1. **State the problem:** Given the function $y = Ae^{-3x} + Be^{x} + Ce^{-x}$, we want to find a differential equation for which this is the general solution. 2. **Recall the form of solutions to linear differential equations:** The general solution to a linear differential equation with constant coefficients is a linear combination of exponential functions whose exponents are roots of the characteristic equation. 3. **Identify the characteristic roots:** From the solution, the exponents are $-3$, $1$, and $-1$. These correspond to roots of the characteristic polynomial. 4. **Form the characteristic polynomial:** The roots $r = -3, 1, -1$ imply the characteristic polynomial is $$ (r + 3)(r - 1)(r + 1) = 0 $$ 5. **Expand the polynomial:** $$ (r + 3)(r^2 - 1) = (r + 3)(r^2 - 1) = r^3 + 3r^2 - r - 3 $$ 6. **Write the differential equation:** Replace $r^n$ by $D^n$ where $D = \frac{d}{dx}$, so the differential operator is $$ D^3 + 3D^2 - D - 3 $$ The differential equation is $$ y''' + 3y'' - y' - 3y = 0 $$ **Final answer:** The differential equation whose general solution is $y = Ae^{-3x} + Be^{x} + Ce^{-x}$ is $$ y''' + 3y'' - y' - 3y = 0 $$