Solving a system of equations in R2014a

What am I doing wrong?:
being "beta_ex" a known function dependent on 't',
syms t r_ex_trial alpha_ex_trial
u1=-r_ex_trial*cos(alpha_ex_trial);
v1=-sin(beta_ex);
u2=r_ex_trial*sin(alpha_ex_trial);
v2=-cos(beta_ex);
eqns=[u1==v1+10, u2==v2+5];
vars=[r_ex_trial, alpha_ex_trial];
[sol1,sol2]=solve(eqns,vars,'ReturnConditions',true);
Why does this happen?:
Warning: 4 equations in 2 variables.
Somehow it does not sepparate the equations from the variables. (I've already tried in different ways)
Thanks in advance!

Respuestas (1)

Alan Weiss
Alan Weiss el 13 de Mayo de 2015
After I played with your system for a couple of minutes I got the following error:
Error using solve (line 258)
The number of output arguments must equal the number of independent variables in a
system plus 2.
So I tried the following:
[sol1,sol2,sol3,sol4]=solve(eqns,'ReturnConditions',true);
It worked.
This also worked:
[sol1,sol2,sol3,sol4]=solve(eqns,vars,'ReturnConditions',true);
Alan Weiss
MATLAB mathematical toolbox documentation

1 comentario

José Sarilho
José Sarilho el 13 de Mayo de 2015
yes, that way it should return 4 constants.
I was looking to obtain r_ex_trial and alpha_ex_trial as functions of beta_ex (3 symbolic variables, where beta_ex is known)

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Preguntada:

el 13 de Mayo de 2015

Editada:

el 13 de Mayo de 2015

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