Least square Equilibrium Solution (ODE)

Question

I've recently encountered a problem stated as follows:

Given that $z\in C([0,+\infty))$ and $a_1,a_2\in\mathbb{R}$, find the least square equilibrium solution of the ODE system: $$\left\{\begin{array}{l}x'=x+2y+z-a_1,\ t>0\\y'=x+2y+2z-a_2,\ t>0\end{array}\right.$$ if it is known that $y(t)=-z(t)$.

So, I made the following thoughts:

1. Plug in the extra condition, so the system may be written, in matrix form, as: $$\mathbb{x}'=\underbrace{\left(\begin{array}{cc}1 & 1\\1 & 0\end{array}\right)}_{C}\mathbb{x}+\underbrace{\left(\begin{array}{c}-a_1\\-a_2\end{array}\right)}_{b},$$ where $\mathbf{x}=(x,y)^t$.
2. At first, solve the homogenous system in the usual way (one may also observe that this system is easy to convert to a second order equation, however I thought solving the system would be easier). The general solution is: $$\psi(t)=c_1\lambda_1e^{\lambda_1t}(\lambda_1,1)^t+c_2\lambda_1e^{\lambda_2t}(\lambda_2,1)^t,$$ where $\lambda_{1,2}=\frac{1\mp\sqrt{5}}{2}$, respectively. So, one of the system's fundamental matrix is: $$\Phi(t)=\left(\begin{array}{cc}\lambda_1e^{\lambda_1t} & \lambda_2e^{\lambda_2t}\\e^{\lambda_1t} & e^{\lambda_2t}\end{array}\right).$$ No, for the non-homogeneous equation, if $\mathbf{y}(t_0)=\mathbf{y}_0$, $_0\geq0$, is some initial condition, then the solution has the general form: $$\mathbf{x}(t)=G(t,t_0)\int_{t_0}^tG(t,s)b(s)ds,$$ where $G(t,t_0)$ is the state transition matrix: $$G(t,t_0)=\Phi(t)\Phi^{-1}(t_0).$$
3. Since this is the soluion of the homogenous, $\lVert L\mathbf{x}-b\rVert_2^2=0$, so the soluion above has minimal square error ($L$ is the linear operator $Lu=u'-Cu$ which describes the system).

At this point it is where I am confused, because, if this was he appropriate way to tackle the question, why not request to simply solve the systme instead of searching for a minimum equilibrium solution?

So, to conclude, my question is: Is this the right way to deal with this problem or have I misunderstood somehing regarding least square error in ODE systems? If the case is the latter, what is this misconception (reference to other works to study is welcome).