Answer to Question #158753 in Quantitative Methods for manish

Question

Use Runge-Kutta method of order 2 to solve y′ = xy, y(1) = 1, in [1, 1.4] by taking step-length h = 0.2

Answer 1

Runge - Kutta method of order 2

We begin with two functions evaluates of the form:

$F_1 = hf(x,y)$

$F_2 = hf(x+\alpha h, y + \beta F_1)$

The $\alpha$ and $\beta$ are unknown quantities. The idea was to take a linear combination of the

$F_1 \space and \space F_2$ terms to obtain an approximation for the y value at $x = x_o+h,$ and to find appropriate values of $\alpha$ and $\beta$.

By comparing the values obtains using Taylor's Series method and the above terms (I will spare you the details here), they obtained the following, which is Runge-Kutta Method of Order 2:

$y(x+h) = y(x) + \frac{1}{2}(F_1+F_2)$

We have: $y'(x, y) = xy$ y(1) = 1 [1, 1.4] h = 0.2

We start with x = 0 and y = 1. We'll find the F values first.

$F_1 = hf(x,y) = 0.2*1*1=0.2$

$x + \frac{h}{2} = 1 + 0.1 = 1.1$

$y + \frac{F_1}{2} = 1+0.1 = 1.1$

$y(x+h) = y(x) + \frac{1}{2}(F_1+F_2) = 1*1+\frac{1}{2}(1.1+1.1) = 2.1$