447655964 = -0{,}5920195120507560,\\[4pt] k_4 &= -0{,}9\cdot\bigl(0{,}7261192084712222+0{,}1k_3\bigr) = -0{,}9\cdot 0{,}6669172572661466 = -0{,}6002255315395320. \end{aligned}
\Delta y\approx -0{,}0591697745107176, portanto
y_9 \approx 0{,}7261192084712222 - 0{,}0591697745107176 = 0{,}6669494339605046.]

Passo 12

Dados: \;x_9=0{,}9,\;y_9\approx 0{,}6669494339605046.
Cálculos:
\begin{aligned} k_1 &= -0{,}9\cdot 0{,}6669494339605046 = -0{,}6002544905644541,\\[4pt] k_2 &= -0{,}95\cdot\bigl(0{,}6669494339605046+0{,}05k_1\bigr) = -0{,}95\cdot 0{,}6369367094322819 = -0{,}6050908739606678,\\[4pt] k_3 &= -0{,}95\cdot\bigl(0{,}6669494339605046+0{,}05k_2\bigr) = -0{,}95\cdot 0{,}6366948902624712 = -0{,}6048601457493476,\\[4pt] k_4 &= -1{,}0\cdot\bigl(0{,}6669494339605046+0{,}1k_3\bigr) = -1{,}0\cdot 0{,}606463419385570 \;=\; -0{,}6064634193855700. \end{aligned}
\Delta y\approx -0{,}0604436658228336, então
y_{10} \approx 0{,}6669494339605046 - 0{,}0604436658228336 = 0{,}6065057681376710.]

Resposta

j=0:\;x_0=0,\;y_0=1,\;k_1=0,\;k_2=-0{,}05,\;k_3=-0{,}049875,\;k_4=-0{,}09950125
j=1:\;x_1=0{,}1,\;y_1\approx 0{,}995012479167,\;k_1\approx -0{,}099501247917,\;k_2\approx -0{,}148505612516,\;k_3\approx -0{,}148138079781,\;k_4\approx -0{,}196039734238
j=2:\;x_2=0{,}2,\;y_2\approx 0{,}980198673054,\;k_1\approx -0{,}196039734611,\;k_2\approx -0{,}242599171581,\;k_3\approx -0{,}242017178619,\;k_4\approx -0{,}286799086558
j=3:\;x_3=0{,}3,\;y_3\approx 0{,}955997481028,\;k_1\approx -0{,}286799244308,\;k_2\approx -0{,}329580131584,\;k_3\approx -0{,}328831466057,\;k_4\approx -0{,}369245733769
j=4:\;x_4=0{,}4,\;y_4\approx 0{,}923116344805,\;k_1\approx -0{,}369246537922,\;k_2\approx -0{,}407094307059,\;k_3\approx -0{,}406242733254,\;k_4\approx -0{,}441246035740
j=5:\;x_5=0{,}5,\;y_5\approx 0{,}882496900567,\;k_1\approx -0{,}441248450284,\;k_2\approx -0{,}473238962929,\;k_3\approx -0{,}472359223831,\;k_4\approx -0{,}501156586910
j=6:\;x_6=0{,}6,\;y_6\approx 0{,}835270210389,\;k_1\approx -0{,}501162126233,\;k_2\approx -0{,}526637868651,\;k_3\approx -0{,}526809905021,\;k_4\approx -0{,}547812453921
j=7:\;x_7=0{,}7,\;y_7\approx 0{,}782672374930,\;k_1\approx -0{,}547870662451,\;k_2\approx -0{,}566459131356,\;k_3\approx -0{,}565762063772,\;k_4\approx -0{,}580876934843
j=8:\;x_8=0{,}8,\;y_8\approx 0{,}726119208471,\;k_1\approx -0{,}580895366777,\;k_2\approx -0{,}592513274113,\;k_3\approx -0{,}592019512051,\;k_4\approx -0{,}600225531540
j=9:\;x_9=0{,}9,\;y_9\approx 0{,}666949433961,\;k_1\approx -0{,}600254490564,\;k_2\approx -0{,}605090873961,\;k_3\approx -0{,}604860145749,\;k_4\approx -0{,}606463419386
j=10:\;x_{10}=1{,}0,\;y_{10}\approx 0{,}606505768138

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