1. **Problem statement:** Solve the differential equation $$y'' - y = f(t)$$ where $$f(t) = |t|$$ for $$-\pi \leq t \leq \pi$$ and $$f(t)$$ is periodic with period $$2\pi$$ using complex Fourier series. 2. **Fourier series setup:** The complex Fourier series for a $$2\pi$$-periodic function $$f(t)$$ is given by: $$ f(t) = \sum_{n=-\infty}^{\infty} c_n e^{i n t}, \quad \text{where} \quad c_n = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(t) e^{-i n t} dt $$ 3. **Calculate coefficients $$c_n$$:** Since $$f(t) = |t|$$ is even, the coefficients simplify: $$ c_n = \frac{1}{2\pi} \int_{-\pi}^{\pi} |t| e^{-i n t} dt = \frac{1}{\pi} \int_0^{\pi} t \cos(n t) dt $$ 4. **Evaluate integral for $$c_n$$:** Use integration by parts: Let $$u = t$$, $$dv = \cos(n t) dt$$, then $$du = dt$$, $$v = \frac{\sin(n t)}{n}$$. $$ \int_0^{\pi} t \cos(n t) dt = \left. t \frac{\sin(n t)}{n} \right|_0^{\pi} - \int_0^{\pi} \frac{\sin(n t)}{n} dt = \frac{\pi \sin(n \pi)}{n} - \frac{1}{n} \int_0^{\pi} \sin(n t) dt $$ Since $$\sin(n \pi) = 0$$ for all integers $$n$$, $$ \int_0^{\pi} t \cos(n t) dt = - \frac{1}{n} \left[ -\frac{\cos(n t)}{n} \right]_0^{\pi} = - \frac{1}{n} \left( -\frac{\cos(n \pi) - \cos(0)}{n} \right) = \frac{1}{n^2} (1 - (-1)^n) $$ 5. **Therefore,** $$ c_n = \frac{1}{\pi} \cdot \frac{1}{n^2} (1 - (-1)^n) = \frac{1 - (-1)^n}{\pi n^2} $$ 6. **Note:** For even $$n$$, $$1 - (-1)^n = 0$$, so $$c_n = 0$$ for even $$n$$. 7. **Coefficient $$c_0$$:** $$ c_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} |t| dt = \frac{1}{\pi} \int_0^{\pi} t dt = \frac{1}{\pi} \cdot \frac{\pi^2}{2} = \frac{\pi}{2} $$ 8. **Solution form:** The differential equation is linear with constant coefficients. The general solution is: $$ y(t) = y_h(t) + y_p(t) $$ where $$y_h$$ solves the homogeneous equation $$y'' - y = 0$$ and $$y_p$$ is a particular solution. 9. **Homogeneous solution:** Characteristic equation: $$r^2 - 1 = 0 \Rightarrow r = \pm 1$$ $$ y_h = C_1 e^t + C_2 e^{-t} $$ 10. **Particular solution using Fourier series:** Express $$y_p$$ as: $$ y_p = \sum_{n=-\infty}^{\infty} Y_n e^{i n t} $$ Substitute into the equation: $$ \sum_{n} (-n^2 - 1) Y_n e^{i n t} = \sum_n c_n e^{i n t} $$ Equate coefficients: $$ (-n^2 - 1) Y_n = c_n \Rightarrow Y_n = \frac{c_n}{-n^2 - 1} = -\frac{c_n}{n^2 + 1} $$ 11. **Calculate $$Y_0$$:** $$ Y_0 = \frac{c_0}{-0 - 1} = -c_0 = -\frac{\pi}{2} $$ 12. **Final solution:** $$ y(t) = C_1 e^t + C_2 e^{-t} - \frac{\pi}{2} + \sum_{n \neq 0, \text{odd}} -\frac{c_n}{n^2 + 1} e^{i n t} $$ where $$c_n = \frac{2}{\pi n^2}$$ for odd $$n$$ (since $$1 - (-1)^n = 2$$ for odd $$n$$). 13. **Summary:** - Compute $$c_n$$ from $$f(t) = |t|$$. - Use $$Y_n = -\frac{c_n}{n^2 + 1}$$ for particular solution. - Add homogeneous solution. This completes the solution by complex Fourier series for problem 4.