Question

The general solution of the differential equation $(1+y^2)dx+(1+x^2)dy=0$ is

• A. $x-y=C(1-xy)$
• B. $x-y=C(1+xy)$
• C. $x+y=C(1-xy)$
• D. $x+y=C(1+xy)$

Answer 1

The correct option is C $x+y=C(1-xy)$

$(1+y^2)dx+(1+x^2)dy=0$
$\Rightarrow \frac{dx}{1+x^2}+\frac{dy}{1+y^2}=0$
On integrating, we get
$ta{n}^{-1}x+ta{n}^{-1}y=C$
$\Rightarrow \frac{x+y}{1-xy}=C$
$\Rightarrow x+y=C(1-xy)$ is the required solution.