Question

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

Answer 1

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

Dividing by $(x^2-1)$ on both sides, we get,

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

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

Compare with $\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$ and using $y{e}^{\int P\left(x\right)dx}=\int Q\left(x\right).e^{P\left(x\right)dx}dx$

Here $P\left(x\right)=\frac{2x}{x^2-1}$

Thus $\int P\left(x\right)dx=\int \frac{2x}{x^2-1}dx=\log (x^2-1)$

Hence, $e^{\int P\left(x\right)dx}=e^{\log (x^2-1)}=x^2-1$

Therefore, $y(x^2-1)=\int \frac{1}{(x^2-1{)}^2}.(x^2-1)dx=\int \frac{1}{1-x^2}dx$

$=\frac{1}{2}\int \frac{(x+1-(x-1\left)\right)}{(x+1)(x-1)}$

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

$=\frac{1}{2}\left[\log \right(x-1)-\log (x+1\left)\right]$

Thus $y(x^2-1)=\frac{1}{2}\log \left(\frac{x-1}{x+1}\right)+C$