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### Example of the Runge-Kutta Method

```
Date: 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/
```
Associated Topics:
College Calculus

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