Matlab微分方程仿真图求解

Question

微分方程是
x(1-x)(550y-1000α+200）与y(1-y)(400x-200)
α=0.2,0.4,0.6,0.8,1. i=0.3 j=0.8 x∈[0,1]步长为0.1 t∈[0 0.06],求x与t的仿真图以及原始代码。

Answer 1

我想题主的意思应该是求解下面的微分方程：
dx/dt = x(1-x)(550y-1000α+200）
dy/dt= y(1-y)(400x-200)
可使用ode45来求解这个常微分方程组，代码如下

x0 = [0.3;0.8];

alpha = 0.2;
dfun = @(t,X)  [X(1)*(1-X(1))*(550*X(2)-1000*alpha+200);  X(2)*(1-X(2))*(400*X(1)-200) ];
[T1,X1] = ode45( dfun,[0 0.06],x0);

alpha = 0.4;
dfun = @(t,X)  [X(1)*(1-X(1))*(550*X(2)-1000*alpha+200);  X(2)*(1-X(2))*(400*X(1)-200) ];
[T2,X2] = ode45( dfun,[0 0.06],x0);

alpha = 0.6;
dfun = @(t,X)  [X(1)*(1-X(1))*(550*X(2)-1000*alpha+200);  X(2)*(1-X(2))*(400*X(1)-200) ];
[T3,X3] = ode45( dfun,[0 0.06],x0);

alpha = 0.8;
dfun = @(t,X)  [X(1)*(1-X(1))*(550*X(2)-1000*alpha+200);  X(2)*(1-X(2))*(400*X(1)-200) ];
[T4,X4] = ode45( dfun,[0 0.06],x0);

alpha = 1.0;
dfun = @(t,X)  [X(1)*(1-X(1))*(550*X(2)-1000*alpha+200);  X(2)*(1-X(2))*(400*X(1)-200) ];
[T5,X5] = ode45( dfun,[0 0.06],x0);

figure
subplot(211)
hold on
plot(T1,X1(:,1));
plot(T2,X2(:,1));
plot(T3,X3(:,1));
plot(T4,X4(:,1));
plot(T5,X5(:,1));
xlabel('Time /s')
ylabel('x')
legend({'\alpha = 0.2','\alpha = 0.4','\alpha = 0.6','\alpha = 0.8','\alpha = 1.0'})

subplot(212)
hold on
plot(T1,X1(:,2));
plot(T2,X2(:,2));
plot(T3,X3(:,2));
plot(T4,X4(:,2));
plot(T5,X5(:,2));
xlabel('Time /s')
ylabel('y')
legend({'\alpha = 0.2','\alpha = 0.4','\alpha = 0.6','\alpha = 0.8','\alpha = 1.0'})

结果图：