# What is the general solution of the differential equation  dy/dx = (x+y)/x ?

Sep 19, 2017

$y = x \ln | x | + C x$

#### Explanation:

We have:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x + y}{x}$ ..... [A]

If we use the suggested substitution, $y = v x$ then differentiating wrt $x$ and applying the product rule we have:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \left(v\right) \left(\frac{d}{\mathrm{dx}} x\right) + \left(\frac{d}{\mathrm{dx}} v\right) \left(x\right)$
$\setminus \setminus \setminus \setminus \setminus = \left(v\right) \left(1\right) + \frac{\mathrm{dv}}{\mathrm{dx}} x$
$\setminus \setminus \setminus \setminus \setminus = v + x \frac{\mathrm{dv}}{\mathrm{dx}}$

Substituting this result into the initial differential equation [A] we get:

$v + x \frac{\mathrm{dv}}{\mathrm{dx}} = \frac{x + v x}{x}$
$\therefore v + x \frac{\mathrm{dv}}{\mathrm{dx}} = 1 + v$
$\therefore x \frac{\mathrm{dv}}{\mathrm{dx}} = 1$
$\therefore \frac{\mathrm{dv}}{\mathrm{dx}} = \frac{1}{x}$

Which has reduced the equation to a trivial First Order separable equation, which we can "separate the variables" to get:

$\int \setminus \mathrm{dv} = \int \setminus \frac{1}{x} \setminus \mathrm{dx}$

And if we integrate we get:

$v = \ln | x | + C$

Restoring the earlier substitution, we get:

$\frac{y}{x} = \ln | x | + C$

$y = x \ln | x | + C x$