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 terms,we get: $\frac{dx}{dy} = \frac{x + 2y^3}{y}$. This can be w

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$\frac{x}{y} = y^2 + c$

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

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

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