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

8 Ansichten (letzte 30 Tage)
%%%%%%%%%%%%%%%% 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 Kommentar
John D'Errico
John D'Errico am 17 Mär. 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.)

Melden Sie sich an, um zu kommentieren.

Antworten (1)

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

Produkte

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by