a.
$\displaystyle y'+\frac{2}{x}y=\frac{y^{3}}{x^{2}} $
b.
$\displaystyle 2y'-\frac{x}{y}=\frac{xy}{x^{2}-1}$
$\displaystyle \displaystyle \frac{dy}{dx} + \frac{2}{x}\,y = \frac{1}{x^2}\,y^3$.
This is a Bernoulli Equation. So make the substitution $\displaystyle \displaystyle v = y^{-2}$.
Then $\displaystyle \displaystyle y = v^{-\frac{1}{2}}$ and
$\displaystyle \displaystyle \frac{dy}{dx} = \frac{d}{dx}(v^{-\frac{1}{2}})$
$\displaystyle \displaystyle = \frac{d}{dv}(v^{-\frac{1}{2}})\,\frac{dv}{dx}$
$\displaystyle \displaystyle = -\frac{1}{2}v^{-\frac{3}{2}}\,\frac{dv}{dx}$.
Then the DE becomes
$\displaystyle \displaystyle \frac{dy}{dx} + \frac{2}{x}\,y = \frac{1}{x^2}\,y^3$
$\displaystyle \displaystyle -\frac{1}{2}v^{-\frac{3}{2}}\,\frac{dv}{dx} + \frac{2}{x}\,v^{-\frac{1}{2}} = \frac{1}{x^2}\,v^{-\frac{3}{2}}$
$\displaystyle \displaystyle \frac{dv}{dx} - \frac{4}{x}\,v = -\frac{2}{x^2}$.
This is now first order linear, so use the integrating factor method.
The integrating factor is
$\displaystyle \displaystyle e^{\int{-\frac{4}{x}\,dx}} = e^{-4\ln{x}} = e^{\ln{(x^{-4})}} = x^{-4}$.
Multiplying through by the integrating factor gives
$\displaystyle \displaystyle x^{-4}\,\frac{dv}{dx} - 4x^{-5}\,v = -2x^{-6}$
$\displaystyle \displaystyle \frac{d}{dx}(x^{-4}\,v) = -2x^{-6}$
$\displaystyle \displaystyle x^{-4}\,v = \int{-2x^{-6}\,dx}$
$\displaystyle \displaystyle x^{-4}\,v = \frac{2}{5}x^{-5}$
$\displaystyle \displaystyle v = \frac{2}{5}x^{-1}$.
Since $\displaystyle \displaystyle v = y^{-2}$
$\displaystyle \displaystyle \frac{2}{5}x^{-1} = y^{-2}$
$\displaystyle \displaystyle \left(\frac{2}{5}x^{-1}\right)^{-\frac{1}{2}} = y$
$\displaystyle \displaystyle y = \left(\frac{5}{2}\right)^{\frac{1}{2}}x^{\frac{1}{ 2}}$
$\displaystyle \displaystyle y = \frac{\sqrt{10x}}{2}$