How can I solve this?
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clc
clear
syms PL c k
eqn = [PL * exp(-c*exp(-k*0)) == 179323 ; PL * exp(-c*exp(-k*10)) == 203302 ; PL * exp(-c*exp(-k*20)) == 226542]
S = solve(eqn , [PL;c;k] )
S =
struct with fields:
PL: [0×1 sym]
c: [0×1 sym]
k: [0×1 sym]
2 Commenti
Muhammad Usman
il 23 Ott 2022
the set of equations are non linear in nature that's why you can't use solve to compute the solution
Risposte (2)
Muhammad Usman
il 23 Ott 2022
% Solve the system of equations starting at the point [0,0,0].
% PL = x(1); c = x(2); k = x(3);
% Initial guess is [0,0,0], you can change it accordingily
fun = @root2d;
x0 = [0,0,0];
x = fsolve(fun,x0)
function F = root2d(x)
F(1) = x(1) * exp(-x(2)*exp(-x(3)*0)) - 179323;
F(2) = x(1) * exp(-x(2)*exp(-x(3)*10)) - 203302;
F(2) = x(1) * exp(-x(2)*exp(-x(3)*20)) - 226542;
end
2 Commenti
Alex Sha
il 24 Ott 2022
There are actually numerical solutions like below:
x1: 446505.431672107
x2: 0.912262916225993
x3: 0.0148006249649759
Torsten
il 24 Ott 2022
syms PL c k
eqn1 = PL * exp(-c*exp(-k*0)) == 179323;
eqn2 = PL * exp(-c*exp(-k*10)) == 203302;
eqn3 = PL * exp(-c*exp(-k*20)) == 226542;
sPL = solve([eqn1 eqn2],[c,k]);
PLsol = solve(subs(eqn3,[c,k],[sPL.c sPL.k]),PL);
sck = solve([subs(eqn1,PL,PLsol),subs(eqn2,PL,PLsol)],[c,k]);
csol = sck.c;
ksol = sck.k;
vpa(PLsol)
vpa(csol)
vpa(ksol)
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