The easiest way is to let the Symbolic Math Toolbox do the heavy lifting:
syms y(t) u(t) t
Dy = diff(y);
D2y = diff(y,2);
D3y = diff(y,3);
DEq = D3y + 6*D2y + 11*Dy + 6*y == 6*u;
[SS,Sbs] = odeToVectorField(DEq)
producing:
SS =
Y[2]
Y[3]
6*u(t) - 6*Y[1] - 11*Y[2] - 6*Y[3]
Sbs =
y
Dy
D2y
The ‘Sbs’ output simply tells you the substitutions the solver made, so for example ‘Y[1]=y’.
I’m sure you can take it from there.
