using gradient with ode45
6 views (last 30 days)
Show older comments
i have this code, where gradient(u,t) should be the derivative of u:
function [xdot]=pid_practica9_ejercicio3_ec_diferencial_prueba2(t,x)
Vp=5;
u=Vp*sin(2*pi*t)+5;
xdot = [
x(2, :);
1.776*0.05252*20*gradient(u,t)-10*x(1, :)-7*x(2, :);
];
%[t,x]=ode45('pid_practica9_ejercicio3_ec_diferencial_prueba2',[0,10],[0,0])
% plot(t,x)
but in the solution i get only zeros (and they shouldn't be)
is it possible to work with a derivative in the definition of a diferential equation like i do?
2 Comments
Answers (1)
Walter Roberson
on 7 Jul 2018
The t and x values passed into your function will be purely numeric, with t being a scalar and x being a vector the length of your initial conditions (so a vector of length 2 in this case.)
You calculate u from the scalar t value, and you pass the scalar u and scalar t into gradient -- the numeric gradient routine. The numeric gradient() with respect to scalar F and scalar H is always 0.
8 Comments
See Also
Categories
Find more on Ordinary Differential Equations in Help Center and File Exchange
Products
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!