second order ODE fun. with polynomial and loop

Question

umang 2014-5-5

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/128281-second-order-ode-fun-with-polynomial-and-loop

编辑： umang 2014-5-8

在 MATLAB Online 中打开

x''=(W/M)+((c*x')/M)+((K*(x*cosd((Thet1))-(Cr+(Hd*(sind((180)/((ph)*(Thet1))))))).^3/2 *(cos(Thet1)))/M)

this is the equation to solve

Di=0.0321;

Do=54.67*10^-3;

D=43.385*10^-3;

d=11.274*10^-3;

z=7;

Cr=11.285*10^-6;

M=3;

c=200;

Om=125.66;

Oc=46.5;

h=37*10^-6;

x0=10^-9;

x'0=10^-9;

ph=0.4623;

W=100;

K=8.37536*10^9;

Pi=3.14;

x(1)=x0; v(1)=v0;

j=1; tt=1;

Hd=0.1*10^-3;

for t=0:h:0.7

    for i=1:7;
        Thet=((46.5)*t)+(((7-i)*(2*3.14)/7)) ;
        Thet2=(Thet*180)/3.14;
        if (Thet2<=360)
            Thet1=Thet2;
        elseif (360<=Thet2)&&(Thet2<=720)
            Thet1=Thet2-360;
        elseif (720<=Thet2)&&(Thet2<=1080)
            Thet1=Thet2-720;
        elseif (1080<=Thet2)&&(Thet2<=1440) 
            Thet1=Thet2-1080;
        elseif (1440<=Thet2)&&(Thet2<=1800)
            Thet1=Thet2-1440;
        elseif (1800<=Thet2)&&(Thet2<=2160)
            Thet1=Thet2-1800;
        elseif (2160<=Thet2)&&(Thet2<=2520)
            Thet1=Thet2-2160;
        elseif (2520<=Thet2)&&(Thet2<=2880)
            Thet1=Thet2-2520;
        else 
            Thet1=Thet2-2880;
        end
         if (0<=Thet1)&&(Thet1<=0.4623)
%Fx(j)=K*(x*cosd((Thet1))-(Cr+(Hd*(sind((180)/((ph)*(Thet1))))))).^3/2 *(cos(Thet1));
            y0=[10^-9;10^-9];
             [t,y]=ode45('line',[t t+h],y0);
         else

% Fx(j)=0; cosd(Thet1);

         end
        j=j+1;
    end

function fx=line(t,y)

Cr=11.285*10^-6;

c=200;

ph=0.4623;

W=100;

K=8.37536*10^9;

Hd=0.1*10^-3;

fx=zeros(2,1);

fx(1)=y(2);

fx(2)=((33.34)+(66.67*y(2))+((2.791*10^9*(y(1)*cosd((Thet1))-(Cr+(Hd*(sind((180)/((ph)*(Thet1))))))).^3/2 *(cos(Thet1))))); pls tell me how to solve this equation