1. The problem involves finding the particular solution $y_p(t)$ given by the series: $$y_p(t) = \frac{\pi}{8} - \frac{2}{\pi} \sum_{n=-\infty}^\infty \frac{e^{i(2n-1)t}}{(2n-1)^2 \left[4 - (2n-1)^2\right]}$$ 2. We need to carefully evaluate the infinite sum and simplify the expression to match the book's final answer. 3. Note the summation index $n$ runs over all integers, and the terms involve complex exponentials and denominators with $(2n-1)^2$ and $4-(2n-1)^2$. 4. The key is to separate the sum into positive and negative $n$ terms and use the property $e^{i\alpha} + e^{-i\alpha} = 2\cos(\alpha)$ to rewrite the sum in terms of cosines: $$\sum_{n=-\infty}^\infty \frac{e^{i(2n-1)t}}{(2n-1)^2 [4-(2n-1)^2]} = 2 \sum_{n=1}^\infty \frac{\cos((2n-1)t)}{(2n-1)^2 [4-(2n-1)^2]}$$ 5. Substitute this back into $y_p(t)$: $$y_p(t) = \frac{\pi}{8} - \frac{4}{\pi} \sum_{n=1}^\infty \frac{\cos((2n-1)t)}{(2n-1)^2 [4-(2n-1)^2]}$$ 6. This matches the form typically found in Fourier series solutions for such problems. 7. To verify or simplify further, one can evaluate the first few terms numerically or check the book's final expression for equivalence. 8. The key formula used is the Euler formula for complex exponentials and the symmetry of cosine. 9. Thus, the corrected and simplified particular solution is: $$\boxed{y_p(t) = \frac{\pi}{8} - \frac{4}{\pi} \sum_{n=1}^\infty \frac{\cos((2n-1)t)}{(2n-1)^2 [4-(2n-1)^2]}}$$ This matches the expected final answer format.