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。
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已结题
Matlab微分方程仿真图求解
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微信扫一扫2条回答 默认 最新
- MatlabFans_Mfun 2021-07-27 21:44已采纳
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：
点赞 打赏 评论分享 - shifenglv 2021-07-27 10:19
你是要绘制这两个方程的曲线还是？
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