The differential equation problem with variable solution by using ode45
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I have four coupled diffrential equation shown bellow :-
In which a ,b ,c ,d ,e,f are constant and x[t] we will get from the solution of this second order diffrential equation .how to write code for it in matlab.plz help
4 Comments
Accepted Answer
darova
on 31 Oct 2019
Here is an idea
[t1,x1] = ode45(@F1,ts,x0); % solve second order
[t2,s] = ode45(@(t,s)F2(t,s,t1,x1(:,1)),ts,s0); % solve system
function ds = F2(t,s,t1,x1)
x = interp1(t1,x1,t); % extract x
ds(1,1) = a*x-b ...
ds(2,1) = c*x*s(1) ...
end
3 Comments
darova
on 31 Oct 2019
Edited: darova
on 31 Oct 2019
Here is how your code should look like
function main
%% your constants
[t1,x1] = ode45(@noscillator,[0 10],[0 1]);
[t2,s1] = ode45(@(t,s) xotss(t,s,t1,x1(:,1)), [0 10], [1 0 0 0]);
plot(t2,s1(:,1))
function xdot=noscillator(t,x)
xdot(1) = x(2);
xdot(2) = -(omega^2)*x(1)-3*(gamma/m)*x(1)^2 - 4*(beta/m)*x(1)^3 + (V/m)*cos(w *t);
xdot=xdot';
end
function dxdt = xotss(t,s,t1,x1)
x = interp1(t1,x1,t);
dxdt(1) = (a*x-b)*s(1) + c*x*s(2);
% ...
dxdt = -1i*dxdt'
end
end
More Answers (6)
abhishek singh
on 31 Oct 2019
1 Comment
darova
on 31 Oct 2019
It means
सदस्यता सूचकांकों को वास्तविक धनात्मक पूर्णांक या तार्किक होना चाहिए।
In your language. Any ideas what the problem it might be?
Walter Roberson
on 31 Oct 2019
[t1,x1] = ode45(@noscillator,[0:100],[0 1]);
[t2,s1] = ode45(@(t,s) xotss(t,s,t1,x1(:,1)), [0:100], [1 0 0 0]);
for ti = 0:1:100
rho11(ti+1)=s1(ti+1,1).*s1(ti+1,1)'-s1(ti+1,3).*s1(ti+1,3)';
rho12(ti+1)=s1(ti+1,1).*s1(ti+1,2)'+s1(ti+1,3).*s1(ti+1,4)';
rho21(ti+1)=s1(ti+1,2).*s1(ti+1,1)'+s1(ti+1,4).*s1(ti+1,3)';
rho22(ti+1)=s1(ti+1,2).*s1(ti+1,2)'-s1(ti+1,4).*s1(ti+1,4)';
end
3 Comments
Walter Roberson
on 1 Nov 2019
Note that s1(ti+1,1)' means the conjugate complex transpose of s1(ti+1,1) . It is, however, a scalar, so transpose does not make any change. The You are also expecting real-valued results, so the conjugate is probably not makeing any changes. I suspect you are doing the equivalent of squaring the value.
I worry that you might have that that s1(ti+1,1)' is the derivative of s1(ti+1,1) .
abhishek singh
on 1 Nov 2019
1 Comment
Rik
on 1 Nov 2019
Please do not post your comments as answer. Their order can change, which makes it confusing.
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