要用的时候总是忘记，这回给把它写在这里！

%%程序1

arg1 = 2;

arg2 = 1;

[T,Y] = ode45('vdp1000',[0 10],[2 0], [], arg1, arg2);

plot(T,Y(:,1),'-o');

%%程序2

function dy = vdp1000(t, y, flag, arg1, arg2)

dy =

zeros(2,1); %

a column vector

dy(1) = y(2);

dy(2) = arg1*(arg2 - y(1)^2)*y(2) - y(1);

%%ode5

function Y = ode5(odefun,tspan,y0,varargin)

%ODE5 Solve differential equations with a non-adaptive method of

order 5.

% Y = ODE5(ODEFUN,TSPAN,Y0) with TSPAN = [T1, T2, T3, ... TN]

integrates

% the system of differential equations y' = f(t,y) by stepping from

T0 to

% T1 to TN. Function ODEFUN(T,Y) must return f(t,y) in a column

vector.

% The vector Y0 is the initial conditions at T0. Each row in the

solution

% array Y corresponds to a time specified in TSPAN.

%

% Y = ODE5(ODEFUN,TSPAN,Y0,P1,P2...) passes the additional

parameters

% P1,P2... to the derivative function as

ODEFUN(T,Y,P1,P2...).

% This is a non-adaptive solver. The step sequence is determined by

TSPAN

% but the derivative function ODEFUN is evaluated multiple times

per step.

% The solver implements the Dormand-Prince method of order 5 in a

general

% framework of explicit Runge-Kutta methods.

%

% Example

% tspan = 0:0.1:20;

% y = ode5(@vdp1,tspan,[2 0]);

% plot(tspan,y(:,1));

% solves the system y' = vdp1(t,y) with a constant step size of

0.1,

% and plots the first component of the solution.

if ~isnumeric(tspan)

error('TSPAN

should be a vector of integration steps.');

end

if ~isnumeric(y0)

error('Y0

should be a vector of initial conditions.');

end

h = diff(tspan);

if any(sign(h(1))*h <= 0)

error('Entries of TSPAN are not in order.')

end

try

f0 =

feval_r(odefun,tspan(1),y0,varargin{:});

catch

msg =

['Unable to evaluate the ODEFUN at t0,y0. ',lasterr];

error(msg);

end

y0 = y0(:); % Make a column vector.

if ~isequal(size(y0),size(f0))

error('Inconsistent sizes of Y0 and f(t0,y0).');

end

neq = length(y0);

N = length(tspan);

Y = zeros(neq,N);

% Method coefficients -- Butcher's tableau

%

% C | A

% --+---

% | B

C = [1/5; 3/10; 4/5; 8/9; 1];

A = [ 1/5, 0, 0, 0, 0

3/40, 9/40, 0, 0, 0

44/45 -56/15, 32/9, 0, 0

19372/6561, -25360/2187, 64448/6561, -212/729, 0

9017/3168, -355/33, 46732/5247, 49/176, -5103/18656];

B = [35/384, 0, 500/1113, 125/192, -2187/6784, 11/84];

% More convenient storage

A = A.';

B = B(:);

nstages = length(B);

F = zeros(neq,nstages);

Y(:,1) = y0;

for i = 2:N

ti =

tspan(i-1);

hi =

h(i-1);

yi =

Y(:,i-1);

% General

explicit Runge-Kutta framework

F(:,1) =

feval_r(odefun,ti,yi,varargin{:});

for stage =

2:nstages

tstage = ti + C(stage-1)*hi;

ystage = yi + F(:,1:stage-1)*(hi*A(1:stage-1,stage-1));

F(:,stage) = feval_r(odefun,tstage,ystage,varargin{:});

end

Y(:,i) = yi

+ F*(hi*B);

end

Y = Y.';