\( y ^ { \prime \prime } - y = 0 \)

To solve the differential equation \[ y'' - y = 0, \] we start by finding the characteristic equation associated with the differential equation. The characteristic equation is obtained by substituting \( y = e^{rt} \), where \( r \) is a constant. This gives us: \[ r^2 e^{rt} - e^{rt} = 0. \] Factoring out \( e^{rt} \) (which is never zero), we get: \[ r^2 - 1 = 0. \] Now, we can solve for \( r \): \[ r^2 = 1 \implies r = \pm 1. \] This means we have two distinct real roots: \( r_1 = 1 \) and \( r_2 = -1 \). The general solution to the differential equation is then given by: \[ y(t) = C_1 e^{t} + C_2 e^{-t}, \] where \( C_1 \) and \( C_2 \) are arbitrary constants determined by initial conditions or boundary conditions if provided. Thus, the solution to the differential equation \( y'' - y = 0 \) is: \[ y(t) = C_1 e^{t} + C_2 e^{-t}. \]