Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

Question Number 99646 by 24224 Opiyo Kamuki last updated on 22/Jun/20

$$\mathit{u}\mathit{s}\mathit{e}\mathit{p}\mathit{o}\mathit{w}\mathit{e}\mathit{r}\mathit{s}\mathit{e}\mathit{r}\mathit{i}\mathit{e}\mathit{s}\mathit{s}\mathit{o}\mathit{l}\mathit{u}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathit{m}\mathit{e}\mathit{t}\mathit{h}\mathit{o}\mathit{d}\mathit{t}\mathit{o}\mathit{s}\mathit{o}\mathit{l}\mathit{v}\mathit{e}\mathit{t}\mathit{h}\mathit{e}\mathit{O}\mathit{D}\mathit{E}$$$${\mathit{y}}^{″}-\mathit{x}\mathit{y}=0$$

Answered by MWSuSon last updated on 22/Jun/20

$$\underset{\mathrm{k}=2}{\sum ^{\mathrm{\infty }}}\mathrm{k}(\mathrm{k}-1){\mathrm{a}}_{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}-2}-\mathrm{x}\underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}{\mathrm{a}}_{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}}=0$$$$\underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}(\mathrm{k}+1)(\mathrm{k}+2){\mathrm{a}}_{\mathrm{k}+2}{\mathrm{x}}^{\mathrm{k}}-\underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}{\mathrm{a}}_{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}+1}=0$$$$\underset{\mathrm{k}=0}{\sum ^{\mathrm{\infty }}}(\mathrm{k}+1)(\mathrm{k}+2){\mathrm{a}}_{\mathrm{k}+2}{\mathrm{x}}^{\mathrm{k}}-\underset{\mathrm{k}=1}{\sum ^{\mathrm{\infty }}}{\mathrm{a}}_{\mathrm{k}-1}{\mathrm{x}}^{\mathrm{k}}=0$$$${2\mathrm{a}}_2+\underset{\mathrm{k}=1}{\sum ^{\mathrm{\infty }}}[(\mathrm{k}+2)(\mathrm{k}+1){\mathrm{a}}_{\mathrm{k}+2}-{\mathrm{a}}_{\mathrm{k}-1}]{\mathrm{x}}^{\mathrm{k}}=0$$$${\mathrm{a}}_2=0$$$$\mathrm{recurrence}\mathrm{relation}\mathrm{for}\mathrm{k}⩾1$$$${\mathrm{a}}_{\mathrm{k}+2}=\frac{{\mathrm{a}}_{\mathrm{k}-1}}{(\mathrm{k}+2)(\mathrm{k}+1)}$$$$\mathrm{input}\mathrm{values}\mathrm{for}\mathrm{k}=1,2,3,4,5,...$$$$\mathrm{and}\mathrm{find}{\mathrm{a}}_{\mathrm{k}}\mathrm{interms}\mathrm{of}{\mathrm{a}}_0\mathrm{and}{\mathrm{a}}_1$$

Answered by mathmax by abdo last updated on 22/Jun/20

$$\mathrm{y}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}\Rightarrow {\mathrm{y}}^{{}^{\prime }}=\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{na}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}-1}\Rightarrow {\mathrm{y}}^{{}^{″}}=\sum _{\mathrm{n}=2}^{\mathrm{\infty }}\mathrm{n}(\mathrm{n}-1){\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}-2}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}(\mathrm{n}+2)(\mathrm{n}+1){\mathrm{a}}_{\mathrm{n}+2}{\mathrm{x}}^{\mathrm{n}}$$$$\mathrm{e}\Rightarrow \sum _{\mathrm{n}=0}^{\mathrm{\infty }}(\mathrm{n}+1)(\mathrm{n}+2){\mathrm{a}}_{\mathrm{n}+2}{\mathrm{x}}^{\mathrm{n}}-\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}+1}=0\Rightarrow$$$$\sum _{\mathrm{n}=0}^{\mathrm{\infty }}(\mathrm{n}+1)(\mathrm{n}+2){\mathrm{a}}_{\mathrm{n}+2}{\mathrm{x}}^{\mathrm{n}}-\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\mathrm{a}}_{\mathrm{n}-1}{\mathrm{x}}^{\mathrm{n}}=0\Rightarrow$$$${2\mathrm{a}}_2+\sum _{\mathrm{n}=1}^{\mathrm{\infty }}\{(\mathrm{n}+1)(\mathrm{n}+2){\mathrm{a}}_{\mathrm{n}+2}-{\mathrm{a}}_{\mathrm{n}-1}\}{\mathrm{x}}^{\mathrm{n}}=0\Rightarrow$$$${\mathrm{a}}_2=0\mathrm{and}{\mathrm{a}}_{\mathrm{n}+2}=\frac{{\mathrm{a}}_{\mathrm{n}-1}}{(\mathrm{n}+1)(\mathrm{n}+2)}(\mathrm{n}⩾1)\Rightarrow {\mathrm{a}}_3=\frac{{\mathrm{a}}_0}{2.3}$$$${\mathrm{a}}_4=\frac{{\mathrm{a}}_1}{3.4},{\mathrm{a}}_5=\frac{{\mathrm{a}}_2}{4.5},....\Rightarrow \mathrm{y}\left(\mathrm{x}\right)={\mathrm{a}}_0+{\mathrm{a}}_1\mathrm{x}+{\mathrm{a}}_2{\mathrm{x}}^2+{\mathrm{a}}_3{\mathrm{x}}^3+{\mathrm{a}}_4{\mathrm{x}}^4+...$$$$={\mathrm{a}}_0+{\mathrm{a}}_1\mathrm{x}+\frac{{\mathrm{a}}_0}{6}{\mathrm{x}}^3+\frac{{\mathrm{a}}_1}{12}{\mathrm{x}}^4+\frac{{\mathrm{a}}_2}{20}{\mathrm{x}}^5+...$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com