The solution of differential equation $(x^2 –1)\frac{dy}{dx}+2xy=\frac{1}{x^2-1}$   is  $y(x^2 –1)=\frac{1}{\lambda }.\log \left|\frac{x-1}{x+1}\right|+c$. Then the value of $\lambda$  is ------
 

The given differential equation is
$\left({\mathrm{x}}^2-1\right)\frac{\mathrm{dy}}{\mathrm{dx}}+2\mathrm{xy}=\frac{1}{{\mathrm{x}}^2-1}\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}+\frac{2\mathrm{x}}{{\mathrm{x}}^2-1}\mathrm{y}=\frac{1}{{\left({\mathrm{x}}^2-1\right)}^2}$
This is a linear differential equation of the form
$\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{Py}=\mathrm{Q},\text{ where }\mathrm{P}=\frac{2\mathrm{x}}{{\mathrm{x}}^2-1}\text{ and }\mathrm{Q}=\frac{1}{{\left({\mathrm{x}}^2-1\right)}^2}$
$\therefore \text{ I.F. }={\mathrm{e}}^{\int \mathrm{Pdx}}={\mathrm{e}}^{\int 2\mathrm{x}/\left({\mathrm{x}}^2-1\right)\mathrm{dx}}={\mathrm{e}}^{\log \left({\mathrm{x}}^2-1\right)}=\left({\mathrm{x}}^2-1\right)$

Multiplying both sides of (i) by I.F. = (x² – 1), we get
$\left({\mathrm{x}}^2-1\right)\frac{\mathrm{dy}}{\mathrm{dx}}+2\mathrm{xy}=\frac{1}{{\mathrm{x}}^2-1}$
Integrating both sides, we get $\mathrm{y}\left({\mathrm{x}}^2-1\right)=\int \frac{1}{{\mathrm{x}}^2-1}\mathrm{dx}+\mathrm{c}$
[Using : $\mathrm{y}\left(\text{ I.F.) }=\int \text{ Q. (I.F.) }\mathrm{dx}+\mathrm{c}\right.$
$\begin{array}{r}\Rightarrow \underset{\_}{\mathrm{y}}\left({\mathrm{x}}^2-1\right)=\frac{1}{2}\log \left|\frac{\mathrm{x}-1}{\mathrm{x}+1}\right|+\mathrm{c}\\ \underset{\_}{\mathrm{y}}\left({\mathrm{x}}^2-1\right)=\frac{1}{2}\log \left|\frac{\mathrm{x}-1}{\mathrm{x}+1}\right|+\mathrm{c}\\ \end{array}$