| An illustration of the Runge-Kutta algorithm appliedto the differential equation `y' = 2y` witha step size of 1 and initial value `y(0) = 1/4`.The predicted value is a linear combination, withcoefficients `1/6, 2/6, 2/6, 1/6` of the values obtained by - following the (blue) slope 1/2 line at
`(0,1/4)`. This gives `y = 3/4`. - following the (green) line with the slope (1) picked up by theblue line at
`x = 1/2`. This gives `y = 5/4`. - following the (orange) line with the slope (3/2) picked up by thegreen line at
`x = 1/2`. This gives `y = 7/4`. - following the (black) line with the slope (7/2) picked up by theorange line at
`x = 1`. This gives `y = 15/4`. The total is `(1/24)(3+10+14+15) = 1.75`.The same computational load would buy 4 iterations of Euler'smethod yielding 1.265. The exact value is `y(1) = (1/4)e`^{2} = 1.85. |