Question

用matlab 解方程 y’ = −20y, y(0) = 1 for 0 ≤ t ≤ 1.

UseRunge-Kutta order 4 and the trapezoidal method. Try first the step size h= 0.05,and graph the exact solution and the two numerical solutions. You should seethat Runge-Kutta provides a better approximation than the trapezoidal method.This is to be expected, since Runge-Kutta is a higher-order method.
Now,increase the step size to h= 0.15.You should see that the approximation obtained by the trapezoidal method isso-so. However, Runge-Kutta behaves really badly! Instead of decreasing tozero, the solution increases! Experiment with other values of halso.
Runge-Kuttaworks much better than other methods when the step size hissmall. If the step size is not small, Runge-Kutta (and the other explicitmethods, like Euler, etc.) can fail to even predict the correct trend, whilethe trapezoidal method still provides a reasonable approximation.
Notethat since the trapezoidal method is implicit, it will in general be hard tosolve for wi+1. It is quite easyin this case though.
求大神 解答

Answer 1

程序一：

syms t

dsolve('Dy+20*y=0','y(0)=1')

结果：

exp(-20*t)

程序二：

t=linspace(0,1,1000);

plot(t,exp(-20*t),'r.-')

结果：