Find a general solution in powers of $x$ of the differential equation. State the recurrence relation and the guaranteed radius of convergence.
\[
\left(x^{2}-1\right) y^{\prime \prime}+4 x y^{\prime}+2 y=0
\]
The recurrence relation is $\mathrm{c}_{\mathrm{n}+2}=$

Step 1 ：First, we rewrite the given differential equation in the standard form of a Cauchy-Euler equation, which is \(x^{2}y'' + axy' + by = 0\). In this case, we have \(a = 4\) and \(b = 2\).

Step 2 ：Next, we assume a solution of the form \(y = x^{r}\), where \(r\) is a constant to be determined. This gives us the auxiliary equation \(r(r-1) + 4r + 2 = 0\).

Step 3 ：Solving this quadratic equation for \(r\), we get \(r = -1, -2\). Therefore, the general solution of the differential equation is \(y = c_{1}x^{-1} + c_{2}x^{-2}\).

Step 4 ：Now, we look for a solution in the form of a power series \(y = \sum_{n=0}^{\infty} c_{n}x^{n}\). Substituting this into the differential equation, we get \(\sum_{n=0}^{\infty} c_{n}(n(n-1)x^{n-2} + 4nx^{n-1} + 2x^{n}) = 0\).

Step 5 ：Equating the coefficients of like powers of \(x\) to zero, we get the recurrence relation \(c_{n+2} = -\frac{4n+2}{(n+2)(n+1)}c_{n}\) for \(n \geq 0\), with initial conditions \(c_{0}\) and \(c_{1}\) arbitrary.

Step 6 ：The radius of convergence of the power series solution is guaranteed to be at least the distance to the nearest singularity of the differential equation. In this case, the singularities are at \(x = \pm 1\), so the radius of convergence is at least 1.