Solve differential equation: $y' = \frac{y^2}{x^3} + 2\frac{y}{x} - x$ I need to solve this differential equation: $y' = \frac{y^2}{x^3} + 2\frac{y}{x} - x$.
My attempt
I found that $y = x^2$ is a solution. Then I tried to put $y = x^2f(x)$, and solved this way:
$$2xf(x) + x^2f'(x) = xf^2(x) + 2xf(x) - x \implies x^2f'(x) = xf^2(x) - x \implies$$
$$xf'(x) = f^2(x) - 1 \implies \frac{df(x)}{f^2(x) - 1} = \frac{dx}{x} \implies$$
$$\frac{1}{2}\ln\left|\frac{f(x) - 1}{f(x) + 1}\right| = \ln|x| + C_* \implies \frac{f(x) - 1}{f(x) + 1} = Cx^2 \implies$$
$$f(x) = \frac{1 + Cx^2}{1 - Cx^2}$$
And we lost a solution $f(x) = -1$. So finally we have
$$y = x^2\frac{1 + Cx^2}{1 - Cx^2}, y = -x^2$$
Now I have 3 questions:
$\quad 1)$ Is my solution correct? I'm not sure that all solutions were found.
$\quad 2)$ When can we use particular solution to find all other solutions? I mean doing something like $y = g(x)h(x)$, where $g(x)$ is a particular solution.
$\quad 3)$ Is there an easier method to solve this equation?

For question 3
$$y' = \frac{y^2}{x^3} + 2\frac{y}{x} - x$$
$$y'- 2\frac{y}{x}  = \frac{y^2}{x^3} - x$$
$$\left ( \dfrac y {x^2} \right )'= \frac{y^2}{x^5}-\frac{1}{x} $$
It's separable.
$$\left ( \dfrac y {x^2} \right )'=\dfrac 1 x \left ( \frac{y^2}{x^4}-1 \right) $$
$$\dfrac {du}{u^2-1}=\dfrac {dx}{x}$$
Where $ u=\dfrac y {x^2}$
Integrate.
I got this
$$y(x)=x^2\dfrac {1+Kx^2}{1-Kx^2}$$
For $K=0$ you have the particular solution you find by inspection :$y=x^2$
