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

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Answer 1 · 2016-12-30T15:20:28

$y = xln|x| + Ax$

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)
$:. \ \ y = xln|x| + Ax$

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