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y''+1=0

ddy/ddt + 1 =0

how can i resolve this equation with runge kutta method

Jim Riggs
on 22 Jan 2020

Ths is the equation for a parabola.

y'' = -1

y' = -y

y = -1/2 y^2

No need for a numerical approximation.

James Tursa
on 22 Jan 2020

Edited: James Tursa
on 23 Jan 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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