1. **Problem statement:** Solve the differential equation $$(1 - x^2) y'' - 2 x y' + 2 y = 0$$ using the Cauchy-Euler (also called Cauchy-Legendre) method. 2. **Recall the Cauchy-Euler equation form:** It typically has the form $$x^2 y'' + a x y' + b y = 0$$ where solutions are sought as $$y = x^m$$. 3. **Rewrite the equation:** The given equation is a Legendre-type equation. We can try a substitution or recognize it as a Legendre differential equation: $$ (1 - x^2) y'' - 2 x y' + 2 y = 0 $$ 4. **Identify it as Legendre's equation:** The standard Legendre equation is $$ (1 - x^2) y'' - 2 x y' + n(n+1) y = 0 $$ Comparing, we have $$n(n+1) = 2$$. 5. **Solve for $n$:** $$ n^2 + n - 2 = 0 $$ Using quadratic formula: $$ n = \frac{-1 \pm \sqrt{1 + 8}}{2} = \frac{-1 \pm 3}{2} $$ So, $$ n = 1 \quad \text{or} \quad n = -2 $$ 6. **General solution:** For Legendre's equation, solutions are Legendre polynomials $P_n(x)$ and $Q_n(x)$ for integer $n$. Here $n=1$ is integer, $n=-2$ is not. 7. **Write the solution:** $$ y = C_1 P_1(x) + C_2 Q_1(x) $$ where $$ P_1(x) = x $$ 8. **Summary:** The solution to the ODE is $$ y = C_1 x + C_2 Q_1(x) $$ where $Q_1(x)$ is the Legendre function of the second kind of order 1. This completes the solution using the Cauchy-Legendre approach.