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

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Dhananjay Verma Kaalkhanday el 17 de Mzo. de 2023
Comentada: John D'Errico el 17 de Mzo. de 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
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John D'Errico el 17 de Mzo. de 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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### Respuestas (1)

Cameron el 17 de Mzo. de 2023
Here is a list of built-in ODE solvers within MATLAB.
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