Question

If integrating factor of $x(1-x^2)dy+(2x^{2}y-y-a{x}^3)dx=0$ is $e^{\int Pdx}$, then P is equal to

• A. $\frac{2x^2-a{x}^3}{x(1-x^2)}$
  
• B. $(2x^2-1)$
  
• C. $\frac{2x^2-1}{a{x}^3}$
  
• D. $\frac{(2x^2-1)}{x(1-x^2)}$

Answer 1

The correct option is D $\frac{(2x^2-1)}{x(1-x^2)}$

$x(1-x^2)dy+(2x^{2}y-y-a{x}^3)dx=0\frac{dy}{dx}+\frac{(2x^2-1)}{x(1-x^2)}y=\frac{a{x}^2}{(1-x^2)},\therefore P=\frac{2x^2-1}{x(1-x^2)}$.