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

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

The correct answer is Option (2) → $y(x^2 - 1) = \frac{1}{2} \ln \left| \frac{x - 1}{x + 1} \right| + C$ ##

Given differential equation is

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

On dividing the above equation by $(x^{2} - 1)$, we get

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

which is a linear differential equation.

On comparing it with $\frac{dy}{dx} + Py = Q$, we get

$P = \frac{2x}{x^{2} - 1}, Q = \frac{1}{(x^{2} - 1)^{2}}$

$\text{I.F.} = e^{\int P dx} = e^{\int \left( \frac{2x}{x^{2} - 1} \right) dx}$

Put $x^{2} - 1 = t \Rightarrow 2x dx = dt$

$∴\text{I.F.} = e^{\int \frac{dt}{t}} = e^{\log t} = t = (x^{2} - 1) \quad [∵e^{\log x} = x]$

The complete solution is

$y \cdot (\text{I.F.}) = \int Q \cdot \text{I.F.} dx + C$

$\Rightarrow y \cdot (x^{2} - 1) = \int \frac{1}{(x^{2} - 1)^{2}} \cdot (x^{2} - 1) dx + C$

$\Rightarrow y \cdot (x^{2} - 1) = \int \frac{dx}{x^{2} - 1} + C$

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