1. **Problem Statement:** Solve the differential equation $$y'' - y = f(t)$$ where $$f(t) = \begin{cases} 1, & 0 < t < \pi \\ 0, & \pi < t < 2\pi \end{cases}$$. 2. **Step 1: Understand the problem.** This is a nonhomogeneous linear differential equation with a piecewise forcing function. We want to find the solution $$y(t)$$ that satisfies the equation. 3. **Step 2: Solve the homogeneous equation.** The associated homogeneous equation is $$y'' - y = 0$$. The characteristic equation is $$r^2 - 1 = 0$$. Solving, $$r = \pm 1$$. So the homogeneous solution is $$y_h = C_1 e^t + C_2 e^{-t}$$. 4. **Step 3: Find a particular solution.** Since $$f(t)$$ is piecewise constant, we solve separately on intervals: - For $$0 < t < \pi$$, $$f(t) = 1$$. - For $$\pi < t < 2\pi$$, $$f(t) = 0$$. 5. **Step 4: Particular solution for $$0 < t < \pi$$.** Try a constant solution $$y_p = A$$. Substitute into the equation: $$y_p'' - y_p = 1 \implies 0 - A = 1 \implies A = -1$$. So $$y_p = -1$$. 6. **Step 5: Particular solution for $$\pi < t < 2\pi$$.** Here $$f(t) = 0$$, so the particular solution is zero. 7. **Step 6: General solutions on intervals:** - For $$0 < t < \pi$$: $$y = C_1 e^t + C_2 e^{-t} - 1$$. - For $$\pi < t < 2\pi$$: $$y = D_1 e^t + D_2 e^{-t}$$. 8. **Step 7: Apply continuity conditions at $$t = \pi$$.** Since $$y$$ and $$y'$$ must be continuous at $$t=\pi$$: $$C_1 e^{\pi} + C_2 e^{-\pi} - 1 = D_1 e^{\pi} + D_2 e^{-\pi}$$ and $$C_1 e^{\pi} - C_2 e^{-\pi} = D_1 e^{\pi} - D_2 e^{-\pi}$$. 9. **Step 8: Solve for constants using initial/boundary conditions if given.** Since none are provided, the solution is expressed in terms of constants. **Final answer:** $$y(t) = \begin{cases} C_1 e^t + C_2 e^{-t} - 1, & 0 < t < \pi \\ D_1 e^t + D_2 e^{-t}, & \pi < t < 2\pi \end{cases}$$ where constants satisfy continuity at $$t=\pi$$.