How to input a "sym" type equation into ODE45 to solve first order differential equation?
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haohaoxuexi1
am 5 Sep. 2021
Kommentiert: Star Strider
am 6 Sep. 2021
% Sample
close all;
clear all;
clc;
syms t tt;
a=1;
b=2;
c=3;
fs_d=50*t^2+60*t^4+70*t^9;
fd_d=diff(fs_d,t);
dTime=1e-6; % time step
Tfinal=timeduration; % time final
zeit=0:dTime:Tfinal; %
% ODE solver
% initial condition
x1_0=0;
dx1_0=0;
%ode45
options = odeset('RelTol',1.e-6);
[tt,dx] = ode45(@(tt,x) Output123(tt, x, a, b, c, fd_d, t), zeit, x1_0, options);
function dx = Output123(tt, x, a, b, c, fd_d, t)
tic;
%dx(1)=x(2);
dx=-a*subs(fd_d,t,tt)*b/(c)-x*(b/(c));
dx = dx'; % output result
toc;
end
The above the a sample code I want to achieve, it is a first order differential equation, the "subs" method works for second order differential equation.
Can anyone help me with this problem?
Thanks,
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Akzeptierte Antwort
Star Strider
am 5 Sep. 2021
I do not understand what you are doing.
However the correct way to use a symbolic different ia equation with the numeric solvers would be something like this —
syms a b c t tt x
fs_d=50*t^2+60*t^4+70*t^9;
fd_d=diff(fs_d,t);
f=-a*subs(fd_d,t,tt)*b/(c)-x*(b/(c))
Output123 = matlabFunction(f, 'Vars',{tt,x,a,b,c})
a=1;
b=2;
c=3;
timeduration = 1;
dTime=1e-6; % time step
Tfinal=timeduration; % time final
zeit=0:dTime:Tfinal; %
x1_0=0;
dx1_0=0;
options = odeset('RelTol',1.e-6);
[tt,x] = ode45(@(tt,x) Output123(tt, x, a, b, c), zeit, x1_0, options);
figure
plot(tt, x)
grid
Make appropriate changes to ger the resullt you want.
.
11 Kommentare
Walter Roberson
am 5 Sep. 2021
Then code it by hand if it is morally important for you.
Reminder: the ode*() functions strictly require that the output is double() or single(), but the output of subs() is always symbolic.
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