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

Given (2y-e^x)dx + xdy = 0
=> 2y-e^x+xdy/dx =0
=> 2y/x -e^x/x +dy/dx=0
=> 2y/x +y'=e^x/x
=> y'+(2/x)y=(e^x)/x
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'+(2/x)y=(e^x)/x--------(1)
y'+p(x)y=q(x)---------(2)
on comparing both we get,
p(x) = (2/x) and q(x)=(e^x)/x
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) *((e^x)/x)) dx +c)/e^(int (2/x) dx)
first we shall solve
e^(int (2/x) dx)=e^(ln(x^2))=x^2     
So proceeding further, we get
y(x) =((int e^(int (2/x) dx) *((e^x)/x)) dx +c)/ e^(int (2/x) dx)
 =(int (x^2 *e^x/x dx) +c)/x^2
=(int xe^xdx +c)/x^2
=(xe^x -e^x +c)/x^2
=(e^x (x- 1) +c)/x^2
y(x) =(e^x (x- 1) +c)/x^2