Which of the following Python code snippets correctly implements the 4th-order Runge-Kutta method?

Question

Answer

def runge_kutta_4th_order(f, x0, y0, h, n):

x = x0

y = y0

for _ in range(n):

k1 = h * f(x, y)

k2 = h * f(x + h, y + k1)

k3 = h * f(x + h, y + k2)

k4 = h * f(x + h, y + k3)

y = y + (k1 + 2*k2 + 2*k3 + k4) / 6

x = x + h

return y

0%

Answer

def runge_kutta_4th_order(f, x0, y0, h, n):

x = x0

y = y0

for _ in range(n):

k1 = h * f(x, y)

k2 = h * f(x + h/2, y + k1/2)

k3 = h * f(x + h/2, y + k2/2)

k4 = h * f(x + h, y + k3)

y = y + (k1 + k2 + k3 + k4) / 4

x = x + h

return y

50%

Answer

def runge_kutta_4th_order(f, x0, y0, h, n):

x = x0

y = y0

for _ in range(n):

k1 = h * f(x, y)

k2 = h * f(x + h, y + k1)

k3 = h * f(x + h, y + k2)

k4 = h * f(x + h, y + k3)

y = y + (k1 + k2 + k3 + k4) / 4

x = x + h

return y

0%

Answer

def runge_kutta_4th_order(f, x0, y0, h, n):

x = x0

y = y0

for _ in range(n):

k1 = h * f(x, y)

k2 = h * f(x + h, y + k1)

k3 = h * f(x + h, y + k2)

k4 = h * f(x + h, y + k3)

y = y + (k1 + k2 + k3 + k4) / 4

x = x + h

return y

0%

Answer

def runge_kutta_4th_order(f, x0, y0, h, n):

x = x0

y = y0

for _ in range(n):

k1 = h * f(x, y)

k2 = h * f(x + h, y + k1)

k3 = h * f(x + h, y + k2)

k4 = h * f(x + h, y + k3)

y = y + (k1 + 2*k2 + 2*k3 + k4) / 6

x = x + h

return y

0%

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Which of the following Python code snippets correctly implements the 4th-order R...

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