#  y' = x/y - x/(1+y)  ? by using separation of variable

Apr 28, 2018

$3 {y}^{2} + 2 {y}^{3} = 3 {x}^{2} + A$

#### Explanation:

We have:

$y ' = \frac{x}{y} - \frac{x}{1 + y}$

Which we can write as:

$y ' = \frac{x \left(1 + y\right) - x y}{y \left(1 + y\right)}$

$\setminus \setminus \setminus = \frac{x + x y - x y}{y \left(1 + y\right)}$

$\setminus \setminus \setminus = \frac{x}{y \left(1 + y\right)}$

$\therefore y + {y}^{2} \setminus \frac{\mathrm{dy}}{\mathrm{dx}} = x$

Which is now a seperal DE, so we can "separate the variables" to get:

$\int \setminus y + {y}^{2} \setminus \mathrm{dy} = \int \setminus x \setminus \mathrm{dx}$

Which we can now readily integrate:

$\frac{1}{2} {y}^{2} + \frac{1}{3} {y}^{3} = \frac{1}{2} {x}^{2} + C$

$\therefore 3 {y}^{2} + 2 {y}^{3} = 3 {x}^{2} + A$

Which is the Implicit General Solution