tricky differential equation I need help with solving this equation:
$$ y'= \frac{y}{3x-y^2} $$
is there any "trick" I don't see? I don't see any first order equation I know of here. 
any hints?
edit:
the question was in the first order linear equation section. can it be arranged to this form?
$$ y'= p(x)y = q(x) $$
 A: $$ \frac{dy}{dx}= \frac{y}{3x-y^2} $$
HINT :
$$ \frac{dx}{dy}=3\frac{x}{y}-y $$
Considering the function $x(y)$, this is a first order linear ODE easy to solve.
To convince you, let $x=Y$ and $y=X$ :
$$Y'=\frac{dY}{dX}=3\frac{Y}{X}-X $$
I suppose that you can continue.
NOTE :
In the first edition of the question, the equation was : $\frac{dy}{dx}= \frac{y}{3x^2-y^2}$ which is also solvable, but more complicated. With the same trick: considering the inverse function $x(y)$, the ODE becomes a Riccati ODE. The classical method to solve this kind of ODE leads to the solution involving Bessel functions. This is an implicit form of solution because, in this case, the inverse function $y(x)$ cannot be expressed on a closed form.
A: $$ \frac{dy}{dx}= \frac{y}{3x-y^2} $$
It can be rearranged as $$ \frac{dx}{dy} = \frac{3x-y^2}{y} $$
Which means $$ \frac{dx}{dy} = \frac{3x}{y} - y $$
It is a linear equation. You can solve it accordingly.
A: this equation can be exact the integration factor is:
$$ \mu(x)= \frac{1}{y^4} $$ 
so after multiplying you get:
$$ \frac{3x-y^2}{y^4}dy - \frac{1}{y^3}dx $$ 
from here it can be solved like an ordinary exact equation.
