solving a system of ODEs in 3x3 matrix form

The following allowed me to solve wit 2x2 matrix , how about 3x3 matrix ? what the code. thanks

Answer 1

[Star Strider]

The easiest way would be to use the expm function.

The expression changes slightly:

A = rand(3);
b = rand(3,1);
t = 0:0.1:10;
for k = 1:numel(t)
    y(:,k) = expm(A*t(k))*b;
end

figure
plot(t, y)
grid

Experiment with your own ‘A’ and ‘b’ arrays.

.

[chen tianle]

i try your code not working. please help

[chen tianle]

example what your code ?

if my A is [ 3 1 1 ; 0 3 -1 ; 0 - 1 3]

and my B= 0 ?

what the Sx, Sy, Sz , value ?

[Star Strider]

The code works. The problem is that ‘b’ is 0 so the output is 0, and there was an error in ‘A’ (spaces are delimiters, so the dimensions did not originally match) —

A = [ 3 1 1 ; 0 3 -1 ; 0 -1 3] ;
b = ones(3,1)*eps;                                          % 'b': Small, Non-Zero Value

t = 0:0.1:1;
for k = 1:numel(t)
    y(:,k) = expm(A*t(k))*b;
end
figure
plot(t, y)
grid

This would also be the situation doing symbolic integration:

syms t x1(t) x2(t) x3(t) x10 x20 x30 
x = [x1(t); x2(t); x3(t)]

x = 

eqn = diff(x) == A*x + b

eqn = 

Y = dsolve(eqn, x1(0)==x10, x2(0)==x20, x3(0)==x30)

Y = struct with fields:
    x2: [1×1 sym]
    x1: [1×1 sym]
    x3: [1×1 sym]

Y.x1

ans = 

Y.x2

ans = 

Y.x3

ans = 

sympref('AbbreviateOutput',false);
Y = vpa(simplify([Y.x1; Y.x2; Y.x3], 500), 3)

Y = 

Setting the intital conditions to 0:

Y = dsolve(eqn, x1(0)==0, x2(0)==0, x3(0)==0)

Y = struct with fields:
    x2: [1×1 sym]
    x1: [1×1 sym]
    x3: [1×1 sym]

% sympref('AbbreviateOutput',false);
Y = vpa(simplify([Y.x1; Y.x2; Y.x3], 500), 3)

Y = 

.