**Update:**One of the numerical example (example 1) has incorrect numerical results. This blog has been edited. The results posted are now correct. Apologizes. - Eddie

1/4/2015

**Introduction**

In this blog we will explore one of the ways to solve the differential equation y' = f(x,y) by numerical methods. One of the common methods is the Runge-Kutta 4th Order Method, probably the most common methods.

With the initial condition (x_n, y_n) and step size h, the next point (x_n+1, y_n+1) is calculated by the following equations:

x_n+1 = x_n + h

y_n+1 = y_n + (k1 + 2*k2 + 2*k3 + k4)/6 with

k1 = h * f(x_n, y_n)

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

k3 = h * f(x_n + h/2, y_n + k2/2)

k4 = h * f(x_n + h, y_n + k3)

(Almost) obviously use either a calculator or mathematical software for this process.

Examples:

1. dy/dx = -0.23 * (y + 1)

Initial Condition: (0, 24)

We want y1 when x1 = 1 and y2 when x2 = 2.

Observe that h = 1 with x0 = 0 and y0 = 24. Note the lack of x term in f(x,y).

Then: x1 = 0 + 1 = 1

k1 ≈ -5.75

k2 ≈ -5.08875

k3 ≈ -5.16479375

k4 ≈ -4.562097438

y1 ≈ 18.86346918

x2 = 1 + 1 = 2

k1 ≈ -4.568597911

k2 ≈ -4.043209151

k3 ≈ -4.103628858

k4 ≈ -3.624763273

y2 ≈ 14.78229631

So our points are:

(0, 24)

(1, 18.86346918)

(2, 14.78229631)

2. y' = y - x with I.C. y(0) = 2

Find y(0.1) and y(0.2).

Note h = 0.1 with x0 = 0, y0 = 2, and f(x,y) = -x + y.

x1 = 0.1

k1 = 0.1 * f(0, 2) = 0.2

k2 = 0.1 * f(0.5 * 0.1, 2 + 0.5 * 0.2) = 0.205

k3 = 0.1 * f(0.5 * 0.1, 2 + 0.5 * 0.205) = 0.20525

k4 = 0.1 * f(0.1, 2 + 0.20525) = 0.210525

y1 = 2 + (0.2 + 2 * 0.205 + 2 * 0.20525 + 0.210525)/6 ≈ 2.205171

x2 = 0.2

k1 ≈ 0.210517

k2 ≈ 0.216043

k3 ≈ 0.216319

k4 ≈ 0.222149

y2 ≈ 2.421203

So our points are:

(0, 2)

(0.1, 2.205171)

(0.2, 2.421203)

Talk to you all next time!

Eddie

This blog is property of Edward Shore. 2013

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