# What is the general solution of the differential equation  dy/dx=(y^3 - yx^2) / (x^3 + y^2x) ?

## (portions of this question have been edited or deleted!)

Sep 22, 2017

${y}^{2} / \left(2 {x}^{2}\right) + \ln | x y | + C = 0$

#### Explanation:

Presumably you require the solution, of the Differential equation rather than the equation, which is of course in the question.

We have:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{y}^{3} - y {x}^{2}}{{x}^{3} + {y}^{2} x}$ ..... [A]

Note, that we cannot isolate terms in $x$ and $y$ alone so let us attempt some manipulation in an attempt to simplify the equation:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y \left({y}^{2} - {x}^{2}\right)}{x \left({y}^{2} + {x}^{2}\right)}$
$\setminus \setminus \setminus \setminus \setminus = \left(\frac{y}{x}\right) \frac{{y}^{2} - {x}^{2}}{{y}^{2} + {x}^{2}} \times \frac{\frac{1}{x} ^ 2}{\frac{1}{x} ^ 2}$
$\setminus \setminus \setminus \setminus \setminus = \left(\frac{y}{x}\right) \frac{{\left(\frac{y}{x}\right)}^{2} - 1}{{\left(\frac{y}{x}\right)}^{2} + 1}$ ..... [B]

So Let us try a substitution, Let:

$v = \frac{y}{x} \implies y = v x \implies \frac{\mathrm{dy}}{\mathrm{dx}} = v + x \frac{\mathrm{dv}}{\mathrm{dx}}$

And substituting into the above DE [B], to eliminate $y$:

$v + x \frac{\mathrm{dv}}{\mathrm{dx}} = \left(v\right) \frac{{v}^{2} - 1}{{v}^{2} + 1}$

$\therefore x \frac{\mathrm{dv}}{\mathrm{dx}} = v \left\{\frac{{v}^{2} - 1}{{v}^{2} + 1} - 1\right\}$

$\therefore x \frac{\mathrm{dv}}{\mathrm{dx}} = v \left\{\frac{\left({v}^{2} - 1\right) - \left({v}^{2} + 1\right)}{{v}^{2} + 1}\right\}$

$\therefore x \frac{\mathrm{dv}}{\mathrm{dx}} = \frac{- 2 v}{{v}^{2} + 1}$

$\therefore \frac{{v}^{2} + 1}{v} \frac{\mathrm{dv}}{\mathrm{dx}} = - \frac{2}{x}$

Which has indeed helped as it has reduced the initial equation [A] to a separable equation, so we can now "seperate the variables" to get:

$\int \setminus \frac{{v}^{2} + 1}{v} \setminus \mathrm{dv} = \int \setminus - \frac{2}{x} \setminus \mathrm{dx}$

$\therefore \int \setminus v + \frac{1}{v} \setminus \mathrm{dv} = \int \setminus - \frac{2}{x} \setminus \mathrm{dx}$

And this is trivial to integrate to get:

$\frac{1}{2} {v}^{2} + \ln | v | = - 2 \ln | x | - C$

And restoring the substitution we get:

$\frac{1}{2} {\left(\frac{y}{x}\right)}^{2} + \ln | \frac{y}{x} | = - 2 \ln | x | - C$
$\therefore {y}^{2} / \left(2 {x}^{2}\right) + \ln | y | - \ln | x | = - 2 \ln | x | - C$
$\therefore {y}^{2} / \left(2 {x}^{2}\right) + \ln | y | + \ln | x | + C = 0$
$\therefore {y}^{2} / \left(2 {x}^{2}\right) + \ln | x y | + C = 0$ QED