The solution of the differential equation (x | vedclass

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(A) Given equation: $(x + 2y^3) \frac{dy}{dx} - y = 0$Rearranging the equation: $(x + 2y^3) = y \frac{dx}{dy}$$\frac{dx}{dy} = \frac{x + 2y^3}{y} = \f

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$x = (c + y^2)y$, where $c$ is the constant of integration

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(A) Given equation: $(x + 2y^3) \frac{dy}{dx} - y = 0$Rearranging the equation: $(x + 2y^3) = y \frac{dx}{dy}$$\frac{dx}{dy} = \frac{x + 2y^3}{y} = \frac{x}{y} + 2y^2$$\frac{dx}{dy} - \frac{1}{y} x = 2y^2$This is a linear differential equation of the form $\frac{dx}{dy} + Px = Q$, where $P = -\frac{1}{y}$ and $Q = 2y^2$.Integrating factor $IF = e^{\int P dy} = e^{\int -\frac{1}{y} dy} = e^{-\ln y} = \frac{1}{y}$.The solution is $x \cdot IF = \int (Q \cdot IF) dy + c$.$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(y^2 + c) = (c + y^2)y$.

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

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