Diferencia entre revisiones de «Logistic equation»
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| Línea 1: | Línea 1: | ||
== Resolución de la Ecuación logística por el método de Euler == | == Resolución de la Ecuación logística por el método de Euler == | ||
| − | * Logistic equation | + | * Logistic equation |
| − | <math> | + | |
| − | + | <math>y_0 = f(t,y) = y\cdot (1-y)</math> | |
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| + | <math>y(t_0) = y_0</math> | ||
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* Numerical scheme | * Numerical scheme | ||
| − | + | ||
| − | + | <math> y(0) = 1/10</math> | |
| + | |||
| + | <math>y(n+1) = y(n) + h\cdot y(n)\cdot(1 - y(n))</math> | ||
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| + | |||
| + | |||
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* MATLAB code | * MATLAB code | ||
{{matlab|codigo= | {{matlab|codigo= | ||
Revisión del 13:33 31 ene 2013
Resolución de la Ecuación logística por el método de Euler
- Logistic equation
[math]y_0 = f(t,y) = y\cdot (1-y)[/math]
[math]y(t_0) = y_0[/math]
- Numerical scheme
[math] y(0) = 1/10[/math]
[math]y(n+1) = y(n) + h\cdot y(n)\cdot(1 - y(n))[/math]
- MATLAB code
% Euler method to solve the logistic equation y'=y(1-y)
clear all;
t0=0; tN=4; % initial and final time
y0=1/10; % value of y at time t=0
N=40; % Number of intervals
h=(tN-t0)/40; % Time step h
yy=y0; % yy -> variable with the solution at each time step
y(1)=yy; % y -> vector where we store the solution
for n=1:N-1
yy=yy+h*yy*(1-yy); % numerical scheme
y(n+1)=yy; % store the solution
end
x=t0:h:tN; % Draw the solution
plot(x,y,'x');
