Dane

a = 6 b = 15 c = 6

Warunki początkowe:

t_k=0 y=1 y'=0 y”=0 dla h=0,01

Układ otwarty

y(s)=x(s)×G₁(s)×G₂(s)

Układ zamknięty

y(s)=[x(s)×G₁(s)×G₂(s)]-[y(t)×G₁(s)×G₂(s)]

y(s)+y(s)×G₁(s)×G₂(s)=x(s)×G₁(s)×G₂(s)

G(s)=

Równanie różniczkowe:

y(s)(s³+16s²+16s+37)=36x(s)

y'”(t)+ 16y”(t)+ 16y'(t)+36y(t)=36x(t)

z' z x

K y' = x

M x' = z

N z' = -16z-16x-37y+36

K₁ = hx_k

M₁ = hz_k

N₁ = h(-16z_k-16x_k-37y_k+36)

K₂ = h(x_k+M₁/2)

M₂ = h(z_k+N₁/2)

N₂ = h[-16(z_k+N₁/2)-16(x_k+M₁/2)-37(y_k+K₁/2)+36]

K₃ = h(x_k+M₂/2)

M₃ = h(z_k+N₂/2)

N₃ = h[-16(z_k+N₂/2)-16(x_k+M₂/2)-37(y_k+K₂/2)+36]

K₄ = h(x_k+M₃)

M₄ = h(z_k+N₃)

N₄ = h[-16(z_k+N₃)-16(x_k+M₃)-37(y_k+K₃)+36]

y_k+1 = y_k+1/6(K₁+2K₂+2K₃+K₄)

x_k+1 = x_k+1/6(M₁+2M₂+2M₃+M₄)

z_k+1 = z_k+1/6(N₁+2N₂+2N₃+N₄)

Wykresy:

y_k = f(t_k) x_k = f(t_k) x_k = f(y_k) z_k = f(y_k)