`(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 `