Question

Solve the ordinary differential equation below using Runge-Kutta 4th order method.

Step size h=0.2

5 dy/dx+xy3= cos(x) ,y(0) = 3

The value ofy(0.2) is (upto two decimal points)

Answer 1

Given :

Differencial equation : 5Dy + xy³ = cos x

y(0) = 3,

step size h = 0.2

To find :

The value of y(0.2) by Runge-Kutta 4th order method.

Solution:

As it is given that,

5Dy + xy³ = cos x,

$Dy = f(x,y) = \frac{1}{5} ( \cos x \: - x {y}^{3} )$

(equation 1)

According to Runga-Kutta 4th order method :

$</p><p> \\ \bf \: y(x+h) = y(x) + \frac{1}{6} (F_1 + 2F_2 + 2F_3 + F_4)</p><p></p><p>$

(equation 2)

where h = step size = 0.2 given,

x = 0 given,

Carl Runge and Wilhelm Kutta wanted to provide a method of approximating a function without use of differentiation of the original equation.

This approach was to simulate so many steps of the Taylor's Series method but only by using evaluation of the original function .

In this case,

$F_1 = hf(x,y)$

(equation 3)

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

(equation 4)

$F_3 = h(x+\frac{h}{2},y+\frac{F_2}{2})$

(equation 5)

$F_4 = h(x+h,y+F_3)$

(equation 6)

So, we have to find all the values of unknowns from equation 2,

as

h = 0.2,

x =0,

y(0) = 3,

putting in equation 3, we get :

$F_1 = 0.2f(0,3) = 0.2 \times ( \frac{1}{5} (\cos(0) - (0 \times {3}^{3))} = 0.2 \times \frac{1}{5} \times 1\\ F_1 = 0.04$

then,

by equation 4, we get :

$F_2 = 0.2f(0.1,3.02) = 0.2 \times ( \frac{1}{5} (\cos(0.1) - (0.1\times {3.02}^{3))} \\ F_2 = 0.04(0.99 - 2.75) \\ F_2 = - 0.07$

then by equation 5, we get :

$F_3 = 0.2f(0.1,2.965) = 0.2 \times ( \frac{1}{5} (\cos(0.1) - (0.1\times {2.965}^{3))} \\ F_3 = 0.04(0.99 - 26.06) \\ F_3 = - 0.0646$

then by equation 6, we get :

$F_4 = 0.2f(0.2,2.9345) = 0.2 \times ( \frac{1}{5} (\cos(0.2) - (0.2\times { 2.9345}^{3))} \\ F_4 = 0.04(0.99 -25.2698312 ) \\ F_4 = 0.9711$

Putting all the vaues in equation 2,

$</p><p> \\ \bf \: y(0 + 0.2) = y(0) + \frac{1}{6} (0.04 + (2 \times - 0.07) +( 2 \times - 0.0646)+ 0.9711)</p><p></p><p>$

we know that

y(0) = 3,

so

$</p><p> \\ \bf \: y( 0.2) = 3+ \frac{1}{6} (0.04 - 0.14 - 0.1292 + 0.9711)</p><p></p><p>$

then

$</p><p> \\ \bf \: y( 0.2) = 3+ (\frac{1}{6} \times 1.1019)</p><p></p><p>$

so

$</p><p> \\ \bf \: y( 0.2) = 3.18365$

only upto two decimal numbers,

y(0.2) = 3.18