$y'=2-\frac{3}{x}y+\frac{2}{x^2}y^2$ (Riccati) $y'=2-\frac{3}{x}y+\frac{2}{x^2}y^2$ (Riccati)
(a) Find the solutions.
(b) $y(x_0)=y_0$, prove two cases:
$$0<y_0<x_0 \implies \text{solution's domain is} [x_0,\infty) $$
$$0<x_0<y_0 \implies \text{solution's domain is} [x_0,x_0+\alpha) , \alpha \in \mathbb {R}.$$
I will be grateful for help in $(b)$.
My solution for (a):
$y_1(x)=x$
Denote $y=x+z$
$$(x+z)'=2-\frac{3}{x}(x+z)+\frac{2}{x^2}(x+z)^2$$
$$1+z'=-1-\frac{3z}{x}+\frac{2}{x^2}(x^2+2xz+z^2)$$
$$z'=\frac{z}{x}+\frac{2z^2}{x^2}$$
This is a Bernoulli differential Equation.
Denote $u=z^{-1} \implies -\frac{z'}{z^2} \implies z'=-u'z^2$
Then,
$$u'=\frac{u}{x}-\frac{2}{x^2}$$
Integration factor is $\mu=x$
$$\int(xu)'=\int-\frac{2}{x} \implies xu=-2\ln|x|+c \implies u=\frac{-2\ln|x|+c}{x} \implies z=\frac{x}{-2\ln|x|+c}$$
The solution is $y=x+\frac{x}{-2\ln|x|+c}$
Is my solution correct ?
Thanks !
 A: $$z'=\frac{z}{x}+\frac{2z^2}{x^2}$$
$$\dfrac {z'}{z^2}=\frac{1}{zx}+\frac{2}{x^2}$$
$$-\left(\dfrac {1}{z}\right)'=\frac{1}{zx}+\frac{2}{x^2}$$
Looks like there is a little sign mistake. But all what you did looks really good.
$$u'=\color{red}{-\frac{u}{x}}-\frac{2}{x^2}$$
$$(ux)'=-\frac{2}{x}$$
A: An alternative approach parametrizes solutions as $y=2\frac{u}{u'}$ to get
$$
y''=2-2\frac{u''u}{u'^2}=2-\frac6{x}\frac{u}{u'}+\frac{8}{x^2}\frac{u^2}{u'^2}
$$
This simplifies to
$$
0=u''-\frac{3}{x}u'+\frac{4}{x^2}u
$$
This now is an Euler-Cauchy equation, with correspondingly simple solutions.
A: The equation is homogeneous, so you can use $v=\frac{y}{x}$ so that $y=vx$ and $y'=v+xv'$. Then
$$v+xv'= 2 -3v +2v^2$$
$$xv' = 2(1-2v +v^2)$$
Then separating variables
$$\int \frac{dv}{(v-1)^2} = \int\frac{2dx}{x}$$
$$-\frac{1}{v-1} = \ln(x^2) -C$$
$$ v-1 = \frac{1}{C-\ln(x^2)}$$
$$ v = 1+\frac{1}{C-\ln(x^2)}$$
Then
$$y= x +\frac{x}{C-\ln(x^2)}$$
So your solution is correct and confirmed by another method.
