Question

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

Answer 1

Given differential equation is
$(x^2-1)\frac{dy}{dx}+2xy=\frac{1}{x^2-1}$
$\Rightarrow \frac{dy}{dx}+\left(\frac{2x}{x^2-1}\right)y=\frac{1}{(x^2-1{)}^2}$
which is a linear differential equation,
On comparing it with $\frac{dy}{dx}+Py=Q,$ we get
$P=\frac{2x}{x^2-1},Q=\frac{1}{(x^2-1{)}^2}IF=e^{\int Pdx}=e^{\int {\left(\frac{2x}{x^2-1}\right)}^{dx}}$
Put $x^2-1=t\Rightarrow 2xdx=dt\therefore IF=e^{\int \frac{dt}{t}}=e^{logt}=t=(x^2-1)$
The complete solution is
$y.IF=\int Q.IF+K\Rightarrow y.(x^2-1)=\int \frac{1}{(x^2-1{)}^2}.(x^2-1)dx)+K\Rightarrow y.(x^2-1)-\int \frac{dx}{(x^2-1)}+K\Rightarrow y.(x^2-1)=\frac{1}{2}log\left(\frac{x-1}{x+1}\right)+K$