how to take the derivative for the retained value from ode45?

I'm using the ode45 to find the velocity, now I need to find the acceleration how can I take the derivative of the retained value.

My ode45 is:

[ts,xs]= ode45(@my_function,[0,10],[5,0])

Thanks in advance!

Answer 1

[Walter Roberson]

gradient(xs(1,:), ts(:))

[Richard Brown]

With gradient you provide the spacing, not the independent variable coordinates. Also, you need to index the column not row of the solution.

If you get the ode solver to return values at uniformly spaced points, you can then use gradient

e.g.

h = 1e-2;
ts = 0:h:10;
[~,xs] = ode45(@my_function, ts, [5, 0]);
gradient(xs(:, 1), h);

assuming velocity is your first variable

edit actually you can provide the coordinates, as Walter points out below). So the only correction needed is

gradient(xs(:, 1), ts)

I'll leave the rest of the comment up to avoid confusion later, but that's all you need

[Walter Roberson]

Thanks, Richard.

[Richard Brown]

No problem. Confusingly the symbolic toolbox gradient works the way you described

[Walter Roberson]

Richard,

gradient(xs(:,1), ts)

will work as well. The use of a coordinate vector instead of a scalar spacing is documented for the 2D case but not for the 1D case, but it works anyhow.

[Richard Brown]

I stand corrected, good to know!

[Jan]

gradient uses a first order method when the spacing is not equidistant. You can use the faster method FEX: DGradient and some other equivalent tools from the FEX, which apply a 2nd order method to get more accurate results for the derivatives.

Answer 2

[Nawal]

Can I do something like

diff [ts,xs]