1. **State the problem:** Solve the differential equation $$(x^2-1)y'' + xy' - y = 0.$$\n\n2. **Identify the type:** This is a second-order linear differential equation with variable coefficients. It resembles the Legendre differential equation but with singularities at $x=\pm 1$.\n\n3. **Rewrite the equation:** Divide both sides by $x^2-1$ (assuming $x \neq \pm 1$):\n$$y'' + \frac{x}{x^2-1} y' - \frac{1}{x^2-1} y = 0.$$\nIntermediate step showing cancellation:\n$$y'' + \frac{\cancel{x}}{\cancel{x^2-1}} y' - \frac{1}{\cancel{x^2-1}} y = 0.$$\n(This is just notation to show division, actual cancellation is not possible here since numerator and denominator differ.)\n\n4. **Method:** Use the Frobenius method or recognize this as a form of the associated Legendre equation. The general solution involves Legendre functions of the first and second kind, $P_\nu(x)$ and $Q_\nu(x)$.\n\n5. **Solution:** The equation matches the Legendre equation with degree $\nu=1$, so the general solution is:\n$$y = C_1 P_1(x) + C_2 Q_1(x),$$\nwhere $P_1(x) = x$ and $Q_1(x)$ is the Legendre function of the second kind of degree 1.\n\n6. **Final answer:**\n$$\boxed{y = C_1 x + C_2 Q_1(x)}.$$\n\nThis completes the solution.