1. **State the problem:** Solve the system of differential equations: $$\frac{d}{dt}x - 4y = 1$$ $$\frac{d}{dt}y + x = 2$$ 2. **Rewrite the system:** Let $x' = \frac{dx}{dt}$ and $y' = \frac{dy}{dt}$. Then: $$x' - 4y = 1 \quad (1)$$ $$y' + x = 2 \quad (2)$$ 3. **Express $x'$ and $y'$:** From (1): $$x' = 1 + 4y$$ From (2): $$y' = 2 - x$$ 4. **Differentiate (1) again to get $x''$:** $$x'' = 4y'$$ Substitute $y'$ from above: $$x'' = 4(2 - x) = 8 - 4x$$ 5. **Form a second order ODE for $x$:** Recall from (1): $$x' = 1 + 4y \Rightarrow y = \frac{x' - 1}{4}$$ But we use the second derivative equation: $$x'' + 4x = 8$$ 6. **Solve the homogeneous equation:** $$x'' + 4x = 0$$ Characteristic equation: $$r^2 + 4 = 0 \Rightarrow r = \pm 2i$$ General solution of homogeneous: $$x_h = C_1 \cos 2t + C_2 \sin 2t$$ 7. **Find a particular solution $x_p$:** Since RHS is constant 8, try $x_p = A$ (constant): $$0 + 4A = 8 \Rightarrow A = 2$$ 8. **General solution for $x$:** $$x = C_1 \cos 2t + C_2 \sin 2t + 2$$ 9. **Find $y$ using $y = \frac{x' - 1}{4}$:** Calculate $x'$: $$x' = -2C_1 \sin 2t + 2C_2 \cos 2t$$ Then: $$y = \frac{-2C_1 \sin 2t + 2C_2 \cos 2t - 1}{4} = -\frac{1}{4} + \frac{C_2}{2} \cos 2t - \frac{C_1}{2} \sin 2t$$ **Final answer:** $$x = C_1 \cos 2t + C_2 \sin 2t + 2$$ $$y = -\frac{1}{4} + \frac{C_2}{2} \cos 2t - \frac{C_1}{2} \sin 2t$$ Constants $C_1$ and $C_2$ are determined by initial conditions.