I need the Coding to solve the simultaneous equations. To find T, tow, TC.

lc
clear all
c0=40;
cd=50;
y0=0.09;
h=0.7;
r=5;
u=0.05;
a=260;
b=0.1;
x='((a*h)/2)+((a*r*T)/3)+((a*h*T*((exp(-u*tow))+b))/3+((a*r*(T^2)*(exp(-u*tow))+b))/8)-(c0/(T^2))+((cd*((3*a*y0*(exp(-u*tow)))-2(2*T*a*b*y0*(exp(-u*tow)))-(2*T*a*(b^2))/6)';
y='1-((u*a*h*(T^2)*y0*(exp(-u*tow)))/6)-((u*a*r*(T^3)*y0*(exp(-u*tow)))/24)-(cd*u*T*y0*(exp(-u*tow))*((3*a-a*b*T)/6))';
TC=(c0/T)+(1/T)*(((a*h*(T^2))/2)+((a*r*(T^3))/6)+((((a*(y0*(exp(-u*tow)))+b)/2))*(((h*(T^3))/3)+((r*(T^4))/12))))+((cd/T)*((((T^2)*y0*(exp(-u*tow))*(3*a-a*b*T))-(a*(b^2)*(T^3)))/6))+tow

Answer 1

[Walter Roberson]

You cannot solve that without fixing your formulas.

syms T tow
c0=40;
cd=50;
y0=0.09;
h=0.7;
r=5;
u=0.05;
a=260;
b=0.1;

A = a;   %??? maybe ?
R = r;   %??? maybe ?
B = b;   %??? maybe ?

x = ((a*h)/2)+((a*r*T)/3)+((a*h*T*((exp(-u*tow))+b))/3+((a*r*(T^2)*(exp(-u*tow))+b))/8)-(c0/(T^2))+((cd*((3*a*y0*(exp(-u*tow)))-2*(2*T*a*b*y0*(exp(-u*tow)))-(2*T*a*(b^2))/6);

Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for mismatched delimiters.

    12   1  0 12     1  0 12      34   5      43  21   23    4   3 4   5      43  21  0 1   2   10 12   34       5   6      543   4           5   6      543 4      5   43  2

You have two more ( than you have )

y = 1-((u*a*h*(T^2)*y0*(exp(-u*tow)))/6)-((u*a*r*(T^3)*y0*(exp(-u*tow)))/24)-(cd*u*T*y0*(exp(-u*tow))*((3*a-a*b*T)/6));

sol = solve([x,y], [T tow]);
T = sol.T;
tow = sol.tow;

TC = (c0/T)+(1/T)*(((a*h*(T^2))/2)+((a*r*(T^3))/6)+((((a*(y0*(exp(-u*tow)))+b)/2))*(((h*(T^3))/3)+((r*(T^4))/12))))+((cd/T)*((((T^2)*y0*(exp(-u*tow))*(3*a-a*b*T))-(a*(b^2)*(T^3)))/6))+tow;

disp(T)
disp(tow)
disp(TC)

vpa(T)
vpa(tow)
vpa(TC)

Answer 2

[shunmugam hemalatha]

Even i couldn't find answer for the same.

clc

clear all

c0=40;

cd=50;

y0=0.09;

h=0.7;

r=5;

u=0.05;

a=260;

b=0.1;

syms T tow

x='((a*h)/2)+((a*r*T)/3)+((a*h*T*((exp(-u*tow))+b))/3+((a*r*(T^2)*(exp(-u*tow))+b))/8)-(c0/(T^2))+((cd*((3*a*y0*(exp(-u*tow)))-2*(2*T*a*b*y0*(exp(-u*tow)))-(2*T*a*(b^2))/6)';

y = '(1-((u*a*h*(T^2)*y0*(exp(-u*tow)))/6)-((u*a*r*(T^3)*y0*(exp(-u*tow)))/24)-(cd*u*T*y0*(exp(-u*tow))*((3*a-a*b*T)/6))';

sol = solve([x,y], [T tow]);

T = sol.T;

tow = sol.tow;

TC = (c0/T)+(1/T)*(((a*h*(T^2))/2)+((a*r*(T^3))/6)+((((a*(y0*(exp(-u*tow)))+b)/2))*(((h*(T^3))/3)+((r*(T^4))/12))))+((cd/T)*((((T^2)*y0*(exp(-u*tow))*(3*a-a*b*T))-(a*(b^2)*(T^3)))/6))+tow;

disp(T)

disp(tow)

disp(TC)

vpa(T)

vpa(tow)

vpa(TC)

[Walter Roberson]

Using solve with quoted strings has not been valid for several releases, and has not been a good idea since 2016 (there were some bugs then that forced using quoted strings for certain differential equations... for nearly any purpose other than differential equations, using quoted strings for symbolic expressions has not been a good idea since r2014a.

I posted the version without quoted strings already and you should have built on that.

[Walter Roberson]

Your code here has the same problem as before, that you have more ( than you have )