Let C be the matrix $\ \left(\begin{array}{cc} -2 & 1\\ 1 & -2 \end{array}\right)$, with sole eigenvalue $-1$. Solve the system $y'=Cy+b(t)$, where $b(t)=e^{-t}\left(\begin{array}{c} 1 \\ 1 \end{array}\right)$, subject to $\ y(0)=\left(\begin{array}{c} 3 \\ 1 \end{array}\right)$
My attempt:
Firstly, I found that \begin{align*} E_{-1}&=\text{ker}\ \left(\begin{array}{cc} -1 & 1\\ 1 & -1 \end{array}\right)\\ &=\text{span}{\left(\begin{array}{c} 1 \\ 1 \end{array}\right)}\\ \end{align*}
Then I make the substitution $\ y=e^{Ct}z(t)$. Differentiating this with respect to t and substituting this into $y'=Cy+b(t)$, i get that $z'(t)=e^{-Ct}e^{-t}\left(\begin{array}{c} 1 \\ 1 \end{array}\right)$. Now, using the theorem which which states $e^{Ct}v=e^{\lambda t}v$, where $\lambda$ is the corresponding eigenvalue for v, I arrive at $$z'(t)=\left(\begin{array}{c} 1 \\ 1 \end{array}\right)$$ Integrating, I get $$z(t)=\left(\begin{array}{c} 1 \\ 1 \end{array}\right)t+C$$ where C is some constant. Putting this back into my initial substitution and using the initial conditions provided in the question, I get \begin{align*} y&=e^{Ct}\left(\begin{array}{c} 1 \\ 1 \end{array}\right)t+\left(\begin{array}{c} 3 \\ 1 \end{array}\right)\\ \end{align*}
My question is, what does this simplify to? I know \begin{align*} e^{Ct}\left(\begin{array}{c} 1 \\ 1 \end{array}\right)t&=e^{-t}\left(\begin{array}{c} 1 \\ 1 \end{array}\right)t \ \ \ \ \ \ \ \text{by the theorem mentioned earlier}\\ \end{align*} But what does $\ e^{Ct}\left(\begin{array}{c} 3 \\ 1 \end{array}\right)$ equal? I don't believe that $\left(\begin{array}{c} 3 \\ 1 \end{array}\right)$, is a generalised eigenvector, so I'm very confused on how to finish this problem.
Let $V= \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}$ and note that $C = V \begin{bmatrix} -1 & 0 \\ 0 & -3 \end{bmatrix} V^{-1}$ and so $e^{Ct} = V \begin{bmatrix} e^{-t} & 0 \\ 0 & e^{-3t} \end{bmatrix} V^{-1}$.
The solution is given by $y(t) = e^{Ct} y(0) + \int_0^t e^{C(t-\tau)} b(\tau) d \tau$.
Note that $e^{C(t-\tau)} b(\tau) = e^{-3(t-\tau) }e^{-\tau} \begin{bmatrix} 1 \\ 1\end{bmatrix}$ and $V^{-1} y(0) = \begin{bmatrix} 2 \\ 1\end{bmatrix}$.