dx/dt=x(1–2y)
dy/dt=–y(2-x)
x(0)=1000,y(0)=500
其中t:时间、x:兔子数量、y:狐狸数量
绘制解的图像并做出解释。
例如要求解
clear;
clc;
close all;
h=1e-5; %步进长度
t=0:h:1; %生成自变量t的向量
%%创建计算结果x,y,z的数组
N=length(t);
x=ones(1,N);
y=ones(1,N);
z=ones(1,N);
%%四阶龙格库塔迭代
for i=2:N
t_n=t(i-1);
x_n=x(i-1);
y_n=y(i-1);
z_n=z(i-1);
kx1=y_n+3*z_n+sin(5*t_n);
ky1=x_n+cos(t_n);
kz1=x_n+z_n-3*cos(3*t_n)*sin(4*t_n);
kx2=(y_n+ky1*h/2)+3*(z_n+kz1*h/2)+sin(5*(t_n+h/2));
ky2=(x_n+kx1*h/2)+cos(t_n+h/2);
kz2=(x_n+kx1*h/2)+(z_n+kz1*h/2)-3*cos(3*(t_n+h/2))*sin(4*(t_n+h/2));
kx3=(y_n+ky2*h/2)+3*(z_n+kz2*h/2)+sin(5*(t_n+h/2));
ky3=(x_n+kx2*h/2)+cos(t_n+h/2);
kz3=(x_n+kx2*h/2)+(z_n+kz2*h/2)-3*cos(3*(t_n+h/2))*sin(4*(t_n+h/2));
kx4=(y_n+ky3*h)+3*(z_n+kz3*h)+sin(5*(t_n+h));
ky4=(x_n+kx3*h)+cos(t_n+h);
kz4=(x_n+kx3*h)+(z_n+kz3*h)-3*cos(3*(t_n+h))*sin(4*(t_n+h));
x(i)=x_n+h/6*(kx1+2*kx2+2*kx3+kx4);
y(i)=y_n+h/6*(ky1+2*ky2+2*ky3+ky4);
z(i)=z_n+h/6*(kz1+2*kz2+2*kz3+kz4);
end
%%画图
figure();
hold on;
plot(t,x,'r');
plot(t,y,'g');
plot(t,z,'b');
legend('x','y','z');
xlabel('t');
title('xyz函数图象');
hold off;