Question

The integrating factor of the first order differential equation $x^{2}\left(x^{2}-1\right) \frac{dy}{dx}+x\left(x^{2}+1\right)y=x^{2}-1$ is

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

Answer 1

The Correct Option is B

We have,
$x^{2}\left(x^{2}-1\right) \frac{d y}{d x}+x\left(x^{2}+1\right) y=x^{2}-1 $
$\frac{d y}{d x}+\frac{x^{2}+1}{x\left(x^{2}-1\right)} y=\frac{1}{x^{2}} $
$\Rightarrow I F=e^{\frac{x^{2}+1}{x\left(x^{2}-1\right)}}$
$\therefore =e^{1} \frac{x^{2}+1}{x(x-1)(x+1)} d x$
Let $\frac{x^{2}+1}{x(x-1)(x+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}$ $\Rightarrow x^{2}+1=A(x-1)(x+1)+B x(x+1)$ $+C x(x-1)$
Put $ x=0, $
$ \therefore 1=-A $
$ \Rightarrow A=-1 $
Put $ x=1, $
$ \therefore 2=2 B$
$ \Rightarrow B=1 $
Put $ x=-1, $
$ \therefore 2=2 C$
$ \Rightarrow C=1$
$\therefore \frac{x^{2}+1}{x(x-1)(x+1)}=\frac{-1}{x}+\frac{1}{x-1}+\frac{1}{x+1}$
$\therefore I F=e^{\int\left(\frac{-1}{x}+\frac{1}{x-1}+\frac{1}{x+1}\right) d x}$
$=e^{[-\log x+\log (x-1)+\log (x+1)]}$
$=e^{\text{log}\left(\frac{x^{2}-1}{x}\right)}=\frac{x^{2}-1}{x}$
$=x-\frac{1}{x}$