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solving Differential Equations

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Vijay Marathe
Vijay Marathe am 12 Okt. 2011
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.

Akzeptierte Antwort

Grzegorz Knor
Grzegorz Knor am 12 Okt. 2011
  2 Kommentare
Vijay Marathe
Vijay Marathe am 12 Okt. 2011
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 am 12 Okt. 2011
You have to reduce to a system to first-order ODEs, because ode solvers solve ony first-order ODEs.

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Weitere Antworten (1)

Walter Roberson
Walter Roberson am 12 Okt. 2011
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 Kommentare
Grzegorz Knor
Grzegorz Knor am 12 Okt. 2011
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 am 12 Okt. 2011
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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