Question

What is the general solution of the differential equation `(x+y)dx-xdy = 0`?

Answer 1

`y = xln|x| + Ax`

Explanation:

We can write the equation `(x+y)dx-xdy = 0` as:

`" " xdy = (x+y)dx`
`:. dy/dx = (x+y)/x`
`:. dy/dx = 1+y/x`

`:. dy/dx - 1/x y = 1`

This is a First Order DE of the form:

`y'(x) + P(x)y = Q(x)`

Which we know how to solve using an Integrating Factor given by:

`IF = e^(int P(x) \ dx)`

And so our Integrating Factor is:

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

If we multiply by this Integrating Factor we will (by its very design) have the perfect differential of a product:

`" " dy/dx - 1/x y = 1`

`:. 1/xdy/dx - 1/x^2 y = 1/x`
`:. \ \ \ \ \ d/dx(1/xy) = 1/x`

Which is now a separable DE, and we can separate the variables to get:

`" "y/x = int \ 1/x \ dx`
`:. y/x = ln|x| + A` (where `A` is an arbitrary constant)
`:. \ \ y = xln|x| + Ax`