Ploting solutions to linear equations

Consider the linear system of equations x' = Ax. Let x(t)=[x1(t), x2(t), x3(t)]T be the unique solution such that x(0) = v1. In the same figure, I want to plot plot each of the functions x1(t), x2(t) and x3(t). I dont think I should have to use ODE 45. I have already looked on the matlab website to no avail.

M = [7 7 7; 2 6 7; 4 3 1];

B = transpose(M); 

A = M+B;

[V,D]=eig(A);

lambda1= -5.9049;
v1= [-0.3429 -0.3208 0.8829];

% part c
syms x(t) y(t) %  I use x(t) and y(t) to represent x1' and x2' 
Y = [x; y];
odes = diff(Y) == A*Y %matlab keeps saying there is a problem with this line and I dont know why 
[xSol(t), ySol(t)] = dsolve(odes);
xSol(t) = simplify(xSol(t));
ySol(t) = simplify(ySol(t));
C = Y(0) == [-0.3429 -0.3208 0.8829];
[xSol(t), ySol(t)] = dsolve(odes,C)
clf
fplot(ySol)
hold on
fplot(xSol)
grid on
savefig('symmetric.fig')

Answer 1

[Star Strider]

The ‘A*Y’ operation attempts to multiply (3x3) matrix ‘A’ by a (2x1) vector ‘Y’. That will just never work!

The ‘Y’ vector must be (3x1) to do that multiplication.

[Mike Chan]

Hello,

I made the code in terms of 3 variables now . The code complies but, matlab is returning [ empty sym ] and a blank graph. What does that mean?

Thanks

M = [7 7 7; 2 6 7; 4 3 1];

B = transpose(M); 

A = M+B;

[V,D]=eig(A);

lambda1= -5.9049;
v1= [-0.3429 -0.3208 0.8829];

% part c
syms x(t) y(t) z(t)
Y = [x; y; z];
odes = diff(Y) == A*Y
[xSol(t), ySol(t),zSol(t)] = dsolve(odes);
xSol(t) = simplify(xSol(t));
ySol(t) = simplify(ySol(t));
zSol(t) = simplify(zSol(t));
C = Y(0) == [-0.3429 -0.3208 0.8829];
[xSol(t), ySol(t),zSol(t)] = dsolve(odes,C)
clf
fplot(ySol)
hold on
fplot(xSol)
grid on
fplot(zSol)
savefig('symmetric.fig')

[Star Strider]

The Symbolic Math Toolbox appears not to be able to solve that, even though it is a linear system, and actually has an analytic solution.

Try this instead:

M = [7 7 7; 2 6 7; 4 3 1];
B = transpose(M); 
A = M+B;
[V,D]=eig(A);
lambda1= -5.9049;
v1= [-0.3429 -0.3208 0.8829];
% part c
syms x(t) y(t) z(t) t Y 
Y = [x; y; z];
odes = diff(Y) == A*Y

[VF,Sbs] = odeToVectorField(odes)
odesfcn = matlabFunction(VF, 'Vars',{'t','Y'})

t = linspace(0, 0.3, 50);
Y0 = [-0.3429 -0.3208 0.8829];
[T,xyz] = ode45(odesfcn, t, Y0);

figure
plot(t,xyz)
grid
legend(string(Sbs))

Parenthetically, I have no idea what the problem is with odeFunction. It used to take a cell array of symbolic variables:

odesfcn = matlabFunction(VF, 'Vars',{t,Y})

without problems, and that is still in the documentation for it. Now, it requires them to be a cell array of character vectors, and will not work without defining them with single quotes, as I did here.