i want to solve that system using 4 order kutta method

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nutpavorn yongvikul
nutpavorn yongvikul 2021-11-18
编辑： Jan 2021-11-18
this is my code
function dxdt = myderiv(t,x)
x1 = x(1);
x2 = x(2);
dx1dt = (-x1 + (0.097).*(1-x1)*exp(x2/(x2/20)+1));
dx2dt = (-x2 + 8.*(0.097).*(1-x1)*exp(x2/(x2/20)+1) - (0.3).*x2);
dxdt = [dx1dt;dx2dt];
end
and
f = @(t,x) myderiv(t,x)
h = 0.00001;
t = (0:h:1)';
t(1)=0;
x(:,1) = [0 ; 0] ;
for i=1:100000
K1 = f( t(i) , x(:,i) );
K2 = f( t(i) + h/2, x(:,i) + K1*h/2 );
K3 = f( t(i) + h/2, x(:,i) + K2*h/2 );
K4 = f( t(i) + h , x(:,i) + K3*h );
x(:,i+1) = x(:,i) + (h/6)*( K1 + 2*K2 + 2*K3 + K4 );
end
plot(t,x,'.b')
I need to see a graph result of system.

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采纳的回答

Jan
Jan 2021-11-18
编辑：Jan 2021-11-18
The output is a simple point at [0, 0].
You start at 0. Then the expression x2/(x2/20) + 1 is NaN, because you divide by 0:
x2/(x2/20) + 1 =
20 * x2 / x2 + 1
All following points are NaN in consequence, so the trajectory is you initial value only.
Replace:
exp(x2 / (x2 / 20) + 1)
% by
exp(x2 / (x2 / 20 + 1))
Avoid using too many parentheses. Including scalar contants in parentheses is clutter. Spaces around operators are useful:
dx1dt = -x1 + 0.097 * (1 - x1) * exp(x2 / (x2 / 20 + 1));
dx2dt = -x2 + 8 * 0.097 * (1 - x1) * exp(x2 / (x2 / 20 + 1)) - 0.3 .* x2;

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