dy/dx + 2y/x = 3x-5 Solve the first-order differential equation

Given dy/dx+2y/x=3x-5
y'+2y/x=3x-5
when the first order linear ordinary Differentian 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'+2y/x=3x-5--------(1)
y'+p(x)y=q(x)---------(2)
on comparing both we get,
p(x) = 2/x and q(x)=3x-5
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 2/x dx) *(3x-5)) dx +c)/e^(int 2/x dx)
first we shall solve
e^(int 2/x dx)=e^(2ln(x)) =x^2     
so
proceeding further, we get
y(x) =((int e^(int 2/x dx) *(3x-5)) dx +c)/e^(int 2/x dx)
=((int x^2 *(3x-5)) dx +c)/x^2
=((int (3x^3 -5x^2) ) dx +c)/x^2
= (3x^4 /4 -5x^3/3+c)/x^2
so y(x)=(3x^4 /4 -5x^3/3+c)/(x^2 )