how to substitute symbolic equation into symbolic equation, and to reorganize symbolic equations

I am trying to learn symbolic equations, but I'm worried they don't work the way I think they work, or are not intended for what I am hoping ot use them for.

My expectation is that I can solve problems by substituting relationships into larger constituent equations and reduce them analytically, then isolate the variable of interest, substitute values and solve for instance:

syms epsilon_33 delta_0 d_33 s_33 f f_bl u k_s k_l n V A L_s t
% known constants
d_33 = 3400e-12;
s_33 = 169e-12;
f = 60;
u = 30e-6;
V = 300;

% Relevent relationships
k_l/k_s == 1
u == delta_0/(1+k_1/k_2)
delta_0 == n*d_33*V

% For this simple example, the first two relationships say that delta_0=2*u, so the third relationship can be solved for n

I can make Matlab spit out the answer but I have to do the analytical solution myself, which kind of defeats the purpose.

Is what I am trying to do outside the scope of this function?

Answer 1

[Sulaymon Eshkabilov]

There are a couple of points in the code to be corrected and then you can get an analytical solution expressed in terms of other symbolic variables, e.g.:

syms epsilon_33 delta_0 d_33 s_33 f f_bl u k_s k_l n V A L_s t
% known constants
d_33 = 3400e-12;
s_33 = 169e-12;
f = 60;
u = 30e-6;
V = 300;
% Relevent relationships
% delta_0=u/(1+k_1/k_2);
% delta_0 == n*d_33*V;
% Presumably k_2 is also symbolic variable, and thus, 
syms k_2
k_1 = 1/k_s;
Solution_n = solve(u/(1+k_1/k_2)-n*d_33*V==0, n)

Solution_n = 

[cdlapoin]

so you have to combine the symbolic equations yourself? That is essentially what I was asking if you can avoid.

Answer 2

[Paul]

syms epsilon_33 delta_0 d_33 s_33 f f_bl u k_s k_l n V A L_s t
% known constants
d_33 = 3400e-12;
s_33 = 169e-12;
f = 60;
u = 30e-6;
V = 300;
% Relevent relationships
eq1 = k_l/k_s == 1

eq1 = 

%u == delta_0/(1+k_1/k_2) % this line appeared to contain two typos
eq2 = u == delta_0/(1+k_l/k_s)

eq2 = 

eq3 = delta_0 == n*d_33*V

eq3 = 

Now we can use solve(). Solving for three variables returns the expected result for n

sol = solve([eq1 eq2 eq3],[n delta_0 k_l])

sol = struct with fields:
          n: 18889465931478580854784/321120920835135859375
    delta_0: 3/50000
        k_l: k_s

For reasons I don't understand, just asking to solve for n alone doesn't work, even though there is a clear solution

sol = solve([eq1 eq2 eq3],n)

 
sol =
 
Empty sym: 0-by-1
 

[Walter Roberson]

Except in some cases involving inequalities, you must solve() for the same number of variables as you have equations

Answer 3

[burak enes kavas]

% Do you want analyticaly reduce  any function use this comand
syms x  
func  =  sin ( x )
diff ( func , x)
ans =  cos( x )
% Make analytical solution
solve ( cos ( x  ) )
ans = pi/2   %  variable type is sybolic
% change the varible type
double(ans)   % variable type is integer

ans = 1.5708

more exaple

syms x
F = x^2 - 4
solve ( F , x )

ans = [ 2 , -2 ]

G= 3*x^2 - 3*x + 3  
diff ( G ,x )

ans = 6*x - 3

i wish it is helpfull