# Non-exact 1st Order Differential Equation? Kindly solve this Non-exact 1st Order Differential Equation by integrating Factor Method. (x+2)sinydx+x cosydy=0

##### 1 Answer
Jan 8, 2018

$y = \arcsin \left({e}^{C - x} / {x}^{2}\right)$

#### Explanation:

We have:

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

Which we can write in standard form as:

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

We note that we do not require the use of an Integrating Factor as this is a separable First Order differential equation, thus we can collect terms and separate the variables to get

$\setminus \setminus \setminus \setminus \setminus \cos \frac{y}{\sin} y \setminus \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{x + 2}{x} = 0$

$\therefore \int \setminus \cot y \setminus \mathrm{dy} = - \int \setminus \frac{x + 2}{x} \setminus \mathrm{dx}$

And we integrate to get:

$\ln \sin y = - x - 2 \ln x + C$

Exponentiating we get:

$\sin y = {e}^{- x - 2 \ln x + C}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = {e}^{C - x} {e}^{- 2 \ln x}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = {e}^{C - x} {e}^{\ln \left(\frac{1}{x} ^ 2\right)}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = {e}^{C - x} \left(\frac{1}{x} ^ 2\right)$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = {e}^{C - x} / {x}^{2}$

And finally:

$y = \arcsin \left({e}^{C - x} / {x}^{2}\right)$