Nonhomogeneous system of linear differential equations What am I supposed to do if I have a system of the linear equations, and its matrix is singular? How do I find the general solutions since I can't find a steady state? 
Example:
$$x'= ax - ay + 1$$
$$y' = -ax + ay - 1$$
 A: One method I find efficient is to decouple the system via diagonalization, which works for a differential equation in the following form, provided that $A$ is diagonalizable: 
$$\frac{d\vec{\mathbf{x}}}{dt}=A\vec{\mathbf{x}}+\vec{\mathbf{f}}(t) \tag{1}$$
In your case, we have: $$A=\begin{bmatrix} a & -a \\ -a & a \end {bmatrix} \qquad \vec{\mathbf{f}}(t)=\begin{bmatrix} 1 \\ -1 \end{bmatrix}\qquad \vec{\mathbf{x}}=\begin{bmatrix} x \\ y \end{bmatrix}$$
We can write $A$ in the form $A=PDP^{-1}$, where $D$ is a diagonal matrix and $P$ is invertible. Hence, by the substitution $\vec{\mathbf{x}}(t)=P\vec{\mathbf{u}}(t)$, we can write $(1)$ as:
$$\frac{d\vec{\mathbf{u}}}{dt}=D\vec{\mathbf{u}}+P^{-1}\vec{\mathbf{f}}(t) \tag{2}$$
Since $D$ is a diagonal matrix, our system will be decoupled.

Diagonalizing $A$, we obtain:
$$P=\begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} \qquad D=\begin {bmatrix} 0 & 0 \\ 0 & 2a \end {bmatrix}\qquad P^{-1}=\begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} \end{bmatrix}$$
Hence, from $(2)$, we must solve the following:
$$\begin{cases} u_1'=0 \\ u_2'=2au_2-1 \end{cases}$$
The first one is essentially trivial, and the second one is a separable ODE. One can obtain $x(t)$ and $y(t)$  by using the substitution we performed, that is $\vec{\mathbf{x}}(t)=P\vec{\mathbf{u}}(t)$:
$$\begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} u_1 \\ u_2 \end {bmatrix}=\begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} k_1 \\ k_2 e^{2at}+\frac{1}{2a} \end {bmatrix}$$
Multiplying the two matrices on the RHS gives us the general solution we require:
$$\begin{align} x(t)=k_1-k_2 e^{2at}-\frac{1}{2a} \\ y(t)=k_1+k_2 e^{2at}+\frac{1}{2a} \end{align}$$
