Solving symbolic Matlab equation in term of two variables (x/Va)

Guys/Girls any input will be great. I'm so stuck been spinning my wheels for hours. so.. here is what I got

All i want to do is solve this equation symbolically, for x/va. (the transfer function) so I want matlab to spit out the answer x/Va = blah blah blah. but I'll I get is weird matrices or errors. Honestly any help is appreciated. Thanks again.

Answer 1

[Walter Roberson]

syms s J B x K kt Va ke Ra La 
La = 0;
eqn = (J*(s^2)*x)+ (B*s*x)+ (K*x)== kt*((Va)-(ke*s*x)/(Ra+(La*s)));
syms F
eqn1 = subs(eqn, x, F*Va);
x/Va == simplify( solve(eqn1, F) )

[Mitch Ott]

is there anyway you could write that out a bit better? i'm new to matlab and having a hard time understanding what your doing. your code works, but I'm not sure what it is doing

[Walter Roberson]

syms s J B x K kt Va ke Ra La 
La = 0;
eqn = (J*(s^2)*x)+ (B*s*x)+ (K*x)== kt*((Va)-(ke*s*x)/(Ra+(La*s)));
syms F
eqn1 = subs(eqn, x, F*Va);
eqn_in_terms_of_F = solve(eqn1, F);
cancel_extra_terms_in_eqn = simplify(eqn_in_terms_of_F);
left_side_for_presentation = x/Va;
desired_eqn = left_side_for_presentation == cancel_extra_terms_in_eqn

[Mitch Ott]

Why are you substituting F/Va for x? All I am trying do is rearrange the whole equation and have x/Va on the Left and all the other stuff on the right. Thanks for your help.

Edit: I did the math by hand and your answer is totally correct, I just for the life of me can't figure out the substitution.

[John D'Errico]

Mitch -

Suppose you decided to define a variable F, such that

F = x/Va

Then it is true that

x = F*Va

Right?

The idea is to put F into the equation. You will need to replace either x or Va. So when you do the substitution for x here, now you can solve for F. But what is F? Yep. F=x/Va. Which happens to be exactly what you wanted to see all alone on the left hand side.

[Walter Roberson]

You want F = x/Va . Multiply through by Va, to get F*Va = x. Substitute that F*Va for x in the formula, which gets rid of all of the x in the formula and puts in the new variable F instead, where we know that F = x/Va because that is how we built F. Solve for the variable F, to get something that is implicitly of the form F = SomeSolution. Now expand back F on the left side because that is what you expect to see, so now you have x/Va = SomeSolution

MATLAB cannot solve() for expressions, only for variables, so this is a very common mathematical trick, to take an existing variable that is involved with the formula you want to solve with respect to, and substitute for that variable an expression in a new variable that is the inverse. It is a "change of variables"

[Mitch Ott]

Walter, John, thank you so much. I appreciate you taking the time to help.