Aula de Revisão Exercícios Resolvidos-2023 1

Aula de Revisão – Resolução de Exercícios – 20/Mai/2023 Solução: r^2 - 4r + 3 = 0 \iff r = \frac{4 \pm \sqrt{16 - 12}}{2} = \frac{4 \pm 2}{2} \Rightarrow r_1 = 3 \ r_2 = 1 \\ \mathcal{F}(r) = (r - 3)(r - 1) \\ \mathcal{F}(r+n)a_n + a_{n-1}(r - n - 1) = 0 \Rightarrow \mathcal{F}(r+n)a_n = (r+n-2)a_{n-1} \\ \Rightarrow a_n = \frac{(r+n-2)a_{n-1}}{\mathcal{F}(r+n)}, \ n \geq 1 \\ c) \ \text{Para} \ n = r_1 = 3, \ \text{temos que} \ \, \mathcal{F}(3+n) = ((3+n) - 3)((3+n) - 1) = n(n+2)\\ a_n = \frac{(n+1)a_{n-1}}{n(n+2)}, \ \text{para} \ n \geq 1 \\ = 1 \Rightarrow a_1 = \frac{2a_0}{1 . 3} = \frac{2a_0}{1 . (1+2)} \\ n=2 \Rightarrow a_2 = \frac{3a_1}{2 . 4} = \frac{3}{2 . 4} \left(\frac{2a_0}{1 . 3}\right) = \frac{3}{4} . \frac{2a_0}{3} = \frac{2a_0}{2(2+2)} \\ n=3 \Rightarrow a_3 = \frac{4a_2}{3 . 5} = \frac{4}{3 . 5} \left(\frac{2a_0}{3 . 4}\right) = \frac{2a_0}{23.5} = \frac{2a_0}{3l(3+2)} \\ n=4 \Rightarrow a_4 = \frac{5a_3}{4 . 6} = \frac{5}{4.6} \left(\frac{2a_0}{23.4.6}\right) = \frac{2a_0}{23.4.6} = \frac{2a_0}{4!(4+2)} \\ n=5 \Rightarrow a_5 = \frac{6a_4}{5 . 7} = \frac{6}{5.7.4(3.4.6)} = \frac{2a_0}{23.4.5.7} = \frac{2a_0}{5!(5+2)} \\ \vdots \\ a_n = \frac{2a_0}{n!(n+2)}, \ \ \text{para} \ n \geq 1 \\ y(x) = x^r \sum_{n=0}^{+\infty} a_nx^n \Rightarrow \tilde{y}(x) = x^3 \sum_{n=0}^{+\infty} \frac{2a_0}{n!(n+2)} x^n = a_0 \tilde{x}^3 \sum_{n=0}^{+\infty} \frac{2}{n!(n+2)} x^n\\ \Rightarrow y(x) = x^3 \sum_{n=0}^{+\infty} \frac{2}{n!(n+2)} x^n