how to solve two equations
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Hello 
This is part of a code and I want to solve this part 
thanks in advance 
=========================================
clear
syms U10 U20 
p=0
f =[2*U20 + sin(U20 - sin(p)) - 2*sin(p);2*U10 + sin(U10 - cos(p)) - 2*cos(p)]
s=subs(f,U10,x(1))
s1=subs(s,U20,x(2))
fun=@(x)[sin(x(2))+2*x(2); sin(x(1)-1)+2*(x(1)-1)];
fun=@(x)f;
[x,fval]=fsolve(fun,[0;0]);
U10=x(1)
U20=x(2)
2 Comments
  Dyuman Joshi
      
      
 on 17 Feb 2023
				It's not clear as to what you are trying to do. I am confused as to why you are using syms variable and solving with fsolve?
What is the purpose of subsitution? And you are over-writing fun
fun=@(x)[sin(x(2))+2*x(2); sin(x(1)-1)+2*(x(1)-1)];
fun=@(x)f;
Answers (2)
  Sai
    
 on 20 Feb 2023
        Hi,
I understand that you are trying to solve two nonlinear equations depending on two variables. 
As the equations are not clear in the data provided, let me simplify and finalize the equations to be solved.
First equation: 2*x(1) + sin(x1)
Second equation: 2*x(2) + sin(x(2)-1) - 2
Now the equations are solved using fsolve command as shown.
Following code is placed in eq_solve.m file
function f = eq_solve(x)
f(1) = 2*x(1) + sin(x(1));
f(2) = 2*x(2) + sin(x(2)-1) - 2;
Following code is placed in eq_solve_test.m file
fun = @eq_solve;
x0 = [0;0];
x = fsolve(fun,x0)
 Now the eq_solve_test.m file has been run to get the values

With Regards,
G. Saikumar
MathWorks Technical Support
  John D'Errico
      
      
 on 20 Feb 2023
        Note that the two equations are fully uncoupled. This means U20 appeared only in one equation. It has absolutely no impact on the other equation, and U10 has no impact on equation 1. Do you see that?
syms p U20 U10
2*U20 + sin(U20 - sin(p)) - 2*sin(p) == 0
2*U10 + sin(U10 - cos(p)) - 2*cos(p) == 0
the value of p does have an impact on these equations, but only in a very minor way. In fact, once you solve one of the equations, you could compute the solution to the other equation.
In this case, of course, with p==0, the problem becomes even simpler. The first equation reduces to
fun = @(U20) 2*U20 + sin(U20);
fplot(fun)
yline(0)
That problem provably has a solution only at U20==0, and the solution will be unique for real values of U20. We don't even need to use a tool like solve or fzero or fsolve to solve the problem.
Similarly, when p==0, the second equation reduces to 
   2*(U10-1) + sin(U10 - 1) == 0
My point is, this equation is identically the same as the first equation, but with a transformation appied
   U20 == U10 -1
So if we have U20==0 as a solution to the first equation, then trivially we know that U10==1.
If p weret o take on some other values, then the problem is no more difficult, since your system of two equations is not only an uncoupled system, but once you solve the problem once, you can solve the second problem directly. There are no rootfinders ever needed in the end.
6 Comments
  Torsten
      
      
 on 6 Mar 2023
				syms U10 U20 
Afk=[2*U10 + sin(U10 - 1) - 2,2*U20 + sin(U20)]
[U10 U20] = vpasolve(Afk)
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