:Problem

Solve the following equation:

$\displaystyle \dfrac{dy}{dx}+2 \, \dfrac{sin(x)}{x} \, y = x y^2$

:Solution

Its Bernoulli's Equation in y

Let $\displaystyle t=y^{-1} \implies \dfrac{dy}{dx}=\dfrac{-1}{t^2} \, \dfrac{dt}{dx}$

Multiply the equation by $\displaystyle y^{-2}$ :

$\displaystyle y^{-2} \, \dfrac{dy}{dx}+2 \, \dfrac{sin(x)}{x} \, y^{-1} = x $

Which is:

$\displaystyle -t^2 \, \dfrac{1}{t^2} \, \dfrac{dt}{dx} + 2 \, \dfrac{sin(x)}{x} \, t = x$

Or:

$\displaystyle \dfrac{dt}{dx} - 2 \dfrac{sin(x)}{x} \, t = -x$

Which is a Linear Equation in t, the problem here is I can't find the Integrating Factor

Since I can't find $\displaystyle \int \dfrac{sin(x)}{x} \, dx$

Maybe I did something wrong?