Second Order Differential Equations
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I have seen all of the documentation of converting second order diffeq's to first order, but what if your equations are coupled...for instance:
y''[t] = 3x''[t] -4y[t];
x''[t] = 2y''[t] + 6x[t];
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回答 (1 件)
Brian B
2014 年 7 月 11 日
編集済み: Brian B
2014 年 7 月 11 日
You can rewrite that system with a constant mass matrix M. That is, the system above is equivalent to
M * d/dt[x1; x2; x3; x4] = [x2; x4; -4 x3; 6 x1]
where
M = [1 0 0 0; 0 0 1 0; 0 -3 0 1; 0 1 0 -2].
Use odeset to specify the mass matrix.
2 件のコメント
Helge
2015 年 2 月 20 日
編集済み: Helge
2015 年 2 月 20 日
Hi Community,
isn't it possible to rewrite the above differential equations, so they aren't coupled anymore in terms of the second derivative? I would do it as follows:
- Insert the 2nd eqn into the first, which gives: y''[t] = 3*(2y''[t] + 6x[t]) - 4y[t] and solve this for y''[t]:: y''[t] = -18/5 * x[t] + 4/5 y[t]
- Re-Insert this in the 2nd eqn from "Leila" above, which gives x''[t] = 2(-18/5 * x[t] + 4/5 * y[t]) + 6 * x[t] and solve for x''[t]:: x''[t] = 8/5 * y[t] - 6/5 * x[t]
- Now these two equations can be brought to State Space Representation and solved with ode45()
I tried to solve my problem this way and now I am unsure if that is even possible or do I have two use the mass matrix M in any case?
Best wishes, Helge
Torsten
2015 年 2 月 20 日
Everything is all right with your way of solving the above system.
Best wishes
Torsten.
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