3(y-4x^2)dx + xdy = 0 Solve the first-order differential equation by any appropriate method

Given 3(y-4x^2)dx + xdy = 0
=> 3y - 12x^2 +xdy/dx=0
=> ( 3y - 12x^2)/x +dy/dx=0
=> 3y/x - 12x +dy/dx=0
=> y'+(3/x)y=12x
when the first order linear ordinary differential equation has the form of
y'+p(x)y=q(x)
then the general solution is ,
y(x)=((int e^(int p(x) dx) *q(x)) dx +c)/ e^(int p(x) dx)
so,
y'+(3/x)y=12x--------(1)
y'+p(x)y=q(x)---------(2)
on comparing both we get,
p(x) = (3/x) and q(x)=12x
so on solving with the above general solution we get:
y(x)=((int e^(int p(x) dx) *q(x)) dx +c)/ e^(int p(x) dx)
=(int e^(int (3/x) dx) *((12x)) dx +c)/e^(int (3/x) dx)
first we shall solve
e^(int (3/x) dx)=e^(ln(x^3))=x^3     
so proceeding further, we get
y(x) =(int e^(int (3/x) dx) *((12x)) dx +c)/e^(int (3/x) dx)
=(int x^3 *((12x)) dx +c)/x^3
=(int 12x^4 dx +c)/x^3
 =(12x^5/5 +c ) /x^3
so y=(12x^5/5 +c )/x^3

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