plese help someone.I need to solve simultaneous differential eqns with multiple dependent variable.

here is what I written to solve these eqns. I'm getting error.I don't have any initial conditions.Also I need to plot y1,y2 and y3.I'm still learning matlab and don't have idea how to solve this problem.

    clc;clear;close;
%write given values
m1=1; m2=2; m3=3;
c1=3; c2=1;
k1=2; k2=2;
f1=5; f3=2;
%call the function named 'func'
[y1 y2 y3]=func(m1,m2,m3,c1,c2,k1,k2,f1,f3)

 function [y1 y2 y3] = 
 func(m1,m2,m3,c1,c2,k1,k2,f1,f3)
    %define variables
    syms y1(t) y2(t) y3(t)
    %write differential equations
    eqn1=m1*diff(y1,t,2)==f1-c1*(diff(y1,t)-diff(y2,t))-k1*(y1-y2);
    eqn2=m2*diff(y2,t,2)==-c1*(diff(y2,t)-diff(y1,t))-k1*(y2-y1)-c2*(diff(y2,t)-diff(y3,t))-k2*(y2-y3);
    eqn3=m3*diff(y3,t,2)==-f3-c2*(diff(y3,t)-diff(y2,t))-k2*(y3-y2);
    %solve these eqns
    y = dsolve(eqn1,eqn2,eqn3);
    y1=y.y1;
    y2=y.y2;
    y3=y.y3;
end

[darova]

Attach original equations and initial conditions

[Walter Roberson]

Tricky. If you dsolve() for any two of eqn1, eqn2, eqn3 then it works, but all three together gives the error message.

[ajeet sahu]

There is no initial conditions for this problem.The given eqns are correctly written.I need to solve in symbolic variable t.Also if this works for 2 eqns then why not 3?is there any other way to do that.please help someone.Thankyou

[Paul]

@ajeet sahu

I don't know why dsolve throws an error, but I doubt that a closed form, symbolic solution can be found for y1(t), y2(t), and y3(t). Do you have a reason to think such a solution can be obtained?

Also, it seems to me that to get any solution one has to assume something about the initial conditions.

[Walter Roberson]

Maple says that there is a closed-form solution. It involves particular computed values (call it S) with terms with exp(S*t)/S and terms with exp(S*t) without division by S, and some other terms. Here, what I referred to as S for explanation purposes, is the set of four roots of a polynomial of degree 4. The roots of a quartic have closed form, so it is possible to write a closed form for the solution... just long.

The full set of roots of a quartic always involve some expressions with sqrt(-1) . In cases where all of the roots are real, then ideally the imaginary terms cancel out... but in practice because of rounding, it is common that when you work symbolically, that you end up with some left-over imaginary components, so you have to be careful.

[Paul]

I was able to get Matlab to return an expression in terms of roots(), and I assumed that if a closed form for those roots existed Matlab would have provided it.