The general solution of the differential equation $(1+y)dx-2xdy =0$ is :

$x=C(1+y)^2 $, where C is constant of integration

The correct answer is Option (1) → $x=C(1+y)^2 $, where C is constant of integration

$(1+y)dx=2xdy$

so $\int\frac{dx}{2x}=\int\frac{dy}{1+y}$

$⇒\frac{\log x}{2}=\log(1+y)+\log c$

so $\log x=2\log(1+y)+\log c'$

$x=c'(1+y)^2$