1. **Stating the problem:** Solve the differential equation $$\frac{dy}{dx} - y = e^x y^2$$. 2. **Identify the type of equation:** This is a nonlinear first-order differential equation because of the $y^2$ term. 3. **Rewrite the equation:** Move all terms to one side: $$\frac{dy}{dx} = y + e^x y^2$$ 4. **Try substitution:** Let $v = \frac{1}{y}$, then $y = \frac{1}{v}$ and $$\frac{dy}{dx} = -\frac{1}{v^2} \frac{dv}{dx}$$ 5. **Substitute into the equation:** $$-\frac{1}{v^2} \frac{dv}{dx} = \frac{1}{v} + e^x \frac{1}{v^2}$$ 6. **Multiply both sides by $-v^2$ to clear denominators:** $$\frac{dv}{dx} = -v - e^x$$ 7. **This is a linear first-order ODE in $v$:** $$\frac{dv}{dx} + v = -e^x$$ 8. **Find integrating factor:** $$\mu(x) = e^{\int 1 dx} = e^x$$ 9. **Multiply entire equation by $e^x$:** $$e^x \frac{dv}{dx} + e^x v = -e^{2x}$$ 10. **Left side is derivative of $v e^x$:** $$\frac{d}{dx} (v e^x) = -e^{2x}$$ 11. **Integrate both sides:** $$v e^x = \int -e^{2x} dx = -\frac{1}{2} e^{2x} + C$$ 12. **Solve for $v$:** $$v = -\frac{1}{2} e^x + C e^{-x}$$ 13. **Recall $v = \frac{1}{y}$, so:** $$\frac{1}{y} = -\frac{1}{2} e^x + C e^{-x}$$ 14. **Final solution:** $$y = \frac{1}{C e^{-x} - \frac{1}{2} e^x}$$ This is the implicit general solution to the original differential equation.