Solve the following differential equation: `((x^2-1)dy)/(dx)+2x y=1/(x^2-1);\ |x|\ !=1`

Explanation

Given different equation is `(x^(2)-1)(dy)/(dx)+2xy=(1)/(x^(2)-1)`
`Rightarrow (dy)/(dx)+((2x)/(x^(2)-1))y=(1)/((x^(2)-1)^(2))`
Which is linear differential equation.
On comparing it with `(dy)/(dx)+Py=Q`, we get
`P=(2x)/(x^(2)-1).Q=(1)/((x^(2)-1)^(2))`
`IF=e^(intpdx)=e^(int((2x)/(x^(2)-1))^(dx))`
`"Put" x^(2)-1=t Rightarrow 2xdx=dt`
`therefore IF=r^(int(dt)/(t))=e^(log t)=t=(x^(2)-1)`
The complete solution is `y.IF=int Q.IF+K`
`Rightarrow y.(x^(2)-1)=int(1)/((x^(2)-1)^(2)).(x^(2)-1)dx+K`
`Rightarrow y.(x^(2)-1)=int(dx)/((x^(2)-1))+K`
`Rightarrow y.(x^(2)-1)=(1)/(2)log ((x-1)/(x+1))+K`