Q. The solution of the differential equation $\frac{dy}{dx} + \frac{2x}{1 + x^{2}}.y = \frac{1}{\left( 1 + x^{2} \right)^{2}}$ is
A) $y\left( 1 - x^{2} \right) = \tan^{- 1}x + c$
B) $y\left( 1 + x^{2} \right) = \tan^{- 1}x + c$
C) $y\left( 1 + x^{2} \right)^{2} = \tan^{- 1}x + c$
D) $y\left( 1 - x^{2} \right)^{2} = \tan^{- 1}x + c$
Explanation
(b)
Given, $\frac{dy}{dx} + \frac{2x}{1 + x^{2}}.y = \frac{1}{\left( 1 + x^{2} \right)^{2}}$
$\therefore$ $IF = e^{\int_{}^{}{\frac{2x}{1 + x^{2}}\ dx}} = e^{\log{(1 + x^{2})}} = 1 + x^{2}$
The complete solution is
$y\left( 1 + x^{2} \right) = \int_{}^{}\left( 1 + x^{2} \right).\frac{1}{\left( 1 + x^{2} \right)^{2}}\ dx + c$
$\Rightarrow$ $y\left( 1 + x^{2} \right) = \tan^{- 1}x + c$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.