Question

The solution of the differential equation $(1+y^2){\tan }^{-1}xdx+y(1+x^2)dy=0$ is

• A. $\log \left(\frac{{\tan }^{-1}x}{x}\right)+y(1+x^2)=c$
• B. $\log (1+y^2)+({\tan }^{-1}x{)}^2=c$
• C. $\log (1+x^2)+\log \left({\tan }^{-1}y\right)+c$
• D. $\left({\tan }^{-1}x\right)(1+y^2)+c=0$

Answer 1

Answer: (B)

$\log (1+y^2)+({\tan }^{-1}x{)}^2=c$

Given differential equation is
$(1+y^2){\tan }^{-1}xdx+y(1+x^2)dy=0$
$\Rightarrow \frac{{\tan }^{-1}x}{1+x^2}dx+\frac{y}{1+y^2}dy=0$
On integrating both sides, we get
$\frac{({\tan }^{-1}x{)}^2}{2}+\frac{1}{2}\log (1+y^2)=\frac{c}{2}$
$\Rightarrow ({\tan }^{-1}x{)}^2+\log (1+y^2)=c$