Q.Solve the differential equation $(x^2-1)\frac{dy}{dx}+2xy=\frac{1}{x^2-1}$.

This is a first-order linear ODE. Written as $\frac{dy}{dx}+\frac{2x}{x^2-1}\,y=\frac{1}{(x^2-1)^2}$, its integrating factor is $e^{\int \frac{2x}{x^2-1}\,dx}=e^{\log|x^2-1|}=x^2-1$ — exactly the coefficient already in front of $\frac{dy}{dx}$.

So the left side of the original equation is already an exact derivative:

$\frac{d}{dx}\big[(x^2-1)y\big]=\frac{1}{x^2-1}.$ …

Spot the structure

The equation $(x^2-1)\frac{dy}{dx}+2xy=\frac{1}{x^2-1}$ is first-order linear. Dividing by $x^2-1$ puts it in standard form $\frac{dy}{dx}+P\,y=Q$ with $P=\frac{2x}{x^2-1}$.

Find the integrating factor

Notice the I.F. is exactly the coefficient already multiplying $\frac{dy}{dx}$ in the original equation. That means the given left side is already $\dfrac{d}{dx}\big[(x^2-1)y\big]$:

Integrate …

Method: Linear first-order DE by the integrating factor

Use this for any equation reducible to $\dfrac{dy}{dx}+P(x)\,y=Q(x)$ — linear in $y$ and its derivative.

Steps

Step 1: Put it in standard linear form

Divide through so the coefficient of $\dfrac{dy}{dx}$ is $1$, and read off $P(x)$ and $Q(x)$.

Step 2: Build the integrating factor

Often the original coefficient in front of $\dfrac{dy}{dx}$ is the integrating factor — a sign that the left side is already an exact derivative. …

Common Mistakes

Mistake 1: Forgetting to divide by $(x^2-1)$ before reading off $P(x)$

Why it's wrong: the standard form needs a coefficient of $1$ on $\frac{dy}{dx}$; skipping this gives the wrong $P$ and IF. Correct approach: write $\frac{dy}{dx}+\frac{2x}{x^2-1}y=\frac{1}{(x^2-1)^2}$.

Mistake 2: Not recognising the left side is already an exact derivative …