The solution of the differential equation (x | vedclass

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

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$x = y(y^2 + c)$

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(B) Given the differential equation: $(x + 2y^3) \frac{dy}{dx} = y$. Rearranging the equation,we get: $\frac{dx}{dy} = \frac{x + 2y^3}{y} = \frac{x}{y} + 2y^2$. 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 ($I$.$F$.) is given by $e^{\int P(y) dy} = e^{\int -\frac{1}{y} dy} = e^{-\ln|y|} = \frac{1}{y}$. The general solution is $x \cdot (I.F.) = \int Q(y) \cdot (I.F.) dy + c$. Substituting the values: $x \cdot \frac{1}{y} = \int 2y^2 \cdot \frac{1}{y} dy + c$. $\frac{x}{y} = \int 2y dy + c$. $\frac{x}{y} = y^2 + c$. Therefore,$x = y(y^2 + c)$.

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

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