1. **State the problem:** Solve the differential equation $$y'' + y' + y = x + 1$$. 2. **Identify the type of equation:** This is a non-homogeneous linear second-order differential equation with constant coefficients. 3. **Solve the homogeneous equation:** $$y'' + y' + y = 0$$. The characteristic equation is $$r^2 + r + 1 = 0$$. Using the quadratic formula: $$r = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2}$$. So the roots are complex conjugates: $$r = -\frac{1}{2} \pm i\frac{\sqrt{3}}{2}$$. The homogeneous solution is: $$y_h = e^{-\frac{x}{2}}\left(C_1 \cos\left(\frac{\sqrt{3}}{2}x\right) + C_2 \sin\left(\frac{\sqrt{3}}{2}x\right)\right)$$. 4. **Find a particular solution:** Since the right side is a polynomial of degree 1, try a particular solution of the form: $$y_p = Ax + B$$. Calculate derivatives: $$y_p' = A$$ $$y_p'' = 0$$. Substitute into the original equation: $$0 + A + (Ax + B) = x + 1$$ Simplify: $$Ax + (A + B) = x + 1$$. Equate coefficients: - Coefficient of $$x$$: $$A = 1$$ - Constant term: $$A + B = 1 \Rightarrow 1 + B = 1 \Rightarrow B = 0$$. So the particular solution is: $$y_p = x$$. 5. **Write the general solution:** $$y = y_h + y_p = e^{-\frac{x}{2}}\left(C_1 \cos\left(\frac{\sqrt{3}}{2}x\right) + C_2 \sin\left(\frac{\sqrt{3}}{2}x\right)\right) + x$$. **Final answer:** The particular solution that fits the right side $$x + 1$$ is $$y_p = x$$, which corresponds to option V. x.