1. **State the problem:** Solve the first-order linear differential equation $y' + e^x y = e^x$. 2. **Identify the type and formula:** This is a linear differential equation of the form $y' + P(x)y = Q(x)$ where $P(x) = e^x$ and $Q(x) = e^x$. 3. **Find the integrating factor (IF):** The integrating factor is given by $$\mu(x) = e^{\int P(x) dx} = e^{\int e^x dx} = e^{e^x}.$$ 4. **Multiply the entire differential equation by the integrating factor:** $$e^{e^x} y' + e^{e^x} e^x y = e^{e^x} e^x.$$ 5. **Recognize the left side as a derivative:** $$\frac{d}{dx} \left(e^{e^x} y \right) = e^{e^x} e^x.$$ 6. **Integrate both sides with respect to $x$:** $$e^{e^x} y = \int e^{e^x} e^x dx + C.$$ 7. **Use substitution to solve the integral:** Let $u = e^x$, then $du = e^x dx$, so $$\int e^{e^x} e^x dx = \int e^u du = e^u + C = e^{e^x} + C.$$ 8. **Substitute back and solve for $y$:** $$e^{e^x} y = e^{e^x} + C \implies y = 1 + C e^{-e^x}.$$ **Final answer:** $$y = 1 + C e^{-e^x}.$$