Aplique o método Heun (para duas iterações do corretor) na busca pela solução aproximada do problema de valor inicial
\begin{aligned} \frac{d \varphi}{d t}+3 \varphi & =e^{-3 t}(1-2 t) \\\ \varphi(0) & =2 \end{aligned}
e encontre as soluções em \mathrm{t}=0,3. Obtenha os valores usando \Delta t=0,1.

Passo 1

Resumindo, queremos aproximar a solução do PVI
\frac{d\varphi}{dt}+3\varphi=e^{-3t}(1-2t), \quad \varphi(0)=2
no ponto t=0,3 usando o método de Heun (preditor-corretor com duas iterações no corretor) e passo \Delta t=0,1. Definimos
f(t,\varphi)=e^{-3t}(1-2t)-3\varphi.

Passo 2

O algoritmo de Heun com duas iterações de corretor é:
\varphi_{n+1}^{(0)}=\varphi_n + h\,f(t_n,\varphi_n)
\varphi_{n+1}^{(k+1)}=\varphi_n + \frac{h}{2}\Bigl(f(t_n,\varphi_n)+f\bigl(t_{n+1},\varphi_{n+1}^{(k)}\bigr)\Bigr), \quad k=0,1
onde h=0,1, t_{n+1}=t_n+h. Após duas correções k=0\to1\to2, tomamos \varphi_{n+1}=\varphi_{n+1}^{(2)}.

Passo 3

Passo de t_0=0 para t_1=0,1:
Calculemos \varphi_0=2,
f_0=f(0,2)=1-3\cdot2=-5.
Preditor:
\varphi_1^{(0)}=2+0,1\cdot(-5)=1,5.
Primeiro corretor k=0\to1:
f_1^{(0)}=f\bigl(0,1+0,1,\,1,5\bigr)=e^{-0,3}\cdot0,8-3\cdot1,5 \approx0,740818\cdot0,8-4,5\approx-3,907346,
\varphi_1^{(1)}=2+\frac{0,1}{2}(-5-3,907346)\approx2-0,445367=1,554633.
Segundo corretor k=1\to2:
f_1^{(1)}=f(0,1+0,1,\,1,554633)\approx0,740818\cdot0,8-3\cdot1,554633 \approx-4,071245,
\varphi_1^{(2)}=2+\frac{0,1}{2}(-5-4,071245)\approx2-0,453562=1,546438.
Logo \varphi(0,1)\approx1,546438.

Passo 4

Passo de t_1=0,1 para t_2=0,2:
f_1=f(0,1,\,1,546438)\approx e^{-0,3}\cdot0,8-3\cdot1,546438\approx-4,046659.
Preditor:
\varphi_2^{(0)}=1,546438+0,1\cdot(-4,046659)=1,141772.
Primeiro corretor:
f_2^{(0)}=f(0,2,\,1,141772)\approx e^{-0,6}\cdot0,6-3\cdot1,141772 \approx-3,095029,
\varphi_2^{(1)}=1,546438+\frac{0,1}{2}(-4,046659-3,095029) \approx1,546438-0,357084=1,189354.
Segundo corretor:
f_2^{(1)}=f(0,2,\,1,189354)\approx e^{-0,6}\cdot0,6-3\cdot1,189354 \approx-3,238775,
\varphi_2^{(2)}=1,546438+\frac{0,1}{2}(-4,046659-3,238775) \approx1,546438-0,364272=1,182166.
Logo \varphi(0,2)\approx1,182166.

Passo 5

Passo de t_2=0,2 para t_3=0,3:
f_2=f(0,2,\,1,182166)\approx e^{-0,6}\cdot0,6-3\cdot1,182166 \approx-3,217212.
Preditor:
\varphi_3^{(0)}=1,182166+0,1\cdot(-3,217212)=0,860445.
Primeiro corretor:
f_3^{(0)}=f(0,3,\,0,860445)\approx e^{-0,9}\cdot0,4-3\cdot0,860445 \approx-2,418707,
\varphi_3^{(1)}=1,182166+\frac{0,1}{2}(-3,217212-2,418707) \approx1,182166-0,281796=0,900370.
Segundo corretor:
f_3^{(1)}=f(0,3,\,0,900370)\approx e^{-0,9}\cdot0,4-3\cdot0,900370 \approx-2,538483,
\varphi_3^{(2)}=1,182166+\frac{0,1}{2}(-3,217212-2,538483) \approx1,182166-0,287785=0,894382.

Resposta

\displaystyle \varphi(0,3)\approx0,894382.

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