Error when solve system of equation using vpasolver

function [BF1,BF2,BF3,BF4,CO,COreq] = solveBF(k,MaxBFvsk,PctBF,MuVO2,COmax,BF1,BF2,BF3,BF4,CO,COreq) 

%UNTITLED5 Summary of this function goes here
%   Detailed explanation goes here

syms BF1(k) BF2(k) BF3(k) BF4(k) CO(k) COreq(k) 

eqn1 = BF3(k)-(0.012*CO(k));                        %ml/min

eqn2 = BF1(k)-(0.92*CO(k));                         %ml/min

eqn3 = BF4(k)-(MaxBFvsk*PctBF*1000);                %ml/min

eqn4 = BF2(k)-((100*MuVO2)/(0.8*210));

eqn5 = COreq(k)-BF1(k)-BF2(k)-BF3(k)-BF4(k);

eqn6 = CO(k)-(0.5*(COreq(k)+COmax));                   %ml/min   CO=HRR*SV is another method

[BF1,BF2,BF3,BF4,CO,COreq] = vpasolve(eqn1,eqn2,eqn3,eqn4,eqn5,eqn6);

end

And the error says:

Error using sym/vpasolve (line 149)
Input with 1 variables and output with 6 variables are inconsistent.

Error in solveBF (line 20)
[BF1,BF2,BF3,BF4,CO,COreq] = vpasolve(eqn1,eqn2,eqn3,eqn4,eqn5,eqn6);

Error in test (line 159)
[BF1,BF2,BF3,BF4,CO,COreq] =
solveBF(k,MaxBFvsk,PctBF,MuVO2,COmax,BF1,BF2,BF3,BF4,CO,COreq);

[Torsten]

Not clear why BF1,BF2,BF3,BF4,CO,COreq appear as input and output arguments in your function solveBF. In my opinion, they are only output values.

Best wishes

Torsten.

Answer 1

[Star Strider]

The easiest way to find out what variables vpasolve is solving for is to re-write it as:

solution = vpasolve(eqn1,eqn2,eqn3,eqn4,eqn5,eqn6);

That will return the variables it finds, in the ‘solution’ structure. You can then see what they are.

Another possibility is to rewrite it as:

[BF1,BF2,BF3,BF4,CO,COreq] = vpasolve([eqn1,eqn2,eqn3,eqn4,eqn5,eqn6], [BF1,BF2,BF3,BF4,CO,COreq]);

See which one works best.

[jin wang]

if I change the way you write it as the second option, the error says

Undefined function 'vpasolve' for input arguments of type 'double'.

Error in solveBF (line 18) [BF1,BF2,BF3,BF4,CO,COreq] = vpasolve([eqn1,eqn2,eqn3,eqn4,eqn5,eqn6], [BF1,BF2,BF3,BF4,CO,COreq]);

Error in test (line 159) [BF1,BF2,BF3,BF4,CO,COreq] = solveBF(k,MaxBFvsk,PctBF,MuVO2,COmax,BF1,BF2,BF3,BF4,CO,COreq);

Answer 2

[Walter Roberson]

When you use

syms BF1(k) BF2(k) BF3(k) BF4(k) CO(k) COreq(k)

you are doing

BF1 = sym('unapply(BF1,k)');
BF2 = sym('unapply(BF2,k)');

or similar. In each case you are disconnecting what was passed in as BF1, BF2, etc., from the local meaning of BF1, BF2, etc., just the same way as the below disconnects "x" from what was passed in.

   function solveBF(x)
     x = 5;

"syms" is an assignment statement, not a type declaration.

syms X

means like

syms = @(Name) assignin('caller', Name, sym(Name));   %remember syms is called in command syntax so parameters are strings

If your input BF1, BF2, and so on are already symbolic functions, then you do not need to syms them: MATLAB will already know their datatype is symbolic, the same way it would know the datatype of "x" in my above example. If the inputs are not already symbolic functions then you need to use different names for your syms and possibly subs() the input BF1, BF2 and so on in to those functions or call the functions with those input values.