how can i resolve this equation Runge kutta method

y''+1=0

ddy/ddt + 1 =0

how can i resolve this equation with runge kutta method

[James Tursa]

You can use the ode45 function.

doc ode45

If you need to code a Runge Kutta method yourself, please show what you have done so far and tell us what problems you are having.

[adem ski]

i want use Runge Kutta i don't know how can i Define the equation

[Jim Riggs]

ode45 is a Runge-Kutta method (Dormand-Prince)

[adem ski]

i want use this

k1=h*feval(f,t,y(m));

k2=h*feval(f,t+(1/2)*h,y(m)+(1/2)*k1);

k3=h*feval(f,t+(1/2)*h,y(m)+(1/2)*k2);

k4=h*feval(f,t+h,y(m)+k3);

y(m+1)=y(m)+(1/6)+(k1+2*k2+2*k3+k4);

[James Tursa]

@adem: You need to show us what you have done so far and then we can help you with your code.

[adem ski]

clear all
h=0.01;
t=0:0.01:10;
y(1)=1;
m=1;
f=inline('............');
for t=0:0.01:10
    k1=h*feval(f,t,y(m));
    k2=h*feval(f,t+(1/2)*h,y(m)+(1/2)*k1);
    k3=h*feval(f,t+(1/2)*h,y(m)+(1/2)*k2);
    k4=h*feval(f,t+h,y(m)+k3);
    y(m+1)=y(m)+(1/6)+(k1+2*k2+2*k3+k4);
    m=m+1;
end 

i don't know how to define the equation?

Answer 1

[Jim Riggs]

Ths is the equation for a parabola.

y'' = -1

y' = -y

y = -1/2 y^2

No need for a numerical approximation.

[adem ski]

i want use Runge Kutta i don't know how can i Define the equation

[Jim Riggs]

ddy = @(t, y) -1;

Answer 2

[James Tursa]

You've got a 2nd order equation, so that means you need a 2-element state vector. The two states will be y and y'. All of your code needs to be rewritten with a 2-element state vector [y;y'] instead of the single state y. I find it convenient to use a column vector for this. So everywhere in your code that you are using y, you will need to use a two element column vector instead. E.g., some changes like this:

y = zeros(2,numel(t)+1);  % pre-allocate the solution, where y(:,m) is the state at time t(m)
    :
y(:,1) = [1;0]; % need to initialize two elements at first time, position and velocity
    :
f = @(t,y)[y(2);-1]; % [derivative of position is velocity; derivative of velocity is acceleration = -1]
    :
    k1 = h*feval(f, t        , y(:,m)         ); % changed y(m) to y(:,m)
    

Also, you have a bug in the line that combines the k's ... you need (1/6)* instead of (1/6)+

Make an effort at implementing these changes and then come back with any problems you continue to have.