`xdy = (x + y + 2)dx , y(1) = 10` Find the particular solution of the differential equation that satisfies the initial condition

Given ` xdy = (x + y + 2)dx `

=> `xdy/dx = (x + y + 2)`

=> `y'=1+y/x+2/x`

=> `y'-y/x = 1+2/x`

=> `y'-y/x =(x+2)/x`

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'-y/x =(x+2)/x--------(1)`

`y'+p(x)y=q(x)---------(2)`

on comparing both we get,

`p(x) = -1/x and q(x)=(x+2)/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 (-1/x) dx) *((x+2)/x)) dx +c)/e^(int (-1/x) dx)`

first we shall solve

`e^(int (-1/x) dx)=e^(ln(1/x)) = (1/x)`

so

proceeding further, we get

y(x) =`((int e^(int (-1/x) dx) *((x+2)/x)) dx +c)/e^(int (-1/x) dx)`

=`((int (1/x) *((x+2)/x)) dx +c)/(1/x)`

=`(int (1/x) *((1+(2)/x)) dx +c)/(1/x)`

=` x(ln(x)-2/x +c)`

so now let us find the particular solution of differential equation at y(1)=10

`y(1) = 1(ln(1)-2/1 +c)`

=> `10 = 0-2+c`

`c=12`

`y(x) =x(ln(x)-2/x +12)` is the solution