1. **State the problem:** Solve the differential equation $$\frac{dy}{dx} + 3y = 3x^2 e^{-3x}$$ using Bernoulli's method. 2. **Identify the type:** This is a linear first-order differential equation, not Bernoulli's form. Bernoulli's equation has the form $$\frac{dy}{dx} + P(x)y = Q(x)y^n$$ with $$n \neq 0,1$$. Here, $$n=0$$ since the right side does not involve $$y$$ raised to a power. 3. **Rewrite the equation:** $$\frac{dy}{dx} + 3y = 3x^2 e^{-3x}$$. 4. **Use integrating factor method:** The integrating factor $$\mu(x) = e^{\int 3 dx} = e^{3x}$$. 5. **Multiply entire equation by $$\mu(x)$$:** $$e^{3x} \frac{dy}{dx} + 3 e^{3x} y = 3x^2 e^{3x} e^{-3x} = 3x^2$$ 6. **Recognize left side as derivative:** $$\frac{d}{dx} \left( y e^{3x} \right) = 3x^2$$ 7. **Integrate both sides:** $$y e^{3x} = \int 3x^2 dx = x^3 + C$$ 8. **Solve for $$y$$:** $$y = e^{-3x} (x^3 + C)$$ **Final answer:** $$y = e^{-3x} (x^3 + C)$$