Second Order ODE basics?

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Michael 2014-8-4

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评论： Star Strider 2014-8-4

Hi there... I'm a bit confused on how to solve 2nd order ODE's in Matlab. I feel like I can correctly reduce the 2nd order system to a 1st order system on pencil and paper, but when I go to Matlab, things don't work out.

For example, I'm trying to solve:

and I have the code:

    f = @(x) [x(2);sin(x)-x(1)-x(2)];
tSpan = [0 10];
initial = [1;0];
[t,y] = ode23(f,tSpan,initial)

but I can't figure out how to finish it. I feel like I'm interpreting the "x(1)" and "x(2)" incorrectly. On paper, those are the subscripts of my "new" variables.

Any help is appreciated! :)

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Answer 1

Star Strider 2014-8-4

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You’re almost there. The ODE function needs to return a value for your independent variable as well as the derivative, and the argument to sin needs to be x(1). Otherwise your essential maths for converting the second-order ODE to two first-order ODEs is correct, so you can relax as far as that goes.

This works:

f = @(t,x) [x(2); sin(x(1))-x(1)-x(2)];