Question

我从二阶ODE获得了两个一阶ODE：

y(0)=1
y'(0)=-1/3

u1'=u2
u2=u/9-(pi*u1*e^(x/3)*(2u2*sin(pi*x)+pi*u1cos(pi*x))

u1(0)=y(0)=1    
u2(0)=y'(0)=-1/3

我的问题是如何建立前进欧拉？我有：

n=[0:0.01:2];    
h=2./n;

Answer 1

我们的等式是：

u1' = u2
u2' = u1/9 - \pi u1 exp(x/3)(2u2 sin(\pi x) + \pi u1 cos(\pi x))

现在解决y'= f（x，y）的欧拉方法是：

y_ {n + 1} = y_ {n} + h * f（x_n，y_n）

作为MATLAB代码，我们可以将其写成：

h = 0.01; % Choose a step size
x = [0:h:2]; % Set up x
u = zeros(length(x),2);
u(1,:) = [1; -1/3]; % Initial Conditions for y

for ii = 2:length(x)
    u(ii,:) = u(ii-1,:) + h * CalculateDeriv(x(ii-1),u(ii-1,:)); % Update u at each step
end

function deriv = CalculateDerivative(x,u)
deriv = zeros(2,1);
deriv(1) = u(2);
deriv(2) = u(1)/9 - pi*u(1)*exp(x/9)*(2*u(2)*sin(pi*x) + pi*u(1)*cos(pi*x))
end

转发Euler在Matlab中求解一阶ODE系统

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