1. **State the problem:** Solve the differential equation $$xy'' - (2 + x)y' + 4y = 0$$ with the initial condition $$y(0) = 0$$. 2. **Identify the type of equation:** This is a linear second-order differential equation with variable coefficients. 3. **Check for singular points:** The equation has a singular point at $$x=0$$ because of the $$xy''$$ term. 4. **Try a power series solution around $$x=0$$:** Assume $$y = \sum_{n=0}^\infty a_n x^n$$. 5. **Compute derivatives:** $$y' = \sum_{n=1}^\infty n a_n x^{n-1}$$ $$y'' = \sum_{n=2}^\infty n(n-1) a_n x^{n-2}$$ 6. **Substitute into the equation:** $$x \sum_{n=2}^\infty n(n-1) a_n x^{n-2} - (2+x) \sum_{n=1}^\infty n a_n x^{n-1} + 4 \sum_{n=0}^\infty a_n x^n = 0$$ 7. **Simplify powers:** $$\sum_{n=2}^\infty n(n-1) a_n x^{n-1} - 2 \sum_{n=1}^\infty n a_n x^{n-1} - \sum_{n=1}^\infty n a_n x^n + 4 \sum_{n=0}^\infty a_n x^n = 0$$ 8. **Rewrite sums to align powers:** $$\sum_{n=1}^\infty [(n+1)n a_{n+1} - 2 n a_n] x^{n-1} - \sum_{n=1}^\infty n a_n x^n + 4 \sum_{n=0}^\infty a_n x^n = 0$$ 9. **Separate terms by powers of $$x$$:** - Coefficient of $$x^0$$: $$2 a_2 - 2 a_1 + 4 a_0 = 0$$ - For $$n \geq 1$$, equate coefficients of $$x^n$$: $$ (n+1)(n+2) a_{n+2} - 2 (n+1) a_{n+1} - n a_n + 4 a_n = 0 $$ 10. **Initial condition:** $$y(0) = a_0 = 0$$. 11. **Solve for coefficients:** From $$x^0$$ term: $$2 a_2 - 2 a_1 + 4 \cdot 0 = 0 \Rightarrow 2 a_2 = 2 a_1 \Rightarrow a_2 = a_1$$ 12. **Recurrence relation for $$n \geq 1$$:** $$ (n+1)(n+2) a_{n+2} - 2 (n+1) a_{n+1} + (4 - n) a_n = 0 $$ 13. **Choose $$a_1$$ as a free parameter (say $$a_1 = C$$), and $$a_0=0$$ from initial condition. Then compute $$a_2 = C$$, and subsequent terms using the recurrence. This gives the power series solution. 14. **Summary:** The solution is a power series with $$a_0=0$$, $$a_1=C$$, and $$a_2=C$$, and higher coefficients from the recurrence. Since the problem only gives one initial condition, the general solution involves one arbitrary constant, consistent with the singular point at zero. **Final answer:** The solution is given by the power series $$y = \sum_{n=0}^\infty a_n x^n$$ with $$a_0=0$$, $$a_1=C$$, $$a_2=C$$, and $$a_{n+2} = \frac{2 (n+1) a_{n+1} - (4 - n) a_n}{(n+1)(n+2)}$$ for $$n \geq 1$$.