differential equation 2DOF

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function homework2 = main2(t,X1,X2)

global m1 c1 k1 m2 c2 k2

M=[m1 0;0 m2];

C=[c1+c2 -c2; -c2 c2];

K=[k1+k2 -k2; -k2 k2];

F=[3*cos(t);10*cos(3*t)];

X=[X1;X2];

homework2=M*X(3)+C*X(2)+K*X(1)-F;

end

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tspan= 0:0.1:50 ;

X0 = [0 0;0 0] ;

[t, X] = ode45(@main2, tspan, X0);

plot(X1,X2)

I do it here and it 's ain't working....

Answer 1

[Birdman]

You may try Symbolic Toolbox to solve the problem. Run the attached script to see the result.

[alsgud qor]

Thank you.but If using matrix is essential, what should I do? originally the purpose was that,ode45...like that.and as far as my knowledge goes, the basic ODE for that kind of thing is

for m1

F1=-k1x1-c1x1'+k2(x2-x1)+c2(x2'-x1')+mx1''

for m2

F2=m2x2''-c2x1'+c2x2'-k2x1+k2x2

so the entire thing should be

(the right side's 0 should turn into [F1 F2])

is it wrong?so I change it from your code to

eq1=F2==m2*diff(x2,2)+c2*(diff(x2,1)-diff(x1,1))+k2*(x2-x1);

eq2=F1==m1*diff(x1,2)+c1*diff(x1,1)+k1*x1;

this.and the result graph was this

I feel awkward cause you gave me something different.I hope it is my mistake

[Birdman]

Sorry, my bad. By using the attached function, run the following code from command line:

yInitial=zeros(1,4); %initial conditions
tSpan=0:0.001:10; %time interval for the equations
[T,Y]=ode45(@(t,y) TwoMassSys(t,y), tSpan, yInitial); %solving by using ode45
plot(T,Y(:,1),T,Y(:,3));legend('X1','X2'); %plotting

[Birdman]

Have you tried my last solution? It seems to be working fine.

[alsgud qor]

um sorry.because of time lag, I just recieved it. let me ask some Q

1. dydt = zeros(4,1); what does it mean?
2. dydt(2) = (-k1/m1)*y(1)+(-c1/m1)*y(2)+F1; this part should be changed into 'F1/m1'? same with dydt(4)

[Birdman]

1. It means that I created a vector of zeros with 4 elements.
2. Yes exactly, I missed that.