Let z=y', then z'=y":
z'+6z=18x2-6x+3.
Let z=Ax2+Bx+C, then z'=2Ax+B and z'+6z=2Ax+B+6Ax2+6Bx+6C=18x2-6x+3.
Therefore, equating like powers of x:
6A=18, 2A+6B=-6, B+6C=3.
So A=3, B=-2, C=⅚⇒z=y'=3x2-2x+⅚.
CHECK z'+6z=6x-2+18x2-12x+5=18x2-6x+3.
Therefore y=x3-x2+⅚x+K where K is a constant.