An "integrating factor" for \(\displaystyle \frac{dy}{dx}+ f(x)y\) is a function, u(x), such that multiplying \(\displaystyle \frac{dy}{dx}+ f(x)y\) by it give an "exact differential".
That is, \(\displaystyle u(x)\frac{dy}{dx}+ u(x)f(x)y= \frac{d(u(x)y)}{dx}\).
Here f(x)= -2/x so we want \(\displaystyle \frac{d(u(x)y}{dx}= u(x)\frac{dy}{dx}+ \frac{-2u(x)}{x}y\).
Since, by the "product rule", \(\displaystyle \frac{d(u(x)y}{dx}= u(x)\frac{dy}{dx}+ \frac{du(x)}{dx}y\), we must have \(\displaystyle u(x)\frac{dy}{dx}+ \frac{du(x)}{dx}y= u(x)\frac{dy}{dx}+ \frac{-2u(x)y}{x}\). So we must have \(\displaystyle \frac{du}{dx}= \frac{-2u(x)}{x}\).
That's a "separable differential equation". We can write
\(\displaystyle \frac{du}{u}= \frac{-2dx}{x}\).
Integrating both sides, \(\displaystyle ln(u)= -2ln(x)+ C= ln(x^{-2})+ C\).
Taking the exponential of both sides, \(\displaystyle u(x)= C'x^{-2}= \frac{C'}{x^2}\) where \(\displaystyle C'= e^C\).
Since we only need one solution we can take C'= 1, which gives "E) \(\displaystyle \frac{1}{x^2}\)".
