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: $y \frac{dx}{dy} = x + 2y^3$ Dividing by $y$, we g

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$x = y^3 + cy$

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(B) Given the differential equation: $(x+2y^3) \frac{dy}{dx} = y$ Rearranging the equation, we get: $y \frac{dx}{dy} = x + 2y^3$ Dividing by $y$, we get: $\frac{dx}{dy} - \frac{1}{y}x = 2y^2$ This is a linear differential equation of the form $\frac{dx}{dy} + Px = Q$, where $P = -\frac{1}{y}$ and $Q = 2y^2$. The integrating factor $(IF)$ is given by: $IF = e^{\int P dy} = e^{\int -\frac{1}{y} dy} = e^{-\ln|y|} = \frac{1}{y}$. The solution is given by: $x \cdot (IF) = \int (Q \cdot IF) dy + c$ Substituting the values: $x \cdot \frac{1}{y} = \int (2y^2 \cdot \frac{1}{y}) dy + c$ $\frac{x}{y} = \int 2y dy + c$ $\frac{x}{y} = y^2 + c$ Therefore, $x = y^3 + cy$.

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

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