1. **Stating the problem:** We are given the differential equation $$y' = e^{x+y} - 1$$ and asked to analyze or solve it. 2. **Rewrite the equation:** Recall that $$y' = \frac{dy}{dx}$$, so the equation is: $$\frac{dy}{dx} = e^{x+y} - 1$$ 3. **Separate variables if possible:** Rewrite $$e^{x+y} = e^x e^y$$, so: $$\frac{dy}{dx} = e^x e^y - 1$$ 4. **Attempt to separate variables:** Move terms to isolate $$y$$ and $$x$$: $$\frac{dy}{dx} + 1 = e^x e^y$$ Rewrite as: $$\frac{dy}{dx} + 1 = e^x e^y$$ Divide both sides by $$e^y$$: $$e^{-y} \frac{dy}{dx} + e^{-y} = e^x$$ 5. **Substitute:** Let $$u = e^{-y}$$, then: $$\frac{du}{dx} = -e^{-y} \frac{dy}{dx} = -u \frac{dy}{dx}$$ Rearranged: $$\frac{dy}{dx} = -\frac{1}{u} \frac{du}{dx}$$ 6. **Rewrite the equation in terms of $$u$$:** Substitute into step 4: $$e^{-y} \frac{dy}{dx} + e^{-y} = e^x$$ Becomes: $$u \frac{dy}{dx} + u = e^x$$ Substitute $$\frac{dy}{dx} = -\frac{1}{u} \frac{du}{dx}$$: $$u \left(-\frac{1}{u} \frac{du}{dx}\right) + u = e^x$$ Simplify: $$-\frac{du}{dx} + u = e^x$$ 7. **Rewrite as a linear ODE:** $$\frac{du}{dx} - u = -e^x$$ 8. **Solve the linear ODE:** The integrating factor is: $$\mu(x) = e^{-x}$$ Multiply both sides: $$e^{-x} \frac{du}{dx} - e^{-x} u = -e^{-x} e^x = -1$$ Left side is derivative: $$\frac{d}{dx} (u e^{-x}) = -1$$ Integrate both sides: $$u e^{-x} = -x + C$$ 9. **Solve for $$u$$:** $$u = e^{x} (-x + C)$$ Recall $$u = e^{-y}$$, so: $$e^{-y} = e^{x} (-x + C)$$ 10. **Solve for $$y$$:** Take natural logarithm: $$-y = \ln \left(e^{x} (-x + C)\right)$$ $$y = -x - \ln(-x + C)$$ **Note:** The solution is valid where $$-x + C > 0$$. **Final answer:** $$y = -x - \ln(C - x)$$ This is the implicit general solution to the differential equation.