1. **State the problem:** Solve the differential equation $$\frac{dx}{x^3} = y^4 \, dy$$. 2. **Rewrite the equation:** We want to separate variables to integrate both sides. Rewrite as: $$\frac{dx}{x^3} = y^4 \, dy$$ which can be written as $$x^{-3} \, dx = y^4 \, dy$$. 3. **Integrate both sides:** Integrate the left side with respect to $x$ and the right side with respect to $y$: $$\int x^{-3} \, dx = \int y^4 \, dy$$. 4. **Compute the integrals:** Recall the integral formula: $$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C, \quad n \neq -1$$. For the left side: $$\int x^{-3} \, dx = \frac{x^{-3+1}}{-3+1} + C = \frac{x^{-2}}{-2} + C = -\frac{1}{2x^2} + C$$. For the right side: $$\int y^4 \, dy = \frac{y^{4+1}}{4+1} + C = \frac{y^5}{5} + C$$. 5. **Combine the results:** $$-\frac{1}{2x^2} = \frac{y^5}{5} + C$$ where $C$ is the constant of integration. 6. **Rewrite the solution:** Multiply both sides by 10 to clear denominators: $$10 \left(-\frac{1}{2x^2}\right) = 10 \left(\frac{y^5}{5} + C\right)$$ $$-\frac{10}{2x^2} = 2 y^5 + 10 C$$ $$-\frac{5}{x^2} = 2 y^5 + K$$ where $K = 10 C$ is a new constant. 7. **Final implicit solution:** $$-\frac{5}{x^2} - 2 y^5 = K$$ or equivalently $$\frac{5}{x^2} + 2 y^5 = -K$$ which represents the implicit general solution to the differential equation.