# How can I plot a second order differential equation with boundary condition using fourth order Runge-Kutta method?

6 vues (au cours des 30 derniers jours)
Commenté : John D'Errico le 17 Mar 2023
%%%%%%%%%%%%%%%% Runga-Kutta%%%%%%%%%%%%%%%%
h=0.0001;
xfinal=d;
x(1)=0;
y(1)=0; % initial value of y
y(xfinal)=0; % final value of y
% Let y' = z (f1) and y" = z' (f2);
f1 = @(x, y, z) z;
f2 = @(x, y, z) ky^2*y-(ky*(-2*W*(pi/d)*tan(2*pi*x/d)+2*u0*((pi/d)^2)*cos(2*pi*x/d))*y)/(OP3-ky*u0*(sin(pi*x/d).^2-1/2)+...
B*(OP3-ky*u0*(sin(pi*x/d).^2-1/2)-A*(-2*W*(pi/d)*tan(2*pi*x/d)+2*u0*((pi/d)^2)*cos(2*pi*x/d)-ky*(OP3-ky*u0*(sin(pi*x/d).^2-1/2))))*(1-...
M*(opi^2)/(M*OP3^2-gi*Ti*ky^2)));
for i=1:ceil(xfinal/h)
x(i+1)=x(i)+h;
K1y = f1(x(i), y(i), z(i));
K1z = f2(x(i), y(i), z(i));
K2y = f1(x(i)+0.5*h, y(i)+0.5*K1y*h, z(i)+0.5*K1z*h);
K2z = f2(x(i)+0.5*h, y(i)+0.5*K1y*h, z(i)+0.5*K1z*h);
K3y = f1(x(i)+0.5*h, y(i)+0.5*K2y*h, z(i)+0.5*K2z*h);
K3z = f2(x(i)+0.5*h, y(i)+0.5*K2y*h, z(i)+0.5*K2z*h);
K4y = f1(x(i)+h, y(i)+K3y*h, z(i)+K3z*h);
K4z = f2(x(i)+h, y(i)+K3y*h, z(i)+K3z*h);
y(i+1) = y(i)+(K1y+2*K2y+2*K3y+K4y)*h/6;
z(i+1) = z(i)+(K1z+2*K2z+2*K3z+K4z)*h/6;
end
plot(x,y,'-','linewidth',1)
hold on
##### 1 commentaireAfficher -1 commentaires plus anciensMasquer -1 commentaires plus anciens
John D'Errico le 17 Mar 2023
It looks like you already solved the ODE, and plotted it. Where is the problem? (Even so, if this were not homework, as it surely is, you should be using an ODE solver, not writing your own code.)

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### Réponses (1)

Cameron le 17 Mar 2023
Here is a list of built-in ODE solvers within MATLAB.
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