linear in x

Problem 04
$(y + 1) \, dx + (4x - y) \, dy = 0$
 

Solution 04
$(y + 1) \, dx + (4x - y) \, dy = 0$

$\dfrac{(y + 1) \, dx}{(y + 1) \, dy} + \dfrac{(4x - y) \, dy}{(y + 1) \, dy} = 0$

$\dfrac{dx}{dy} + \dfrac{4x - y}{y + 1} = 0$

$\dfrac{dx}{dy} + \dfrac{4x}{y + 1} - \dfrac{y}{y + 1} = 0$

$\dfrac{dx}{dy} + \left( \dfrac{4}{y + 1} \right)x = \dfrac{y}{y + 1}$       → linear in x
 

Hence,
$P = \dfrac{4}{y + 1}$

$Q = \dfrac{y}{y + 1}$
 

Integrating factor: