1. **State the problem:** We are given the differential equation $$y'' - y = 0$$ with the general solution $$y = c_1 e^x + c_2 e^{-x}$$. We need to find the particular solution satisfying the initial conditions $$y(-1) = 5$$ and $$y'(-1) = -5$$. 2. **Find the derivative:** $$y' = c_1 e^x - c_2 e^{-x}$$. 3. **Apply initial conditions:** At $$x = -1$$, $$y(-1) = c_1 e^{-1} + c_2 e^{1} = 5$$ $$y'(-1) = c_1 e^{-1} - c_2 e^{1} = -5$$ 4. **Set up the system of equations:** $$\begin{cases} c_1 e^{-1} + c_2 e^{1} = 5 \\ c_1 e^{-1} - c_2 e^{1} = -5 \end{cases}$$ 5. **Add the two equations to eliminate $$c_2$$:** $$2 c_1 e^{-1} = 0 \implies c_1 = \cancel{\frac{0}{2 e^{-1}}}0$$ 6. **Substitute $$c_1 = 0$$ into the first equation:** $$0 + c_2 e^{1} = 5 \implies c_2 = \frac{5}{e}$$ 7. **Write the particular solution:** $$y = 0 \cdot e^x + \frac{5}{e} e^{-x} = 5 e^{-x - 1}$$ **Final answer:** $$\boxed{y = 5 e^{-x - 1}}$$