Question

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

Answer 1

$(1+x)(1+y^2)dx+(1+y)(1+x^2)dy=0$

$(1+x)(1+y^2)dx=-(1+y)(1+x{)}^{2}dy$

$\int \frac{(1+x)}{(1+x^2)}dx=-\int \frac{(1+y)}{(1+y^2)}dy$

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

${\tan }^{-1}x+{\tan }^{-1}y+\frac{1}{2}\left[log\right(1+x^2)+log(1+y^2\left)\right]=c$

${\tan }^{-1}\left(\frac{x+y}{1-xy}\right)+\frac{1}{2}log\left[\right(1+x^2\left)\right(1+y^2\left)\right]=c$.