# Solution of differential equation : $\large\frac{dx}{dy}=\frac{y+x-2}{y+x+4}$

$(a)\;(y+x+4)^2-12x=c \\ (b)\;(y+x+4)^2+12x=c \\ (c)\;(y+x-2)^2+12x=c \\ (d)\;(y+x-2)^2-12x=c$

$y+x+4=t$
$\large\frac{dy}{dx}$$+1=\large\frac{dt}{dx} \large\frac{dt}{dx}=\frac{t-6}{t}$$-1$
$\qquad= \large\frac{t-6-t}{t}$
$\int t dt=- \int 6 dx$