$$x'=-y+z-e{-t}$$ $$y'=3x+y-3z$$ $$z'=2x-y-z-2e^{-t}$$
First we have to find the homegene part of the system. We find the fundamental system of solutions: $$(e^{t},e^{-2t},te^{t})$$ from the eigenvector we have that $$u=(0,1,1)$$ which are the values of the constants (c2,c5,c8) of the system $$x=C1e^{t}+C2te^{-2t}+C3te^{t}$$ $$x=C4e^{t}+C5te^{-2t}+C6te^{t}$$ $$x=C7e^{t}+C8te^{-2t}+C9te^{t}$$ so I replace these equations in the original system, solve it and I find that: $$C1=C7;C2=0;C3=-C4=C9;C5=C8;C6=0$$ And so the solution of the homegene part of this equation is : $$x0=c1e^t+C3te^{-t}$$ $$y0=-C3e^t+C5e^{-2t}$$ $$z=c1e^t+C5e^{-2t}+C3te^{-t}$$
But that is not right what am I doing wrong?
After that I have to find the non-homegene part of this system we have that m=0;alpha=-1;beta=0;s=0 That means that the solution is: $$x=ae^{-t}$$ $$y=be^{-t}$$ $$z=ce^{-t}$$ and then we replace in the original system and we have: $$-a+b-c+1=0$$ $$-b-3a-b+3=0$$ $$-2a+b+2=0$$ $$=>a=1;b=c=0$$ and we find that the non homegene part is equal to: $$x1=1e^{-t}$$ $$x2=0e^{-t}$$ $$x3=0e^{-t}$$
But that is also not correct according to wolfram, what am I doing wrong here ?
For the homogeneous part, the matrix has two eigenvalues, one is $-2$, with the eigenvector you found. The other is $1$, which you did not mention at all. You should find the eigenvector $k_1$ corresponding to the eigenvalue $\lambda=1$, and another generalized eigenvector $k_2$, since $\lambda=1$ has multiplicity $2$. The solution of the homogeneous part should be $$\vec{x}=c_1\vec{u}e^{-2t}+c_2(\vec{k}_1te^t+\vec{k}_2e^t).$$
Your set up for the nonhomogeneous part is correct. The matrix is also correct. The solution is not. You can just double check that.