particular solution guessing of a Riccati equation 
I can construct the general solution of a Riccati equation if a particular solution is known. Is there any guessing trick do exist for example:
  $$\text{For }y'+y^2=\frac{2}{x^2} \text{ seek for a P.S. in the form }y=\frac{c}{x}$$
$$\text{For }x^3y'+x^2y-y^2=2x^4 \text{ seek for a P.S. in the form }y=cx^2$$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\vdots$

Why seeking for a particular solution$(\text{P.S.})$ in that form$?$ And how the knew that will be work$?$ I really appreciated any kind of help.
 A: Most textbook and course exercises will resolve into a usefull form when substituting $y=\frac{u'}{au}$ where $a$ is the coefficient of $y^2$ in $y'+ay^2=...$. Note that in advanced exercises and more practical applications, this will give a general second order linear ODE without a "nice" solution. In some cases the original equation will have poles, while the solution for $u$ remains bounded, the poles of $u$ correspond to roots of $u$ which might be easier to handle numerically.

For example, your first equation then transforms using $y=\frac{u'}u$ to $$x^2u''-2u=0$$ which is a Cauchy-Euler equation with basis solutions $x^m$ where $0=m(m-1)-2=(m-2)(m+1)$, so that $u=Ax^2+Bx^{-1}$.
In the second equation, the recipe gives $y=-\frac{x^3u'}{u}$ so that
$$
2x^4=-x^3\frac{x^3u''+3x^2u'}{u}+\color{blue}{x^3\frac{x^3u'^2}{u^2}}-x^2\frac{x^3u'}u\color{blue}{-\frac{x^6u'^2}{u^2}}
\\
0=x^2u''+4xu'+2u
$$
which is again Euler-Cauchy with $0=m(m-1)+4m+2=(m+2)(m+1)$, thus $u=Ax^{-1}+Bx^{-2}$.
A: $$y'+y^2=\frac{2}{x^2} $$
You can try a solution like $y=ax^m$ then:
$$amx^{m-1}+a^2x^{2m}=2x^{-2}$$
We need to have all the eponents be equals so
$$m-1=2m \implies m=-1$$
You have that:
$$-a+a^2=2 \implies a=-1,a=2$$
It's the same trick for the second equation try $ax^m$
These equations you are given are really particular ones. And they are hard to solve if you only change the exponent on the right side. 
