1. **State the problem:** Solve the differential equation $$4x^2y'' - 4xy' + 3y = 0$$ using the Cauchy-Euler method. 2. **Recall the Cauchy-Euler form:** Equations of the form $$x^2y'' + axy' + by = 0$$ can be solved by assuming a solution of the form $$y = x^m$$. 3. **Apply the substitution:** Let $$y = x^m$$, then $$y' = mx^{m-1}$$ $$y'' = m(m-1)x^{m-2}$$ 4. **Substitute into the equation:** $$4x^2(m(m-1)x^{m-2}) - 4x(m x^{m-1}) + 3x^m = 0$$ Simplify powers of $$x$$: $$4m(m-1)x^m - 4m x^m + 3x^m = 0$$ 5. **Factor out $$x^m$$:** $$x^m [4m(m-1) - 4m + 3] = 0$$ Since $$x^m \neq 0$$ for $$x \neq 0$$, solve the characteristic equation: $$4m(m-1) - 4m + 3 = 0$$ 6. **Simplify the characteristic equation:** $$4m^2 - 4m - 4m + 3 = 0$$ $$4m^2 - 8m + 3 = 0$$ 7. **Solve the quadratic:** Using the quadratic formula $$m = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 4 \cdot 3}}{2 \cdot 4}$$ $$= \frac{8 \pm \sqrt{64 - 48}}{8} = \frac{8 \pm \sqrt{16}}{8}$$ $$= \frac{8 \pm 4}{8}$$ 8. **Find roots:** $$m_1 = \frac{8 + 4}{8} = \frac{12}{8} = \frac{3}{2}$$ $$m_2 = \frac{8 - 4}{8} = \frac{4}{8} = \frac{1}{2}$$ 9. **Write the general solution:** Since roots are real and distinct, the general solution is $$y = C_1 x^{\frac{3}{2}} + C_2 x^{\frac{1}{2}}$$ **Final answer:** $$\boxed{y = C_1 x^{\frac{3}{2}} + C_2 x^{\frac{1}{2}}}$$