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Question Number 120631 by 77731 last updated on 01/Nov/20

Answered by mathmax by abdo last updated on 01/Nov/20

$$\mathrm{let}\mathrm{determine}\mathrm{a}\mathrm{solution}\mathrm{at}\mathrm{formy}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}$$$${\mathrm{y}}^{{}^{\prime }}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{na}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}-1}\mathrm{and}{\mathrm{y}}^{{}^{″}}=\sum _{\mathrm{n}=2}^{\mathrm{\infty }}\mathrm{n}(\mathrm{n}-1){\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}-2}$$$$\mathrm{e}\Rightarrow 2\sum _{\mathrm{n}=2}^{\mathrm{\infty }}\mathrm{n}(\mathrm{n}-1){\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}-2}-(\mathrm{x}-1)\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}=0\Rightarrow$$$$2\sum _{\mathrm{n}=0}^{\mathrm{\infty }}(\mathrm{n}+2)(\mathrm{n}+1){\mathrm{a}}_{\mathrm{n}+2}{\mathrm{x}}^{\mathrm{n}}-\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}+1}+\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}=0\Rightarrow$$$$\sum _{\mathrm{n}=0}^{\mathrm{\infty }}2(\mathrm{n}+1)(\mathrm{n}+2){\mathrm{a}}_{\mathrm{n}+2}{\mathrm{x}}^{\mathrm{n}}-\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{a}}_{\mathrm{n}-1}{\mathrm{x}}^{\mathrm{n}}+\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}=0\Rightarrow$$$$\{\begin{array}{l}2(\mathrm{n}+1)(\mathrm{n}+2){\mathrm{a}}_{\mathrm{n}+2}-{\mathrm{a}}_{\mathrm{n}-1}+{\mathrm{a}}_{\mathrm{n}}=0\mathrm{\forall }\mathrm{n}⩾1\\ {4\mathrm{a}}_2+{\mathrm{a}}_0=0\end{array}$$$$\Rightarrow \{\begin{array}{l}{\mathrm{a}}_2=-\\ 2(\mathrm{n}+1)(\mathrm{n}+2){\mathrm{a}}_{\mathrm{n}+2}={\mathrm{a}}_{\mathrm{n}-1}-{\mathrm{a}}_{\mathrm{n}}\end{array}$$$$\mathrm{we}\mathrm{have}{\mathrm{a}}_{\mathrm{n}-1}-{\mathrm{a}}_{\mathrm{n}}=2(\mathrm{n}+1)(\mathrm{n}+2){\mathrm{a}}_{\mathrm{n}+2}\Rightarrow$$$$\sum _{\mathrm{k}=1}^{\mathrm{n}}({\mathrm{a}}_{\mathrm{k}-1}-{\mathrm{a}}_{\mathrm{k}})=2\sum _{\mathrm{k}=1}^{\mathrm{n}}(\mathrm{k}+1)(\mathrm{k}+2){\mathrm{a}}_{\mathrm{k}+2}\Rightarrow$$$${\mathrm{a}}_0-{\mathrm{a}}_{\mathrm{n}}=2\sum _{\mathrm{k}=1}^{\mathrm{n}}(\mathrm{k}+1)(\mathrm{k}+2){\mathrm{a}}_{\mathrm{k}+2}\Rightarrow$$$${\mathrm{a}}_{\mathrm{n}}={\mathrm{a}}_0-2\sum _{\mathrm{k}=1}^{\mathrm{n}}(\mathrm{k}+1)(\mathrm{k}+2){\mathrm{a}}_{\mathrm{k}+2}.....\mathrm{be}\mathrm{contonued}...$$

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