1. **State the problem:** Solve the differential equation $$x^2 y'' - 3x y' + 3y = 0$$ using the Cauchy-Euler method. 2. **Recall the Cauchy-Euler form:** The equation is of the form $$x^2 y'' + a x y' + b y = 0$$ where $$a = -3$$ and $$b = 3$$. 3. **Assume a solution:** Try $$y = x^m$$ where $$m$$ is a constant to be determined. 4. **Compute derivatives:** $$y' = m x^{m-1}$$ $$y'' = m (m-1) x^{m-2}$$ 5. **Substitute into the equation:** $$x^2 (m (m-1) x^{m-2}) - 3x (m x^{m-1}) + 3 x^m = 0$$ Simplify powers of $$x$$: $$m (m-1) x^m - 3 m x^m + 3 x^m = 0$$ 6. **Factor out $$x^m$$ (which is nonzero for $$x \neq 0$$):** $$x^m [m (m-1) - 3 m + 3] = 0$$ 7. **Form the characteristic equation:** $$m (m-1) - 3 m + 3 = 0$$ Simplify: $$m^2 - m - 3 m + 3 = 0$$ $$m^2 - 4 m + 3 = 0$$ 8. **Solve the quadratic:** $$m = \frac{4 \pm \sqrt{16 - 12}}{2} = \frac{4 \pm 2}{2}$$ So, $$m_1 = \frac{4 + 2}{2} = 3$$ $$m_2 = \frac{4 - 2}{2} = 1$$ 9. **Write the general solution:** $$y = C_1 x^{3} + C_2 x^{1} = C_1 x^{3} + C_2 x$$ where $$C_1$$ and $$C_2$$ are arbitrary constants. **Final answer:** $$y = C_1 x^{3} + C_2 x$$