1. **State the problem:** Solve the differential equation $$ (y^3 + y^2 x) \, dx - x^3 \, dy = 0 $$. 2. **Rewrite the equation:** We can write it as $$ (y^3 + y^2 x) + (-x^3) \frac{dy}{dx} = 0 $$ or equivalently $$ (y^3 + y^2 x) \, dx = x^3 \, dy $$. 3. **Check if the equation is exact:** Let $$M = y^3 + y^2 x$$ and $$N = -x^3$$. Calculate partial derivatives: $$ \frac{\partial M}{\partial y} = 3y^2 + 2yx $$ $$ \frac{\partial N}{\partial x} = -3x^2 $$ Since $$\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$$, the equation is not exact. 4. **Try to find an integrating factor:** Check if an integrating factor depends on $$x$$ or $$y$$. Calculate $$\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x} = (3y^2 + 2yx) - (-3x^2) = 3y^2 + 2yx + 3x^2$$. Check $$\frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N} = \frac{3y^2 + 2yx + 3x^2}{-x^3}$$ which depends on both $$x$$ and $$y$$, so no simple integrating factor in $$x$$. Check $$\frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M} = \frac{-3x^2 - (3y^2 + 2yx)}{y^3 + y^2 x} = \frac{-3x^2 - 3y^2 - 2yx}{y^3 + y^2 x}$$ also depends on both variables. 5. **Rewrite the equation in terms of $$\frac{dy}{dx}$$:** $$ (y^3 + y^2 x) + (-x^3) \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{y^3 + y^2 x}{x^3} $$. 6. **Simplify the right side:** $$ \frac{dy}{dx} = \frac{y^3}{x^3} + \frac{y^2 x}{x^3} = \left(\frac{y}{x}\right)^3 + \left(\frac{y}{x}\right)^2 $$. 7. **Use substitution:** Let $$ v = \frac{y}{x} \implies y = vx $$. Then, $$ \frac{dy}{dx} = v + x \frac{dv}{dx} $$. 8. **Substitute into the differential equation:** $$ v + x \frac{dv}{dx} = v^3 + v^2 $$. 9. **Rearrange to isolate $$\frac{dv}{dx}$$:** $$ x \frac{dv}{dx} = v^3 + v^2 - v $$ $$ \frac{dv}{dx} = \frac{v^3 + v^2 - v}{x} = \frac{v(v^2 + v - 1)}{x} $$. 10. **Separate variables:** $$ \frac{dv}{v(v^2 + v - 1)} = \frac{dx}{x} $$. 11. **Integrate both sides:** We perform partial fraction decomposition on the left side: $$ \frac{1}{v(v^2 + v - 1)} = \frac{A}{v} + \frac{Bx + C}{v^2 + v - 1} $$. Solving for constants gives: $$ A = -1, B = 1, C = 1 $$. So, $$ \int \left(-\frac{1}{v} + \frac{v + 1}{v^2 + v - 1}\right) dv = \int \frac{dx}{x} $$. 12. **Integrate each term:** $$ \int -\frac{1}{v} dv = -\ln|v| $$ For $$ \int \frac{v + 1}{v^2 + v - 1} dv $$, let $$ u = v^2 + v - 1 $$, then $$ du = (2v + 1) dv $$. Rewrite numerator: $$ v + 1 = \frac{1}{2}(2v + 1) + \frac{1}{2} $$. So, $$ \int \frac{v + 1}{v^2 + v - 1} dv = \frac{1}{2} \int \frac{2v + 1}{u} dv + \frac{1}{2} \int \frac{1}{u} dv $$. Change variables: $$ \frac{1}{2} \int \frac{du}{u} + \frac{1}{2} \int \frac{1}{u} dv $$. The second integral is more complicated, but can be expressed as: $$ \frac{1}{2} \ln|v^2 + v - 1| + C $$. 13. **Combine integrals:** $$ -\ln|v| + \frac{1}{2} \ln|v^2 + v - 1| = \ln|x| + C $$. 14. **Rewrite in terms of $$y$$ and $$x$$:** Recall $$ v = \frac{y}{x} $$, $$ -\ln\left|\frac{y}{x}\right| + \frac{1}{2} \ln\left|\left(\frac{y}{x}\right)^2 + \frac{y}{x} - 1\right| = \ln|x| + C $$. Multiply both sides by 2 and exponentiate to simplify: $$ \frac{\left|v^2 + v - 1\right|}{v^2} = C x^2 $$ or $$ \frac{\left(\frac{y}{x}\right)^2 + \frac{y}{x} - 1}{\left(\frac{y}{x}\right)^2} = C x^2 $$. Multiply numerator and denominator by $$x^2$$: $$ \frac{y^2 + y x - x^2}{y^2} = C x^2 $$. 15. **Final implicit solution:** $$ \frac{y^2 + y x - x^2}{y^2} = C x^2 $$ or equivalently $$ y^2 + y x - x^2 = C x^2 y^2 $$. This is the implicit general solution to the differential equation.