1. The problem involves solving the system of differential equations: $$Dx - 4y = 1$$ $$x + Dy = 2$$ where $D$ denotes differentiation with respect to $t$. 2. From the first equation, express $y$ in terms of $x$ and its derivatives: $$y = \frac{1}{4}Dx - \frac{1}{4}$$ 3. Substitute $y$ into the second equation: $$x + D\left(\frac{1}{4}Dx - \frac{1}{4}\right) = 2$$ 4. Simplify the derivative: $$x + \frac{1}{4}D^2x - 0 = 2$$ which gives $$(D^2 + 1)x = 2$$ 5. This is a nonhomogeneous second-order linear differential equation. The complementary (homogeneous) equation is: $$(D^2 + 1)x = 0$$ with characteristic equation: $$r^2 + 1 = 0 \implies r = \pm i$$ 6. The general solution to the homogeneous equation is: $$x_h = c_1 \cos t + c_2 \sin t$$ 7. For the particular solution $x_p$, since the right side is constant 2, try a constant solution: $$x_p = A$$ Substitute into the equation: $$(0 + 1)A = 2 \implies A = 2$$ 8. Therefore, the general solution for $x$ is: $$x = c_1 \cos t + c_2 \sin t + 2$$ 9. Substitute $x$ back into the expression for $y$: $$y = \frac{1}{4}Dx - \frac{1}{4} = \frac{1}{4}\left(-c_1 \sin t + c_2 \cos t\right) - \frac{1}{4}$$ 10. Simplify $y$: $$y = -\frac{1}{4} c_1 \sin t + \frac{1}{4} c_2 \cos t - \frac{1}{4}$$ Thus, the solutions are: $$x = c_1 \cos t + c_2 \sin t + 2$$ $$y = -\frac{1}{4} c_1 \sin t + \frac{1}{4} c_2 \cos t - \frac{1}{4}$$ This completes the solution.