# How do I solve this differential equation?  (x^2 + y^2) dx + xydy = 0

Feb 8, 2018

The GS is:

${y}^{2} = \frac{A - {x}^{4}}{2 {x}^{2}}$

Or, alternatively:

$y = \pm \frac{\sqrt{A - {x}^{4}}}{\sqrt{2} x}$

#### Explanation:

We have:

$\left({x}^{2} + {y}^{2}\right) \setminus \mathrm{dx} + x y \setminus \mathrm{dy} = 0$

Which we can write in standard form as:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{{x}^{2} + {y}^{2}}{x y}$ ..... [1]

Which is a non-separable First Order Ordinary Differential Equation. A suggestive substitution would be to perform a substitution of the form:

$y = x v \implies \frac{\mathrm{dy}}{\mathrm{dx}} = v + x v ' \setminus \setminus \setminus$ where $v = v \left(x\right)$

Then if we substitute into the DE [1], we get

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

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

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

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

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

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

Which has transformed the initial DE [1] into a separable, DE, so we can Manipulate further, and "separate the variables":

$\frac{1}{4} \setminus \int \setminus \frac{4 v}{1 + 2 {v}^{2}} \setminus \mathrm{dv} = - \setminus \int \setminus \frac{1}{x} \setminus \mathrm{dx}$

And we can now integrate to get:

$\frac{1}{4} \ln | 1 + 2 {v}^{2} | = - \ln | x | + C$

And, noting that $1 + 2 {v}^{2} > 0 \forall x \in \mathbb{R}$, and using the properties of logarithms, then

$\ln \left(1 + 2 {v}^{2}\right) = - 4 \ln | x | + 4 C$
$\therefore \ln \left(1 + 2 {v}^{2}\right) + 4 \ln | x | - \ln A = 0 \setminus \setminus \setminus$, say
$\therefore \ln \left(1 + 2 {v}^{2}\right) + \ln {x}^{4} - \ln A = 0$
$\therefore \ln \left({x}^{4} / A \left(1 + 2 {v}^{2}\right)\right) = 0$
$\therefore {x}^{4} / A \left(1 + 2 {v}^{2}\right) = {e}^{0}$

And restoring the substitution, we can now write:

$\therefore {x}^{4} / A \left(1 + 2 {\left(\frac{y}{x}\right)}^{2}\right) = 1$
$\therefore 1 + 2 {\left(\frac{y}{x}\right)}^{2} = \frac{A}{{x}^{4}}$
$\therefore 2 {\left(\frac{y}{x}\right)}^{2} = \frac{A}{{x}^{4}} - 1$
$\therefore 2 {y}^{2} / {x}^{2} = \frac{A - {x}^{4}}{x} ^ 4$

$\therefore {y}^{2} = \frac{A - {x}^{4}}{2 {x}^{2}}$

Or, alternatively:

$y = \pm \frac{\sqrt{A - {x}^{4}}}{\sqrt{2} x}$