1. **Stating the problem:** Simplify or understand the expression $$\frac{1}{D^2+1} \cos x$$ where $D$ represents the differentiation operator with respect to $x$. 2. **Understanding the operator:** Here, $D = \frac{d}{dx}$, so $D^2 = \frac{d^2}{dx^2}$ is the second derivative operator. 3. **Applying the operator:** The expression $$\frac{1}{D^2+1} \cos x$$ means applying the inverse operator of $D^2 + 1$ to $\cos x$. 4. **Recall the differential equation:** The operator $D^2 + 1$ corresponds to the differential equation $$y'' + y = f(x)$$. 5. **Solving for $y$:** We want to find $y$ such that $$y'' + y = \cos x$$. 6. **Find the particular solution:** Since the right side is $\cos x$, try a particular solution of the form $$y_p = A x \sin x + B x \cos x$$ because $\cos x$ is a solution to the homogeneous equation. 7. **Compute derivatives:** $$y_p = A x \sin x + B x \cos x$$ $$y_p' = A (\sin x + x \cos x) + B (\cos x - x \sin x)$$ $$y_p'' = A (2 \cos x - x \sin x) + B (-2 \sin x - x \cos x)$$ 8. **Plug into the equation:** $$y_p'' + y_p = A (2 \cos x - x \sin x) + B (-2 \sin x - x \cos x) + A x \sin x + B x \cos x$$ Simplify: $$= 2 A \cos x - A x \sin x - 2 B \sin x - B x \cos x + A x \sin x + B x \cos x$$ $$= 2 A \cos x - 2 B \sin x$$ 9. **Set equal to $\cos x$:** $$2 A \cos x - 2 B \sin x = \cos x$$ Equate coefficients: $$2 A = 1 \Rightarrow A = \frac{1}{2}$$ $$-2 B = 0 \Rightarrow B = 0$$ 10. **Particular solution:** $$y_p = \frac{1}{2} x \sin x$$ 11. **General solution:** $$y = y_p + C_1 \cos x + C_2 \sin x$$ 12. **Interpretation:** Applying $$\frac{1}{D^2 + 1}$$ to $\cos x$ yields the particular solution $$\frac{1}{2} x \sin x$$ plus any solution to the homogeneous equation. **Final answer:** $$\frac{1}{D^2 + 1} \cos x = \frac{1}{2} x \sin x + C_1 \cos x + C_2 \sin x$$