$\qquad (20 \text{ points}) \text{ Use power series to solve the initial-value problem}$ \newline $\qquad \text{Answer:} \quad y = \sum_{n=0}^{8} \text{________}$ \newline $\qquad y'' + 3xy' + 6y = 0, \quad y(0) = 0, \quad y'(0) = 1 \quad .$ \newline $\qquad \sum_{n=0}^{8} \frac{(-3^n)}{(2^n)n!} x^{2n+1}$ \newline $\qquad \text{Note: You can earn partial credit on this problem.}$
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Then y' = sum_{n=1}^\infty n a_n x^{n-1} = sum_{n=0}^\infty (n+1) a_{n+1} x^n, y'' = sum_{n=2}^\infty n(n-1) a_n x^{n-2} = sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n. Show more…
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