1. **State the problem:** Solve the differential equation $$xy' + y = x^4 y^2$$ with the initial condition $$y(1) = 1$$. 2. **Rewrite the equation:** Divide both sides by $$x$$ (assuming $$x \neq 0$$) to get $$y' + \frac{y}{x} = x^3 y^2$$. 3. **Identify the type:** This is a Bernoulli differential equation of the form $$y' + P(x) y = Q(x) y^n$$ where $$P(x) = \frac{1}{x}$$, $$Q(x) = x^3$$, and $$n = 2$$. 4. **Substitution:** Let $$v = y^{1-n} = y^{-1}$$. Then, $$v' = -y^{-2} y'$$. 5. **Rewrite the original equation in terms of $$v$$:** From $$y' + \frac{y}{x} = x^3 y^2$$, multiply both sides by $$-y^{-2}$$: $$-y^{-2} y' - \frac{y^{-1}}{x} = -x^3$$, which becomes $$v' - \frac{v}{x} = -x^3$$. 6. **Solve the linear differential equation for $$v$$:** The integrating factor is $$\mu(x) = e^{-\int \frac{1}{x} dx} = e^{-\ln|x|} = x^{-1}$$. Multiply both sides by $$x^{-1}$$: $$x^{-1} v' - x^{-2} v = -x^{2}$$, which is $$(x^{-1} v)' = -x^{2}$$. 7. **Integrate both sides:** $$\int (x^{-1} v)' dx = \int -x^{2} dx$$ $$x^{-1} v = -\frac{x^{3}}{3} + C$$. 8. **Solve for $$v$$:** $$v = -\frac{x^{4}}{3} + C x$$. 9. **Recall substitution:** $$v = y^{-1}$$, so $$y = \frac{1}{v} = \frac{1}{-\frac{x^{4}}{3} + C x} = \frac{1}{C x - \frac{x^{4}}{3}}$$. 10. **Apply initial condition $$y(1) = 1$$:** $$1 = \frac{1}{C \cdot 1 - \frac{1^{4}}{3}} = \frac{1}{C - \frac{1}{3}}$$, which implies $$C - \frac{1}{3} = 1 \Rightarrow C = \frac{4}{3}$$. 11. **Final solution:** $$y = \frac{1}{\frac{4}{3} x - \frac{x^{4}}{3}} = \frac{3}{4x - x^{4}}$$. **Answer:** $$\boxed{y = \frac{3}{4x - x^{4}}}$$