Finding particular solution to 3x3 matrix using undetermined coefficients

Hi all,

I am trying to find the particular solution to the following system of equations:

function xprime = A(t,x)
eqn1 = - x(1)*(15/4) == -1575;
eqn2 = (15/4)*x(1) - (5/12)*x(2);
eqn3 = (5/12)*x(2) - (5/14)*x(3);

Below I went about finding my general solution:

end
A = sym([-15/4,0,0;15/4,-5/12,0;0,5/12,-5/14]);
>> Id3 = sym([1,0,0;0,1,0;0,0,1]);
>> syms lambda
>> B = lambda*Id3 - A
 
B =
 
[ lambda + 15/4,             0,             0]
[         -15/4, lambda + 5/12,             0]
[             0,         -5/12, lambda + 5/14]
 
>> p=det(B)
 
p =
 
lambda^3 + (95*lambda^2)/21 + (1025*lambda)/336 + 125/224
 
>> evs = solve(p)
 
evs =
 
 -15/4
 -5/12
 -5/14
 
 >> null(evs(1)*Id3-A)
 
ans =
 
 152/21
  -57/7
      1
 

And now, since the system is non-homogenous...this would give me

[0,0,0] = [(-15/4), 0, 0 ; (15/4), (-5/12), 0 ; 0, (5/12) , (-5/14)] x [a1, b1, c1] + [1575, 0, 0]

This results in the numbers that I was looking for.... Xp: 420, 3780, 4410.

My question is...how can I go about this above equation by way of undetermined coefficients in matlab?

Thanks for any advice you have

Answer 1

[madhan ravi]

Read about fsolve():

Roots=vpa(fsolve(@solls,[0;0;0]))   %function call

function eqn=solls(x)              %function definition
eqn(1) = - x(1)*(15/4) + 1575;
eqn(2) = (15/4)*x(1) - (5/12)*x(2);
eqn(3) = (5/12)*x(2) - (5/14)*x(3);
end

[spcrooks]

Ah...I was looking in all the wrong places. fsolve makes sense, now that I am looking at it.

Thank you for your time, madhan.

[madhan ravi]

Anytime :)

Answer 2

[Bruno Luong]

Guys, this is a linear system, why using FSOLVE?

>> [-15/4 0 0; 15/4 -5/12 0; 0 5/12 -5/14]\[-1575; 0; 0]

ans =

         420
        3780
        4410

>> 

[madhan ravi]

oops you are right ! +1