Second Order Laplace solving doesn't work ('Unable to find explicit solution')

2 Ansichten (letzte 30 Tage)
I'm trying to solve an ODE using Laplace method, but I'm stuck on solving the equation
Here's my code:
syms t x(t) s X(s);
% PARAMETERS (tried to do symbolically but it was more diffcult)
m = 1;
k = 0.5;
xi = 1.2;
c = xi*2*sqrt(k*m)
c = 1.6971
f0 = 1;
w = 0.1;
dx = diff(x, t, 1);
ddx = diff(x, t, 2);
% INITIAL CONDITIONS
x0 = 0;
dx0 = 0;
newton = m*ddx+ c*dx +k*x;
f = f0*cos(w*t);
lteqn = laplace(newton, t, s)
lteqn = 
lefteqn = subs(lteqn,{laplace(x(t), t, s), x(0),dx(0)},{X(s), x0, dx0})
lefteqn = 
F_s = laplace(f, t, s);
simplify(solve(lefteqn == F_s, X(s)))
Warning: Unable to find explicit solution. For options, see help.
ans = Empty sym: 0-by-1
I can't believe MATLAB cannot solve this easy equation. I think I'm missing something.
Thank you guys

Akzeptierte Antwort

Aquatris
Aquatris am 13 Aug. 2024
Bearbeitet: Aquatris am 13 Aug. 2024
Not sure the underlying reason behind it but you are defining the X as a function of s and solve function seems to be having trouble solving it. Define X as a standalone and it has no trouble
syms t x(t) s X;
% PARAMETERS (tried to do symbolically but it was more diffcult)
m = 1;
k = 0.5;
xi = 1.2;
c = xi*2*sqrt(k*m)
c = 1.6971
f0 = 1;
w = 0.1;
dx = diff(x, t, 1);
ddx = diff(x, t, 2);
% INITIAL CONDITIONS
x0 = 0;
dx0 = 0;
newton = m*ddx+ c*dx +k*x;
f = f0*cos(w*t);
lteqn = laplace(newton, t, s);
lefteqn = subs(lteqn,{laplace(x(t), t, s), x(0),dx(0)},{X, x0, dx0});
F_s = laplace(f, t, s);
simplify(solve(lefteqn == F_s, X))
ans = 

Weitere Antworten (1)

Sam Chak
Sam Chak am 13 Aug. 2024
Do you want to analytically solve the ODE like this?
syms s t X
%% original parameters
m = 1;
k = 0.5;
xi = 1.2;
c = xi*2*sqrt(k*m);
f0 = 1;
w = 0.1;
%% Test parameters -> should return x(t) = 1/2·(sin(t) - t·e^(-t))
% m = 1;
% k = 1;
% xi = 1;
% c = xi*2*sqrt(k*m);
% f0 = 1;
% w = 1;
%% Main
eqn = m*s^2*X + c*s*X + k*X == laplace(f0*cos(w*t), t, s);
X = solve(eqn, X);
x = ilaplace(X, s, t)
x = 

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R2024a

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