Considere os seguintes problemas de valor inicial:
Resolva-os pela regra do ponto médio:
\left\{\begin{array}{l}y^{\prime}=-x y \quad ; \\\ y(0)=2\end{array} \quad 0 \leq x \leq 0.3 \quad ; \quad h=0.1\right.

Passo 1

O enunciado pede para aproximar a solução do problema de valor inicial
\begin{cases} y' = -x\,y,\\\ y(0)=2, \end{cases}
no intervalo 0\le x\le0.3 usando a regra do ponto médio (um método de Runge–Kutta de segunda ordem) com passo h=0.1. Vamos empregar a fórmula
\begin{aligned} k_1&=f(x_n,y_n),\\\ x_{n+\tfrac12}&=x_n+\tfrac h2,\quad y_{n+\tfrac12}=y_n+\tfrac h2\,k_1,\\\ k_2&=f\bigl(x_{n+\tfrac12},\,y_{n+\tfrac12}\bigr),\\\ y_{n+1}&=y_n+h\,k_2, \end{aligned}
onde f(x,y)=-x\,y.

Passo 2

Em n=0 temos x_0=0 e y_0=2. Calculamos
k_1=f(x_0,y_0)=-0\cdot2=0,
x_{0+\tfrac12}=0+0.05=0.05, \quad y_{0+\tfrac12}=2+0.05\cdot0=2,
k_2=f(0.05,2)=-0.05\cdot2=-0.1,
y_1=y_0+0.1\cdot(-0.1)=2-0.01=1.99.

Passo 3

Em n=1 temos x_1=0.1 e y_1=1.99. Calculamos
k_1=f(0.1,1.99)=-0.1\cdot1.99=-0.199,
x_{1+\tfrac12}=0.1+0.05=0.15, \quad y_{1+\tfrac12}=1.99+0.05\cdot(-0.199)=1.99-0.00995=1.98005,
k_2=f(0.15,1.98005)=-0.15\cdot1.98005=-0.2970075,
y_2=1.99+0.1\cdot(-0.2970075)=1.99-0.02970075=1.96029925.

Passo 4

Em n=2 temos x_2=0.2 e y_2\approx1.96029925. Calculamos
k_1=f(0.2,1.96029925)=-0.2\cdot1.96029925=-0.39205985,
x_{2+\tfrac12}=0.2+0.05=0.25, \quad y_{2+\tfrac12}=1.96029925+0.05\cdot(-0.39205985)=1.96029925-0.0196029925=1.9406962575,
k_2=f(0.25,1.9406962575)=-0.25\cdot1.9406962575=-0.485174064375,
y_3=1.96029925+0.1\cdot(-0.485174064375) =1.96029925-0.0485174064375 \approx1.9117818436.

Resposta

Os valores aproximados em cada nó são:
y(0.1)\approx1.99,\quad y(0.2)\approx1.9603,\quad y(0.3)\approx1.9118.

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