`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 )`