how can i resolve this equation Runge kutta method
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y''+1=0
ddy/ddt + 1 =0
how can i resolve this equation with runge kutta method
6 comentarios
James Tursa
el 22 de En. de 2020
@adem: You need to show us what you have done so far and then we can help you with your code.
Respuestas (2)
Jim Riggs
el 22 de En. de 2020
Ths is the equation for a parabola.
y'' = -1
y' = -y
y = -1/2 y^2
No need for a numerical approximation.
2 comentarios
James Tursa
el 22 de En. de 2020
Editada: James Tursa
el 23 de En. de 2020
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.
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