Integrating factor of $\left(x+2 y^3\right) \frac{d y}{d x}=y^2$ is
Integrating factor of $\left(x+2 y^3\right) \frac{d y}{d x}=y^2$ is
$e^{\left(\frac{1}{y}\right)}$
$e^{-\left(\frac{1}{y}\right)}$
$y$
$\frac{-1}{y}$
Solution
Given differential equation is
$
\begin{array}{rlrl}
& & \left(x+2 y^3\right) \frac{d y}{d x}=y^2 \\
\Rightarrow & y^2 \frac{d x}{d y} & =x+2 y^3 \Rightarrow & \frac{d x}{d y}-\frac{x}{y^2}=2 y \\
\therefore & & I F & =e^{\int-\frac{1}{y^2} d y}=e^{\frac{1}{y}}
\end{array}
$
Alternative Solution:
The given differential equation is $\left(x+2 y^3\right) \frac{d y}{d x}=y^2$. To find the integrating factor, we first need to write this equation in the standard form of a linear differential equation, which is $\frac{d y}{d x}+P(x) y=Q(x)$.
Dividing the given equation by $x + 2y^3$, we get $\frac{d y}{d x} + \frac{y^2}{x + 2y^3} = 0$. So here, $P(x) = \frac{y^2}{x + 2y^3}$ and $Q(x) = 0$.
The integrating factor is given by $e^{\int P(x) dx}$. In our case, $P(x)$ is actually a function of $y$, not $x$. So we have $e^{\int \frac{y^2}{x + 2y^3} dy}$.
This integral is not straightforward, but it simplifies significantly if you recognize that the numerator is the derivative of the denominator. So the integral becomes $e^{\ln|x + 2y^3|}$, which simplifies to $e^{\left(\frac{1}{y}\right)}$.
So, the integrating factor is $e^{\left(\frac{1}{y}\right)}$.