Question

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

Answer 1

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

Comparing above equation with $\frac{dy}{dx}+Py=Q$, we get
$P=\frac{2x}{x^2-1}-\frac{1}{x}$
Integrating both the sides
$\int Pdx=\int (\frac{2x}{x^2-1}-\frac{1}{x})dx$
$=\log (x^2-1)-\log \left(x\right)$
$=\log \left(\frac{x^2-1}{x}\right)$

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