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solving Differential Equations

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Vijay Marathe
Vijay Marathe 2011 年 10 月 12 日
I have following Differential Equations
a1*x2+b2*x1-c*cos(int(x3))*x4=d
-c*cos(int(x3))*x2-a2*sin(int(x3))+b2*x4=0
where a1,b2,c,d,a2,b2 are constants and x1=theta_dot; x2=theta_ddot; x3=alpha_dot; x4=alpha_ddot
I want to solve these equations for x1,x2,x3,x4 and want to plot with time.

採用された回答

Grzegorz Knor
Grzegorz Knor 2011 年 10 月 12 日
  2 件のコメント
Vijay Marathe
Vijay Marathe 2011 年 10 月 12 日
thanks, but I don't want to reduce the order of system.
how to solve for given input Vm, it may be ramp, step, const. input
eqs are
a1*x2+b2*x1-c*cos(int(x3))*x4=d*Vm
-c*cos(int(x3))*x2-a2*sin(int(x3))+b2*x4=0
I want plot x1,x2,x3,x4 with respect time.
Grzegorz Knor
Grzegorz Knor 2011 年 10 月 12 日
You have to reduce to a system to first-order ODEs, because ode solvers solve ony first-order ODEs.

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その他の回答 (1 件)

Walter Roberson
Walter Roberson 2011 年 10 月 12 日
Do you mean:
a1*x2(t)+b2*x1(t)-c*cos(int(x3(t), t))*x4(t) = d*Vm,
-c*cos(int(x3(t), t))*x2(t)-a2*sin(int(x3(t), t))+b2*x4(t) = 0
If so then there is no solution, or perhaps no solution without further information. 2 equations in 4 unknowns is seldom enough to be able to what the functions are, let alone the boundary conditions.
  2 件のコメント
Grzegorz Knor
Grzegorz Knor 2011 年 10 月 12 日
Author in his first post wrote, that there are dependencies between x1 and x2 and between x3 and x4. But still there is lack of the initial and boundary conditions.
Walter Roberson
Walter Roberson 2011 年 10 月 12 日
If x1 and x2 are functions of theta, then one cannot solve for them directly: one would have to express the functions in full first, solve the DE, and then construct x1 and x2 (and x3 and x4) from the results.

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