Question

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

$x^2(x^2-1)\frac{dy}{dx}+x(x^2+1)y=x^2-1$

$⟹\frac{dy}{dx}+\frac{x(x^2+1)}{x^2(x^2-1)}y=x^2-1$ which is in the form of linear D.E.

$I.F=e^{\int \frac{x(x^2+1)}{x^2(x^2-1)}dx}$
$=e^{\int \frac{x^2-1}{x(x^2-1)}+\frac{2}{x(x+1)(x-1)}dx}$

Consider, $\frac{2}{x(x+1)(x-1)}=\frac{A}{x}+\frac{B}{x+1}+\frac{C}{x-1}$

$⟹2=A(x+1)(x-1)+Bx(x-1)+Cx(x+1)$

$⟹2=A{x}^2-A+B{x}^2-Bx+C{x}^2+Cx$

Equating the coefficients we get

$A+B+C=0,-B+C=0$ and $-A=2$

So, we get $A=-2⟹C=1,B=1$

$\therefore \frac{2}{x(x+1)(x-1)}=\frac{-2}{x}+\frac{1}{x+1}+\frac{1}{x-1}$

$\therefore I.F=e^{\left(\int \frac{1}{x}+\frac{-2}{x}+\frac{1}{x+1}+\frac{1}{x-1}dx\right)}$

$\therefore I.F=e^{\left(\int \frac{-1}{x}+\frac{2x}{x^2-1}dx\right)}$

$=e^{(-\ln x+\ln (x^2-1\left)\right)}$

$=e^{\ln \left(\frac{x^2-1}{x}\right)}$

$=\frac{x^2-1}{x}$

$=x-\frac{1}{x}$