4th order Runge-Kutta Problem in Special ranges

I want to solve this question below in Matlab but i didn't do it. This is simple question but i can't do it. If someone will help me, i will be very happy.

Question: Write a Matlab code that finds the approximate solution of the initial-value problem y '= - 2x-y, y (0) = - 1 using the 4th order Runge-Kutta method for the h = 0.1 step interval in the range [0.0,0.6].

Note: The 4th Order Runge-Kutta method is expressed as follows;

..................................................................................................................................................

Original Question:

I tried this code block, but it didn't work:

clc; % Clears the screen

clear all;

h=0.1; % step size

x = 0:h:3; % Calculates upto y(3)

y = zeros(1,length(x));

%y(0) = [0.0;0.6]

y(0) = -1; % redo with other choices here. % initial condition

F_xy = @(x,y) -2*x-y; % change the function as you desire

for i=1:(length(x)-1) % calculation loop

k_1 = F_xy(x(i),y(i));

k_2 = F_xy(x(i)+0.5*h,y(i)+0.5*h*k_1);

k_3 = F_xy((x(i)+0.5*h),(y(i)+0.5*h*k_2));

k_4 = F_xy((x(i)+h),(y(i)+k_3*h));

y(i+1) = y(i) + (1/6)*(k_1+2*k_2+2*k_3+k_4)*h; % main equation

end

Answer 1

[Alan Stevens]

Change

y(0) = -1;  

to

y(1) = -1;  

Matlab indices start at 1 not zero.

[Rooter Boy]

I did it but doesn't work.

[James Tursa]

Please explain "doesn't work" for us. Did you get an error? An answer you didn't expect? Or ...?

According to the assignment, the x values should range between 0.0 and 0.6, so maybe you should be using this:

x = 0:h:0.6; 

[Rooter Boy]

ans =

Empty state-space model.

[Rooter Boy]

Sir, I changed it but still same answer. x = 0:h:0.6;

[James Tursa]

I don't know what your "Empty state-space model" message means in the context of the code you have posted thus far.

When I make these changes to your code:

y(1) = -1;
x = 0:h:0.6;

and run your code, this

plot(x,y)

produces this:

[Rooter Boy]

It comes out in the same way, but the answer must be numerical. For example answer is: =28.9306

[James Tursa]

The answer is in the y vector, which is numerical. I still don't understand your problem. E.g.,

>> x
x =
         0    0.1000    0.2000    0.3000    0.4000    0.5000    0.6000
>> y
y =
   -1.0000   -0.9145   -0.8562   -0.8225   -0.8110   -0.8196   -0.8464

[Rooter Boy]

Sir, I added example about it.

[James Tursa]

This is a different DE than you originally posted. Also your arithmetic is wrong. E.g., your very first k1:

k1 = h * f(x0,y0) = 1 (x0 + 2y0 + 1) = 1*(0+2*2+1) = 5

But you have erroneously written 1*(0+2+1) = 3

So you need to re-examine all of your arithmetic for this.

[Rooter Boy]

Sir, i added another different example.

[James Tursa]

So when I make these changes

h=0.1;
x = 0:h:0.6; 
y(1) = 0.2;
F_xy = @(x,y) y-y^2;

and run your code I get the following:

>> y(2)
ans =
    0.2165

So, again, it looks like things are working to me.

[Rooter Boy]

Sir, thank you so much. The answer is probably like you said.