Question

The intergrating factor of $(1+y^2)\frac{dx}{dy}+x={\tan }^{-1}\left(y\right)$ is

• A. ${\tan }^{-1}y$
• B. $e^{{\tan }^{-1}y}$
• C. $1+y^2$
• D. $e^{1+a{y}^2}$

Answer 1

The correct option is B $e^{{\tan }^{-1}y}$

$\left(1+y^2\right)\frac{dy}{dx}+x=e^{{\tan }^{-1}y}$

$\Rightarrow \frac{dx}{dy}+\frac{x}{1+y^2}=\frac{e^{{\tan }^{-1}y}}{1+y^2}$

Here , $P=\frac{1}{1+y^2},Q=\frac{e^{{\tan }^{-1}y}}{1+y^2}$

$I.F=e^{\int pdy}=e^{\int \frac{1}{1+y^2}dy}=e^{{\tan }^{-1}y}$