Solve $x y' = 3 y - 6 x, \qquad y(1) = -6$.

(a) Identify the integrating factor, $\alpha(x)$.
$\alpha(x)$ =

(b) Find the general solution.
$y(x) =$
Note: Use $C$ for the arbitrary constant.

(c) Solve the initial value problem $y(1) = -6$.
$y(x) =$