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(B) Given the differential equation: $(x + 2y^3) \frac{dy}{dx} - y = 0$.Rearranging the equation,we get: $\frac{dx}{dy} = \frac{x + 2y^3}{y}$.This can

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$x = y^3 + cy$

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(B) Given the differential equation: $(x + 2y^3) \frac{dy}{dx} - y = 0$.Rearranging the equation,we get: $\frac{dx}{dy} = \frac{x + 2y^3}{y}$.This can be written as: $\frac{dx}{dy} - \frac{1}{y}x = 2y^2$.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 $(IF)$ is given by: $IF = e^{\int P(y) dy} = e^{\int -\frac{1}{y} dy} = e^{-\ln y} = \frac{1}{y}$.The general solution is: $x \cdot (IF) = \int Q(y) \cdot (IF) dy + c$.Substituting the values: $x \cdot \frac{1}{y} = \int 2y^2 \cdot \frac{1}{y} dy + c$.$\frac{x}{y} = \int 2y dy + c$.$\frac{x}{y} = y^2 + c$.$x = y^3 + cy$.

The general solution of the differential equation $(x+2y^3) \frac{dy}{dx} - y = 0, y > 0$ is

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