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how to implement without using "function-end" command?

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Hoon
Hoon am 12 Mär. 2016
Kommentiert: Steven Lord am 13 Mär. 2016
I'll start with an example. All the codes are from this website, http://12000.org/my_notes/matlab_ODE/
I'm trying to figure out how to use ode45
function first_oder_ode
t=0:0.001:5; % time scalex
initial_x=0;
[t,x]=ode45( @rhs, t, initial_x);
plot(t,x);
xlabel('t'); ylabel('x');
function dxdt=rhs(t,x)
dxdt = 3*exp(-t);
end
end
Now I want to avoid using
function
end
Because I wanna see what's inside the function in the work space.
then I can simply write like
f=@(t,x)3*exp(-t)+x; %%%define first order ode
t=0:0.001:5; %%%time scalex
x_initial=0; %%%x initial condition
[t,x]=ode45(f,t,x_initial); %%%solving ODE
plot(t,x);
This is easy one for getting numerical solution for 1st order ODE
The problem is, applying this into ODE system.
The Matlab code is (it's already in the website)
function second_oder_ode
% SOLVE d2x/dt2+5 dx/dt - 4 x = sin(10 t)
% initial conditions: x(0) = 0, x'(0)=0
t=0:0.001:3; % time scale
initial_x = 0;
initial_dxdt = 0;
[t,x]=ode45( @rhs, t, [initial_x initial_dxdt] );
plot(t,x(:,1));
xlabel('t'); ylabel('x');
function dxdt=rhs(t,x)
dxdt_1 = x(2);
dxdt_2 = -5*x(2) + 4*x(1) + sin(10*t);
dxdt=[dxdt_1; dxdt_2];
end
end
How do I write this in another way without using "function - end"? is it possible?
  1 Kommentar
Steven Lord
Steven Lord am 13 Mär. 2016
Rather than making the outer function into a script, I recommend giving it some output arguments. This way you don't clutter the caller's workspace with temporary variables that are only needed to help the code compute the variables in which you're interested.

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Akzeptierte Antwort

Walter Roberson
Walter Roberson am 12 Mär. 2016
rhs = @(t, x) [x(2); -5*x(2) + 4*x(1) + sin(10*t)];
  2 Kommentare
Hoon
Hoon am 12 Mär. 2016
so, I wrote like this
t=0:0.001:3; % time scale
initial_x = 0;
initial_dxdt = 0;
rhs = @(t, x) [x(2); -5*x(2) + 4*x(1) + sin(10*t)];
[t,x]=ode45( @rhs, t, [initial_x initial_dxdt] );
plot(t,x(:,1));
xlabel('t'); ylabel('x');
It doesn't work
Hoon
Hoon am 12 Mär. 2016
OH!!! it works!!! I have to loose @!

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