Mathcad fails to solve symbolic system of equations and put it in a function!

Forum|Forum|7 years ago
February 14, 2019
4 replies
4906 views

Hello.

Before anything, I got to say that I searched the whole forum for similar questions (which existed), and tried to apply them to my case, but none worked!

I attached files (one is Mathcad15 and the other is Prime 5 file format) with system of equations that I need to solve symbolically, and then attribute this solution to a function.

The system of equations has x's, a's, z's and a y variable. The goal is to define y as a function of x's and a's, but no z's.

I failed to do it multiple times: sometimes Mathcad just crashs, other times it just says some variables are not defined...

Could someone please help me? This is the first thing I'm ever trying to do using Mathcad!

Thank you very much!

2_eqSys.zip

1_eqSys.zip

Best answer by waverectifier

So I tried to solve a 20x20 similar linear system in both Mathcad and Matlab.

Matlab was able to solve it under 5 minutes. I let Mathcad running for 8 hours and nothing...

L

LucMeekes

23-Emerald IV

I may have made a mistake in retyping, otherwise your system is not complete. You need to solve for z1 through z8 AND y, you only have 8 equations. You need one more to solve the 9 unknowns:

(Note that the picture is chopped off at the right side. It's very very wide....)

Success!
Luc

1_LM_20190214_Solve8eq.zip

W

waverectifier

Author

1-Visitor

The strange thing is that this is a model for a real world system. I cannot find any other equation/relationship as to maintain the system's fidelity.

I could find a solution, but it was doing:

z1 = y/2

z2 = -y/2

But i really don't know the value of z1 and z2, I only know that their difference (z1-z2) is equal y.

W

Werner_E

25-Diamond I

> The system of equations has x's, a's, z's and a y variable. The goal is to define y as a function of x's and a's, but no z's.

This would mean that you want to solve your system for the eight z's and for y.

That makes nine variables to solve for but only eight equations!

This means that you have one degree of freedom. In other words - the solution(s) probably aren't unique!

To define the Function Y you are looking for, you don't need a symbolic solution of your system. If you are just looking for one of the multiple solutions you may use a parametrized numeric solve block - you'll have to provide suitable guess values if you do so.