# How do you solve the differential dy/dx=4x+(4x)/sqrt(16-x^2)?

Sep 27, 2017

$y = 2 {x}^{2} - 4 \sqrt{16 - {x}^{2}} + C$

#### Explanation:

Separate variables:

$\mathrm{dy} = \left(4 x + \frac{4 x}{\sqrt{16 - {x}^{2}}}\right) \mathrm{dx}$

Integrate. On the right side, a $u$ substitution can be helpful if the integral cannot be "eyeballed":

$\int \mathrm{dy} = \int \left(4 x + \frac{4 x}{\sqrt{16 - {x}^{2}}}\right) \mathrm{dx}$
$\int \mathrm{dy} = \int 4 x \mathrm{dx} + \int \frac{4 x}{\sqrt{16 - {x}^{2}}} \mathrm{dx}$
$\int \mathrm{dy} = \int 4 x \mathrm{dx} + \int \left(4 x\right) {\left(16 - {x}^{2}\right)}^{- \frac{1}{2}} \mathrm{dx}$

If we let $u = 16 - {x}^{2}$, then $\mathrm{du} = - 2 x \mathrm{dx}$, or conversely:

$2 x \mathrm{dx} = - \mathrm{du} \implies 4 x \mathrm{dx} = - 2 \mathrm{du}$

Substituting and integrating:

$\int \mathrm{dy} = \int 4 x \mathrm{dx} + \int {u}^{- \frac{1}{2}} \left(- 2 \mathrm{du}\right)$
$\int \mathrm{dy} = \int 4 x \mathrm{dx} - 2 \int {u}^{- \frac{1}{2}} \mathrm{du}$
$y = 2 {x}^{2} - 4 {u}^{\frac{1}{2}} + C$

Lastly, substitute back:

$y = 2 {x}^{2} - 4 {\left(16 - {x}^{2}\right)}^{\frac{1}{2}} + C = 2 {x}^{2} - 4 \sqrt{16 - {x}^{2}} + C$