Solutions of Riccati equation $y'+0.5y^2+0.5x^{-2}=0$ I tried to solve Riccati equation $$y'+0.5y^2+0.5x^{-2}=0$$ using the software Mathematica, and it offered me only one solution, $1/u$. I found it weird, because usually due to constant, we have a whole family of solutions. I tried to solve it by myself, using the supstitution $y=1/z$ because I found it in the literature, but I got stuck. I get
$$z'=-0.5-0.5(x^2z^2)^{-1}$$
and dont know what to do next. Can someone give me a push?
 A: In order to solve the Riccati equation
$$y'(x)+\frac{y(x)^2}{2}+\frac{1}{2x^2}=0$$ 
let $z(x)=xy(x)$, then the ODE becomes
$$\frac{z'(x)}{x}-\frac{z(x)}{x^2}+\frac{1}{2}\frac{z(x)^2}{x^2}+\frac{1}{2x^2}=0$$
that is
$$xz'(x)=z(x)-\frac{z(x)^2}{2}-\frac{1}{2}$$
where we can separate the variables.
One solution the stationary solution $z(x)=1$. If $z(x)\not=1$ then
$$-\frac{z'(x)}{(z(x)-1)^2}=\frac{1}{2x}.$$
and by integration we get
$$\frac{1}{z(x)-1}=\frac{\ln|x|}{2}+C.$$
where $C$ is an arbitrary constant.
Therefore the solutions are $y(x)=1/x$ and, after letting $c=2C$,
$$y(x)=\frac{1}{x}\left(1+\frac{2}{\ln|x|+c}\right)=\frac{\ln|x|+c+2}{x\ln|x|+cx}.$$
We can recover the special solution $y(x)=1/x$ from the above formula by letting $C\to \infty$.
A: Using the other solution method for Riccati equations, set $y=2\frac{u'}{u}$, then
$$
0=2\frac{u''}{u}-2\frac{u'^2}{u^2}+\frac12\cdot 4\frac{u'^2}{u^2}+\frac12x^{-2}
=2\frac{u''}{u}+\frac12x^{-2}
\\~\\
\implies 4x^2u''+u=0
$$
This last one is an Euler-Cauchy equation with characteristic polynomial $0=4m(m-1)+1=(2m-1)^2$ giving the double root $m=\frac12$. Thus the general solution is
$$
u=A\sqrt{x}+B\sqrt{x}\ln|x|
\implies 
y=2\frac{\frac{A}{2\sqrt{x}}+\frac{B\ln|x|}{2\sqrt{x}}+\frac{B\sqrt{x}}{x}}{A\sqrt{x}+B\sqrt{x}\ln|x|}
=\frac{A+B(\ln|x|+2)}{Ax+Bx\ln|x|}.
$$
This gives identical solutions for pairs $(A,B)$ having the same ratio. For $B=0$ you get the solution $y=\frac1x$.
