1. **State the problem:** Solve the differential equation $$xy^2y' + y^3 = x \cos x$$ for $y(x)$. 2. **Rewrite the equation:** The given equation is $$xy^2 \frac{dy}{dx} + y^3 = x \cos x$$. 3. **Isolate $y'$:** Move $y^3$ to the right side: $$xy^2 \frac{dy}{dx} = x \cos x - y^3$$ Divide both sides by $xy^2$ (assuming $x \neq 0$ and $y \neq 0$): $$\frac{dy}{dx} = \frac{x \cos x - y^3}{x y^2} = \frac{\cos x}{y^2} - \frac{y}{x}$$ 4. **Rewrite as:** $$\frac{dy}{dx} + \frac{y}{x} = \frac{\cos x}{y^2}$$ 5. **Substitution:** Let $v = y^3$. Then: $$\frac{dv}{dx} = 3y^2 \frac{dy}{dx}$$ From step 3, multiply both sides by $3y^2$: $$3y^2 \frac{dy}{dx} + 3 \frac{y^3}{x} = 3 \frac{\cos x}{y^2} y^2$$ Simplify: $$\frac{dv}{dx} + \frac{3v}{x} = 3 \cos x$$ 6. **This is a linear ODE in $v$:** $$\frac{dv}{dx} + \frac{3}{x} v = 3 \cos x$$ 7. **Find integrating factor:** $$\mu(x) = e^{\int \frac{3}{x} dx} = e^{3 \ln x} = x^3$$ 8. **Multiply both sides by $x^3$:** $$x^3 \frac{dv}{dx} + 3x^2 v = 3x^3 \cos x$$ 9. **Left side is derivative:** $$\frac{d}{dx} (x^3 v) = 3x^3 \cos x$$ 10. **Integrate both sides:** $$x^3 v = 3 \int x^3 \cos x \, dx + C$$ 11. **Integral $\int x^3 \cos x \, dx$ can be solved by parts (not shown here for brevity). Let $I = \int x^3 \cos x \, dx$** 12. **Final solution:** $$v = \frac{3I + C}{x^3}$$ Recall $v = y^3$, so $$y^3 = \frac{3I + C}{x^3}$$ 13. **Therefore,** $$y = \sqrt[3]{\frac{3 \int x^3 \cos x \, dx + C}{x^3}}$$ This is the implicit general solution to the differential equation.