Study of a Cauchy Problem. I'm in trouble with this problem. The equation is:

$$y'(x)=\frac{y(x)}{x-1}+x(y(x))^2$$

With the initial condition $y(0)=a$
With "a" a real number.
I have to discuss local existence and uniqueness of solutions and find explicitly the solution, depending on a.
I think that it would be easy to prove local existance of solutions. I really don't know how I have to start to find explicitely the solutions. I've tried by separation of variables with no good results. Could someone give me a hint to start? Thank you very Much :).
 A: It's a Bernouilli's equation
$$y'(x)=\frac{y(x)}{x-1}+x(y(x))^2$$
Divide by $y^2 \ne 0$
$$\frac {y'}{y^2}-\frac 1{y(x-1)}=x$$
Substitute $z=\frac 1 {y}$
$$z'+\frac z {x-1}=-x$$
It's simple to solve now...
$$z'x-z'+z=-x(x-1)$$
$$(zx)'-z'=-x(x-1)$$
Just integrate
$$z(x-1)=-\int x(x-1)dx$$
$$z=-\frac 1 {(x-1)} \int x^2-xdx$$
$$\frac 1 y=-\frac 1 {(x-1)} ( \frac {x^3}3-\frac {x^2}2 +K)$$
$$ y(x)=\frac  {(1-x)} {( \frac {x^3}3-\frac {x^2}2 +K)}$$
$$ \boxed {y(x)=\frac  {6(1-x)} {( {2x^3}- {3x^2} +K)}}$$
I let you finish. find the constante K with the cauchy condition..$y(0)=a$
A: For the domain of $y(x)$ I have to study 
$2x^{3}-3x^{2}+\frac{6}{a}=0$ with $a\neq0$.
If $a=0$  , $y(x)=0$ is a solution.
For finding the other discontinuity points I have two idea: one is to try to scompose by Ruffini theorem the polinomia of grade 3 to obtain 3 solutions, the other is considering 
$y(x,a)=2x^{3}-3x^{2}+\frac{6}{a}$
and computing the $\nabla{f}(x,a)=0$ to find and eventually classify the discontinuity point.
I want to obtain the values of $a$ such that $y(x)$ is defined on $(-\infty,1)$. 
I think that "defined" means that for a certain values of $a$, $y(x)$ mustn't have discontinuity points on $(-\infty,1)$.
Thank you very much!!. .
