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Given the differential equation: $$(x + 2y^3) dy = y dx$$ Divide both sides by $dy$ and $y$ to express it in the form $\frac{dx}{dy} + P(y)x = Q(y)$: $$\frac{dx}{dy} = \frac{x + 2y^3}{y}$$ $$\frac{dx}{dy} = \frac{x}{y} + 2y^2$$ $$\frac{dx}{dy} - \frac{1}{y}x = 2y^2$$
This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$, where $P(y) = -\frac{1}{y}$ and $Q(y) = 2y^2$. The Integrating Factor is given by: $$IF = e^{\int P(y) dy} = e^{\int -\frac{1}{y} dy} = e^{-\ln|y|} = e^{\ln|y^{-1}|} = \frac{1}{y}$$
The general solution is given by $x \cdot (IF) = \int Q(y) \cdot (IF) dy + C$: $$x \cdot \frac{1}{y} = \int (2y^2) \cdot \frac{1}{y} dy + C$$ $$\frac{x}{y} = \int 2y dy + C$$ $$\frac{x}{y} = y^2 + C$$ $$x = y^3 + Cy$$
Final Answer: x = y^3 + Cy
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