1. **Problem statement:** Find a series solution about the regular singular point $x=0$ for the differential equation $$x^2 y''(x) + x y'(x) + x^2 y(x) = 0, \quad x > 0.$$ 2. **Method:** We use the Frobenius method since $x=0$ is a regular singular point. Assume a solution of the form: $$y(x) = x^r \sum_{n=0}^\infty a_n x^n = \sum_{n=0}^\infty a_n x^{n+r}$$ where $a_0 \neq 0$ and $r$ is to be determined. 3. **Derivatives:** $$y'(x) = \sum_{n=0}^\infty a_n (n+r) x^{n+r-1}$$ $$y''(x) = \sum_{n=0}^\infty a_n (n+r)(n+r-1) x^{n+r-2}$$ 4. **Substitute into the equation:** $$x^2 y'' + x y' + x^2 y = 0$$ becomes $$x^2 \sum_{n=0}^\infty a_n (n+r)(n+r-1) x^{n+r-2} + x \sum_{n=0}^\infty a_n (n+r) x^{n+r-1} + x^2 \sum_{n=0}^\infty a_n x^{n+r} = 0$$ Simplify powers of $x$: $$\sum_{n=0}^\infty a_n (n+r)(n+r-1) x^{n+r} + \sum_{n=0}^\infty a_n (n+r) x^{n+r} + \sum_{n=0}^\infty a_n x^{n+r+2} = 0$$ 5. **Combine first two sums:** $$\sum_{n=0}^\infty a_n \big[(n+r)(n+r-1) + (n+r)\big] x^{n+r} + \sum_{n=0}^\infty a_n x^{n+r+2} = 0$$ Simplify inside brackets: $$(n+r)(n+r-1) + (n+r) = (n+r)^2 - (n+r) + (n+r) = (n+r)^2$$ So equation is: $$\sum_{n=0}^\infty a_n (n+r)^2 x^{n+r} + \sum_{n=0}^\infty a_n x^{n+r+2} = 0$$ 6. **Rewrite second sum to match powers:** Let $m = n+2$ in second sum: $$\sum_{n=0}^\infty a_n x^{n+r+2} = \sum_{m=2}^\infty a_{m-2} x^{m+r}$$ 7. **Rewrite entire equation:** $$\sum_{n=0}^\infty a_n (n+r)^2 x^{n+r} + \sum_{n=2}^\infty a_{n-2} x^{n+r} = 0$$ 8. **Combine sums starting from $n=2$:** $$a_0 r^2 x^r + a_1 (1+r)^2 x^{1+r} + \sum_{n=2}^\infty \big[a_n (n+r)^2 + a_{n-2}\big] x^{n+r} = 0$$ 9. **Set coefficients of each power to zero:** - For $x^r$: $$a_0 r^2 = 0 \implies r^2 = 0 \implies r=0$$ - For $x^{r+1}$: $$a_1 (1+0)^2 = a_1 = 0 \implies a_1 = 0$$ - For $n \geq 2$: $$a_n (n+0)^2 + a_{n-2} = 0 \implies a_n = -\frac{a_{n-2}}{n^2}$$ 10. **Recurrence relation:** $$a_n = -\frac{a_{n-2}}{n^2}$$ with initial conditions $a_0$ arbitrary and $a_1=0$. 11. **Calculate coefficients:** - $a_0$ arbitrary - $a_1=0$ - $a_2 = -\frac{a_0}{2^2} = -\frac{a_0}{4}$ - $a_3 = 0$ (since $a_1=0$) - $a_4 = -\frac{a_2}{4^2} = -\frac{-a_0/4}{16} = \frac{a_0}{64}$ - $a_5=0$ - $a_6 = -\frac{a_4}{6^2} = -\frac{a_0/64}{36} = -\frac{a_0}{2304}$ 12. **Series solution:** $$y(x) = a_0 \left[1 - \frac{x^2}{4} + \frac{x^4}{64} - \frac{x^6}{2304} + \cdots \right]$$ 13. **Interpretation:** The solution is an even power series with coefficients given by the recurrence. **Final answer:** $$\boxed{y(x) = a_0 \sum_{k=0}^\infty (-1)^k \frac{x^{2k}}{(2^{k} k!)^2}}$$ which matches the pattern of coefficients found.