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Runge-Kutta's Method

Example: Use the Runge-Kutta Method to find an approximate solution to

dy/dx = 2 y , y(0) = 0.5 Use step size, h = 0.25 and y_exact = 2 e^2y

Strategy: Use the Runge-Kutta Algorithm as given below

y_n+1= y_n + h k f(x,y) = 2 y where h = step size

k = (1/6) [ k₁ + 2k₂ + 2k₃ + k₄ ]

and k₁ = f(x_n, y_n) , k₂ = f( x_n + h/2 , y_n + hk₁ / 2 )

k₃ = f( x_n + h/2 , y_n + hk₂ / 2 ) , k₄ = f( x_n+1 , y_n + hk₃ )

y₁= y₀ + h k y₀= 0.5 f(x,y) = 2 y

k₁ = f(x₀, y₀) = 2 y₀ = 2 (0.5) = 1.0

k₂ = f( x₀ + h/2 , y₀ + hk₁ / 2 ) = 2[0.5 + 0.25(1.0/2)] = 1.25

k₃ = f( x₀ + h/2 , y₀ + hk₂ / 2 ) = 2[0.5 + 0.25(1.25/2)] = 1.3125

k₄ = f( x₁ , y₀ + hk₃ ) = [0.5 + 0.25(1.3125)] = 1.65625

So ∑k_i = 1.0 + 2(1.25) + 2(01.3125) + 1.65625 = 7.78125

k = ∑k_i / 6 = 1.296875

y₁ = y₀ + h k = 0.5 + 0.25[1.296875] = 0.82421875 at x = 0.25

y₂= y₁ + h k y₁= 0.82421875

k₁ = f(x₁, y₁) = 2 y₁ = 2 (0.82421875) = 1.6484375

k₂ = f( x₁ + h/2 , y₁ + hk₁ / 2 ) = 2[(0.82421875 + 0.25((0.82421875/2)] = 2.060546875

k₃ = f( x₁ + h/2 , y₁ + hk₂ / 2 ) = 2[(0.82421875 + 0.25(2.060546875/2)] = 2.163574219

k₄ = f( x₂ , y₁ + hk₃ ) = 2[(0.82421875 + 0.25(2.163574219)] = 2.73022469

k = (1/6) [ k₁ + 2k₂ + 2k₃ + k₄ ] y_exact = 2 e^2(0.5)= 1.359140914

So k = [1.6484375+ 2(2.060546875) + 2(2.163574219) + 2.73022469]/6 = 2.137817383

y₂ = y₁ + h k = 0.82421875 + 0.25 [2.137817383 ] = 1.358673096 at x = 0.5

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