Calcule, pelo método de Euler, a diferencial y' = 4x + 2y, com y(1) = 0, no intervalo [1, 2] com n = 4???

Se puder curtir minhas postagens fico grato ;)

resposta passo a passo

h= b - a / N      h= 2 - 1 / 4           h= 0,25

x0 = 1

x1= x0 +h        x1 = 1 + 0,25           x1 = 1,25

x2 = x1 +h       x2 = 1,25 + 0,25      x2 = 1,50

x3 = x2 +h       x3 = 1,50 + 0,25      x3 = 1,75

x4 = x3 +h       x4 = 1,75 + 0,25      x4 = 2

y0 = 0

y1= y0 + h . f(x0,y0)       y1= 0 + 0,25 . (4.x0 + 2.y0)               y1= 0 + 0,25 . (4.1 + 2.0)                              y1= 1

y2= y1 + h . f(x1,y1)       y2= 1 + 0,25 . (4.x1 + 2.y1)               y2= 1 + 0,25 . (4.1,25 + 2.1)                        y2= 2,75

y3= y2 + h . f(x2,y2)       y3= 2,75 + 0,25 . (4.x2 + 2.y2)         y3= 2,75 + 0,25 . (4.1,50 + 2.2,75)             y3= 5,625

y4= y3 + h . f(x3,y3)       y4= 5,625 + 0,25 . (4.x3 + 2.y3)      y4= 5,625 + 0,25 . (4.1,75 + 2.5,625)       y4= 10,1875