Find the general solution of $(x + 2y^3) \frac{dy}{dx} = y$.

$x = y^3 + Cy$

The correct answer is Option (1) → $x = y^3 + Cy$ ##

Given that, $(x + 2y^3) \frac{dy}{dx} = y$

$\Rightarrow y \cdot \frac{dx}{dy} = x + 2y^3$

$\Rightarrow \frac{dx}{dy} = \frac{x}{y} + 2y^2 \quad \text{[dividing throughout by } y \text{]}$

$\Rightarrow \frac{dx}{dy} - \frac{x}{y} = 2y^2$

which is a linear differential equation.

On comparing it with $\frac{dx}{dy} + Px = Q$, we get

$P = -\frac{1}{y}, Q = 2y^2$

$\text{I.F.} = e^{\int -\frac{1}{y} dy} = e^{-\int \frac{1}{y} dy}$

$∴= e^{-\log y} = e^{\log \left( \frac{1}{y} \right)} = \frac{1}{y} \quad [∵e^{\log x} = x]$

$x \cdot \text{I.F.} = \int Q \cdot \text{I.F.} dy + C$

The general solution is, $x \cdot \frac{1}{y} = \int 2y^2 \cdot \frac{1}{y} dy + C$

$\Rightarrow \frac{x}{y} = \frac{2y^2}{2} + C$

$\Rightarrow \frac{x}{y} = y^2 + C$

$\Rightarrow x = y^3 + Cy$