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1-Visitor
February 5, 2019
Question

How to solve symbolic ODE system

  • February 5, 2019
  • 3 replies
  • 2123 views

Hi everyone,

I'm trying to solve this EDO system, finding the solution for x(t) and p(t) with {k1,k2,b} symbolic variables and m=535:

Sistema.jpgSystem Equations:

m*x''(t)=-k1*(x(t)-p(t))-b*(x'(t)-p'(t))

k2*p(t)=k1*(x(t)-p(t))+b*(x'(t)-p'(t))

 

Initial Conditions would be:

x(0)=0

p(0)=0

x'(0)=6.3

 

Already tried with Wolfram Alpha and Matlab, both giving solutions with terms "Root[#1^3+ (k1 m + k2 m) #1^2 + b^2 k2 m #1 + b^2 k1 k2 m^2&, 1]" in which I don't understand what is represented. Examples attached in following pictures:Wolfram.JPG

 

Matlab:

matlab.JPG

 

 

 

3 replies

25-Diamond I
February 5, 2019

Mathcad can solve ODEs only numerically, not symbolically.

 

The "Root" you see in the symbolic solutions of Matlab and Wolfram possibly is the solution of a cubic equation which cannot be given in a closed form as its form depends on the values of the various constants.

 

14-Alexandrite
February 5, 2019

Seems like Maple will give an explicit solution without any Root's

Here's the commands

de.png

 

The full result in in the attached PDF

23-Emerald IV
February 5, 2019

Here's how you could (try to) solve it:

LM_20190205_OdeSetSolve0.png

LM_20190205_OdeSetSolve1.png

So now we have X and P...

LM_20190205_OdeSetSolve2.png

And that's where it becomes a little problematic. You have to take the inverse Laplace transform of a ratio of polynomes where the order of the denomonator polynome is 3. This is not easy, but apparently modern Maple can do it, as Mvenich shows. I've entered one of the two expressions in WolframAlpha, to find a similar expression involving the root() function:

LM_20190205_OdeSetSolve3.png

Conclusion: there's no simple closed form.

However...

LM_20190205_OdeSetSolve4.png

Success!
Luc

P.S. I suppose this timeline is more interesting:

LM_20190205_OdeSetSolve5.png