ODE Piecewise Linear Function Help

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Braulio Diaz
Braulio Diaz el 27 de Feb. de 2014
Editada: Walter Roberson el 31 de Ag. de 2019
Not sure what's wrong but this is what I have to do:
The listing of a Matlab script file which uses dsolve() to solve the following ODE and plot the results from t=0 to t=3. dx/dt + 2 x = f(t) with x(0) = 1 and where
f(t) = exp(-t) 0 <= t <= ln(2) f(t) = 4 for ln(2) < t <=ln(3) f(t) = 0 otherwise
SHOW: The output generated by the script file The plot generated by the script file
THIS IS WHAT I HAVE..
+++++++++++++++++++++++++++++
%last modified: 2/27/2014
%Matlab
%for piecewise continuous input
%of first order linear systems
clear all
clc
format compact
% Example
% dx/dt + 2x = f(t)
% f(t) = 1 for 0<=t<=1 and 0 otherwise
% Symbolic approach
% Find x in the first interval
% Note the use of pure symbolics instead
% of a character string solution like
% x1 = dsolve('Dx1+x1=1','x1(0)=0')
syms t x1(t)
dx1 = diff(x1);
t = 1;
x1 = dsolve(dx1+2*x1==exp(-t), x1(0)==1);
display(['x1 = ', char(vpa(x1,3))])
%Use solutions for first interval to find
%IC for second interval
x2_IC = subs(x1);
display(['x2_IC = ', char(vpa(x2_IC,3))])
%Find solution in second interval using
%x2_IC as an IC
syms t x2(t)
dx2 = diff(x2);
x2 = dsolve(dx2+2*x2==4, x2(log(2))==x2_IC);
x3_IC = subs(x2);
display(['x3_IC = ', char(vpa(x3_IC,3))])
syms x3(t)
dx3 = diff(x3);
x3 = dsolve(dx3+2*x3==0, x3(log(3))==x3_IC);
display(['x3 = ', char(vpa(x3,3))])
%Plot the results
t = 0:0.01:1;
xx1 = subs(x1);
plot(t,xx1,'linewidth', 3,'color','red')
t = 1:0.01:3;
xx2 = subs(x2);
plot(t,xx2,'linewidth',3,'color','green')
t = 3:0.01:5;
xx3 = subs(x3);
hold on
plot(t,xx3,'linewidth',3,'color','blue')
grid on
xlabel('t','FontSize',14)
ylabel('x','FontSize',14)
title('x vs time','FontSize',14)
hold off
  2 comentarios
Mischa Kim
Mischa Kim el 27 de Feb. de 2014
Do you need to solve it using symbolic math (rather than doing it numerically)?
Braulio Diaz
Braulio Diaz el 27 de Feb. de 2014
Editada: Braulio Diaz el 27 de Feb. de 2014
Actually, either one is okay..

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Respuesta aceptada

Mischa Kim
Mischa Kim el 27 de Feb. de 2014
In this case have a look at:
function my_DE()
x0 = 1;
tspan = linspace(0,3,1000);
[T,X] = ode45(@DE, tspan, x0);
plot(T,X)
grid
end
function dX = DE(t,x)
dX = -2*x + f(t);
end
function fval = f(t)
if (t <= log(2))
fval = exp(-t);
elseif (t > log(2)) && (t <= log(3))
fval = 4;
else
fval = 0;
end
end
...not including the color coding in the plot.
  2 comentarios
Braulio Diaz
Braulio Diaz el 27 de Feb. de 2014
Nice! Thanks a lot Mischa!
Braulio Diaz
Braulio Diaz el 27 de Feb. de 2014
Mischa, how would I do this using symbolic math?

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