How to find roots of a quartic function with unknown constant coefficients?

The polynomial is:
x^4 + 10.1ax^3 + 1.81x^2 + 9ax + 0.81 = 0
where a is an unkown constant coefficient.
I have to find the corresponding x values to further solve an ODE.

Antworten (2)

James Tursa
James Tursa am 28 Nov. 2023
Bearbeitet: James Tursa am 28 Nov. 2023
You can look here:
But the nature of the roots is going to depend on the value of a. Do you know anything at all about a? What is the context of the overall problem you are solving? What is the ODE?

1 Kommentar

No it just represents a dampening coefficient for a spring mass system. The ode models the movement of a building. Essentially a fbd with 2 masses, 2 springs, and 2 dampeners. I get two odes that model the forces on the mass block. I am trying to solve the simultaneous set of ODE's using cramers rule than MUC because the odes all have constant coefficients of which we have a value for every constant except the dampening coefficients C1 and C2.
The FBD:

Melden Sie sich an, um zu kommentieren.

Torsten
Torsten am 28 Nov. 2023
Bearbeitet: Torsten am 28 Nov. 2023
syms x a
p = x^4 + 10.1*a*x^3 + 1.81*x^2 + 9*a*x + 0.81;
sol = solve(p==0,'MaxDegree',4)
sol = 
vpa(subs(sol,a,1)) % E.g. for a = 1
ans = 

1 Kommentar

The roots get complicated and fragile
filename = fullfile(tempdir, 'sol.m');
addpath(tempdir);
syms x a
p = x^4 + 10.1*a*x^3 + 1.81*x^2 + 9*a*x + 0.81;
sol = solve(p==0);
solF = matlabFunction(sol, 'file', filename)
solF = function_handle with value:
@sol
A = linspace(-1/2,1/2,200);
Y = cell2mat(arrayfun(solF, A, 'uniform', 0));
figure();
plot(A, real(Y(1,:)), A, real(Y(2,:)), A, real(Y(3,:)), A, real(Y(4,:)));
legend({'1 real', '2 real', '3 real', '4 real'})
figure();
plot(A, imag(Y(1,:)), A, imag(Y(2,:)), A, imag(Y(3,:)), A, imag(Y(4,:)));
legend({'1 imag', '2 imag', '3 imag', '4 imag'})

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Numerical Integration and Differential Equations finden Sie in Hilfe-Center und File Exchange

Tags

Gefragt:

am 27 Nov. 2023

Bearbeitet:

am 29 Nov. 2023

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by