Example of the Runge-Kutta MethodDate: 02/10/2001 at 04:37:22 From: Joel Mitre Subject: Runge-Kutta Method Please explain the Runge-Kutta method for solving ordinary differential equations of the form y' = f(x,y). Date: 02/10/2001 at 05:48:56 From: Doctor Anthony Subject: Re: Runge-Kutta Method The Runge-Kutta method was developed as a step-by-step method for solving differential equations without recourse to Taylor series and the frequently very awkward calculation of first, second, third and higher derivatives. The method is best illustrated by an example. Use the 4th order Runge-Kutta method with h = 0.1 to find the approximate solution for y(1.1), working to 4 decimal places, for the initial value problem: dy/dx = 2xy, y(1) = 1 We have dy/dx = f(x,y) = 2xy. If you require the 4th order approximation the formula will be: y(x0 + h) = y(x0) + (1/6)[w1 + 2w2 + 2w3 + w4] where: w1 = h*f(x0.y0) = 0.1(2)(1)(1) = 0.2 w2 = h*f(x0 + h/2, y0 + w1/2) = 0.1(2)(1.05)(1.1) = 0.231 w3 = h*f(x0 + h/2, y0 + w2/2) = 0.1(2)(1.05)(1.1155) = 0.234255 w4 = h*f(x0 + h, y0 + w3) = 0.1(2)(1.1)(1.234255) = 0.2715361 and so: y(1.1) = 1 + (1/6)[0.2 + 2(0.231) + 2(0.234255) + 0.2715361] = 1.23367435 We can compare this with the exact solution to the problem. dy/y = 2x dx and integrating: ln(y) = x^2 + C and y = 1 when x = 1 0 = 1 + C and so C = -1. Therefore: ln(y) = x^2 - 1 y = e^(x^2-1) When x = 1.1 this gives: y = e^(1.1^2 -1) = e^0.21 = 1.23367806 which is same as Runge-Kutta up to the 5th decimal place. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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