# Question #59b3a

Sep 22, 2016

$y = \pm \sqrt{{x}^{2} + C x}$

#### Explanation:

Another approach

With equations like these, it is often good to check for exactness

An equation in this form:

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

....can be related to level surface $\setminus \Phi \left(x , y\right) = c o n s t$ such that $d \setminus \Phi = 0 = \setminus {\Phi}_{x} \mathrm{dx} + \setminus {\Phi}_{y} \mathrm{dy}$

So in this case

$\setminus {\Phi}_{x} = {x}^{2} + {y}^{2} , \setminus {\Phi}_{x y} = 2 y$

$\setminus {\Phi}_{y} = - 2 x y , \setminus {\Phi}_{y x} = - 2 y$

But because $\setminus {\Phi}_{x y} \ne \setminus {\Phi}_{y x}$, the equation is not exact

We can then re-arrange the equation to say:

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

This is a homogeneous equation, ie

$\frac{\mathrm{dy}}{\mathrm{dx}} = f \left(x , y\right) = f \left(k x , k y\right)$ for any constant $k$

in this case we make a substitution $v \left(x\right) = \frac{y}{x} , y = v x$, so that $y ' = v ' x + v$ and the new equation becomes separable

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

$\implies v ' x + v = \frac{1}{2 v} + \frac{v}{2}$

$v ' x = \frac{1 - {v}^{2}}{2 v}$

$\frac{2 v}{1 - {v}^{2}} v ' = \frac{1}{x}$

We integrate wrt x:
$\int \left(\frac{2 v}{1 - {v}^{2}} v ' = \frac{1}{x}\right) \mathrm{dx}$

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

Or

$- \ln \left\mid 1 - {v}^{2} \right\mid = \ln \left\mid x \right\mid + C \setminus \setminus \setminus \left[= \ln C \left\mid x \right\mid\right]$ where, throughout, C is a generic constant

$\implies - \frac{1}{\left\mid 1 - {v}^{2} \right\mid} = C \left\mid x \right\mid$

For suitable v and x

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

$1 - {\left(\frac{y}{x}\right)}^{2} = \frac{1}{C x}$

${y}^{2} = {x}^{2} + C x$

$y = \pm \sqrt{{x}^{2} + C x}$