# How do you solve the differential dy/dx=(x-4)/sqrt(x^2-8x+1)?

Jan 10, 2017

$y = \sqrt{{x}^{2} - 8 x + 1} + C$

#### Explanation:

$d \frac{y}{\mathrm{dx}} = \frac{x - 4}{\sqrt{{x}^{2} - 8 x + 1}}$

Is a First Order separable DE which we can sole by integrating:

$\therefore y = \int \setminus \frac{x - 4}{\sqrt{{x}^{2} - 8 x + 1}} \setminus \mathrm{dx}$

Let $u = {x}^{2} - 8 x + 1 \implies \frac{\mathrm{du}}{\mathrm{dx}} = 2 x - 8 = 2 \left(x - 4\right)$

Substituting into the RHS integral we get:

$y = \int \setminus \frac{\frac{1}{2}}{\sqrt{u}} \setminus \mathrm{du}$
$\setminus \setminus = \frac{1}{2} \int \setminus {u}^{- \frac{1}{2}} \setminus \mathrm{du}$
$\setminus \setminus = \frac{1}{2} {u}^{\frac{1}{2}} / \left(\frac{1}{2}\right) + C$
$\setminus \setminus = \sqrt{u} + C$
$\setminus \setminus = \sqrt{{x}^{2} - 8 x + 1} + C$

which is the General Solution