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

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

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