Solve for y using D-operator methods from the system of equations: (D+1)x + (D−1)y = e^t and (D+2)x + (2D−1)y = 1.

Multiply equation (1) by $(D+2)$ and equation (2) by $(D+1)$.
Equation (1) becomes:

$$(D+2)(D+1)x + (D+2)(D-1)y = (D+2)e^t$$ $$(D^2+3D+2)x + (D^2+D-2)y = De^t + 2e^t$$

Since $De^t = e^t$, this simplifies to:

$$(D^2+3D+2)x + (D^2+D-2)y = 3e^t \quad (3)$$

Equation (2) becomes:

$$(D+1)(D+2)x + (D+1)(2D-1)y = (D+1)1$$ $$(D^2+3D+2)x + (2D^2+D-1)y = D(1) + 1(1)$$

Since $D(1) = 0$, this simplifies to:

$$(D^2+3D+2)x + (2D^2+D-1)y = 1 \quad (4)$$

Subtracting the equations eliminates the $x$ term:

$$[(D^2+3D+2)x + (D^2+D-2)y] - [(D^2+3D+2)x + (2D^2+D-1)y] = 3e^t - 1$$ $$(D^2+D-2 - (2D^2+D-1))y = 3e^t - 1$$ $$(D^2+D-2 - 2D^2-D+1)y = 3e^t - 1$$ $$(-D^2 - 1)y = 3e^t - 1$$ $$(D^2+1)y = -(3e^t - 1)$$ $$(D^2+1)y = 1 - 3e^t$$

The auxiliary equation is $m^2+1=0$.

$$m^2 = -1$$ $$m = \pm i$$

Thus, the complementary function is:

$$y_c = C_1 \cos t + C_2 \sin t$$

We need to find $y_p = \frac{1}{D^2+1}(1 - 3e^t)$.
We can split this into two parts:

$$y_p = \frac{1}{D^2+1}(1) - \frac{1}{D^2+1}(3e^t)$$

For the first term, $\frac{1}{D^2+1}(1)$, we treat 1 as $e^{0t}$ and substitute $D=0$:

$$\frac{1}{0^2+1}(1) = \frac{1}{1}(1) = 1$$

For the second term, $\frac{1}{D^2+1}(3e^t)$, we substitute $D=1$:

$$\frac{1}{1^2+1}(3e^t) = \frac{1}{1+1}(3e^t) = \frac{3}{2}e^t$$

So, the particular integral is:

$$y_p = 1 - \frac{3}{2}e^t$$

The general solution for $y$ is $y = y_c + y_p$:

$$y = C_1 \cos t + C_2 \sin t + 1 - \frac{3}{2}e^t$$