Reducible Second Order Equation

Use the Reducible Second Order Equation to find a general solution (Assume x, y, and/or y' positive where helpful)

(x^2)y'' + 3xy'=2

This is what I have but came up with the wrong answer

can someone explain what I need to do (Thanks)

y is missing

Let p= y'= dy/dx then let p'=y''=dp/dx

(x^2) dp/dx +3xp= 2

dp/dx + (3/x)p = (2/x^2)

Now i used the linear first-order equation and found

Dx + (3x^2)p = 2x

I then got

(3x^2)p = x^2 + C

Dividing by (3x^2) I got

p= (1/3) + C/(3x^2)

This is not the right answer, the right answer is ln(x) + Ax^-2 + B

I do not have a clue how they got this can someone please help me???

Thanks