a question on ode23 solver

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cgo 2017-6-1

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https://ww2.mathworks.cn/matlabcentral/answers/342813-a-question-on-ode23-solver

编辑： Jan 2017-6-1

采纳的回答： Jan

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Hi, I am using ODE23 solver in Matlab.

After inputting the following

t = linspace(0,10); 
y0 = 0.5;
u = 1.0; 
z = ode23(@(t,y)first_order(t,y,u),t,y0);
time = z.x;
y = z.y; 
plot(time,y)

and

function dydt = first_order(t,y,u)
tau = 5; 
K = 2.0;
dydt = (-y+K*u)/tau; 
end

The output is the plot of time versus y. I wanted to have a plot of y' and t so that I can make an analysis of the ODE without having to solve it. Is there a way of doing it?

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Answer 1

Jan 2017-6-1

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https://ww2.mathworks.cn/matlabcentral/answers/342813-a-question-on-ode23-solver#answer_269157

编辑：Jan 2017-6-1

在 MATLAB Online 中打开

You have y and t from the integration the function first_order calculates the needed y' already. Then simply use it after the integration:

dy = first_order(time, y, u);

Then plotting is easy.

But you asked for "make an analysis of the ODE without having to solve it". This is not possible, because you do not have the required trajectory of y without an integration. All you have is the equation in first_order(), which allows to obtain y' depending on t, y and the initial value y0. But you cannot use this without solving the ODE, because then you do not have y(t).