Step-by-Step Solution

The given differential equation is \((x+2y^{2})\frac{dy}{dx}=y\).

We can rewrite this as \(\frac{dx}{dy} = \frac{x+2y^2}{y}\).

Further simplifying, we get \(\frac{dx}{dy} = \frac{x}{y} + 2y\).

Rearranging the terms, we have \(\frac{dx}{dy} - \frac{1}{y}x = 2y\).

This is a linear differential equation of the form \(\frac{dx}{dy} + P(y)x = Q(y)\), where \(P(y) = -\frac{1}{y}\) and \(Q(y) = 2y\).

The integrating factor (IF) is given by \(e^{\int P(y) dy}\).

So, \(IF = e^{\int -\frac{1}{y} dy} = e^{-\int \frac{1}{y} dy} = e^{-\ln|y|} = e^{\ln|y^{-1}|} = \frac{1}{y}\) (since y > 0).