# `dy/dx + y/x = 6x + 2` Solve the first-order differential equation Given  `dy/dx +y/x=6x+2`

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,

`dy/dx +y/x=6x+2--------(1)`

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

on comparing both we get,

`p(x) = 1/x and q(x)=6x+2`

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) *(6x+2)) dx +c)/e^(int 1/x dx)`

first we shall solve

`e^(int (1/x) dx)=e^ln(x) =|x|`      as we know `int (1/x)dx = ln(x)`

so then we get ` e^(int (1/x) dx) =x`

since x must be greater than 0, or else`ln(x)` is undefined.

Proceeding further with

y(x) =`((int e^(int (1/x) dx) *(6x+2)) dx +c)/e^(int 1/x dx)`

=` (int x *(6x+2) dx +c)/x`

= `(int (6x^2 +2x) dx+c)/x`

=` (6x^3/3 +2x^2/2 +c)/x`

= `(2x^3+x^2+c)/x`

So, ` y= (2x^3+x^2+c)/x`

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