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

Answer: (B)

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