Question

An integrating factor for the $DE:(1+y^2)dx-({\tan }^{-1}y-x)dy=0$ is

• A. ${\tan }^{-1}y$
• B. $e^{{\tan }^{-1}y}$
• C. $\frac{1}{1+y^2}$
• D. $\frac{1}{x(1+y^2)}$

Answer 1

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

We have,

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

$Hence,optionBisthecorrectanswer.$