Q. If integrating factor of $x\left( 1 - x^{2} \right)dy + \left( 2x^{2}y - y - ax^{3} \right)dx = 0$ is $e^{\int_{}^{}{Pdx}}$, then $P$ is equal to
A) $\frac{2x^{2} - ax^{3}}{x(1 - x^{2})}$
B) $2x^{2} - 1$
C) $\frac{2x^{2} - 1}{ax^{3}}$
D) $\frac{2x^{2} - 1}{x(1 - x^{2})}$
Explanation
(d)
Given, $x\left( 1 - x^{2} \right)dy + \left( 2x^{2}y - y - ax^{3} \right)dx = 0$
$\Rightarrow$ $\frac{dy}{dx} + \frac{(2x^{2} - 1)}{x(1 - x)}y = \frac{ax^{3}}{(1 - x^{2})}$
Here, $P = \frac{2x^{2} - 1}{x(1 - x^{2})}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.