# What is the general solution of the differential equation (x^2 + y^2) \ dx - xy \ dy = 0?

Dec 4, 2017

${y}^{2} = {x}^{2} \left(2 \ln x + c\right)$

#### Explanation:

We can rewrite this Ordinary Differential Equation in differential form:

$\left({x}^{2} + {y}^{2}\right) \setminus \mathrm{dx} - x y \setminus \mathrm{dy} = 0$ ..... [A]

as follows:

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

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

Leading to a suggestion of a substitution of the form:

$u = \frac{y}{x} \iff y = u x$

And differentiating wrt $x$ whilst applying the product rule:

$\frac{\mathrm{dy}}{\mathrm{dx}} = u + x \frac{\mathrm{du}}{\mathrm{dx}}$

Substituting into the DE [ B] we have

$u + x \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{u} + u$

$\therefore x \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{u}$

This is now a First Order Separable ODE, so we can rearrange and "separate the variables" as:

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

This is now trivial to integrate, and doing so gives us:

$\setminus \setminus \setminus {u}^{2} / 2 = \ln x + {c}_{1}$

$\therefore {u}^{2} = 2 \ln x + c$

And we restore the earlier substitution to get:

${\left(\frac{y}{x}\right)}^{2} = 2 \ln x + c$

$\therefore {y}^{2} / {x}^{2} = 2 \ln x + c$

$\therefore {y}^{2} = {x}^{2} \left(2 \ln x + c\right)$