Question Number 34311 by prof Abdo imad last updated on 03/May/18
$${let}\:{give}\:{the}\:{d}.{e}.\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}^{''} \:+\mathrm{3}{xy}^{'} \:+{y}\:=\mathrm{0}{find} \\ $$$${a}\:{solution}\:{y}\left({x}\right)\:{deveppable}\:{at}\:{integr}\:{serie}\: \\ $$$${with}\mid{x}\mid<\mathrm{1}\:. \\ $$
Answered by candre last updated on 04/May/18
$${y}=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}} {x}^{{i}} \\ $$$${y}'=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}+\mathrm{1}} \left({i}+\mathrm{1}\right){x}^{{i}} \\ $$$${y}''=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}+\mathrm{2}} \left({i}+\mathrm{2}\right)\left({i}+\mathrm{1}\right){x}^{{i}} \\ $$$$\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}+\mathrm{2}} \left({i}+\mathrm{2}\right)\left({i}+\mathrm{1}\right){x}^{{i}} +\mathrm{3}{x}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}+\mathrm{1}} \left({i}+\mathrm{1}\right){x}^{{i}} +\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}} {x}^{{i}} =\mathrm{0} \\ $$$$\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}+\mathrm{2}} \left({i}+\mathrm{2}\right)\left({i}+\mathrm{1}\right){x}^{{i}} +\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}+\mathrm{2}} \left({i}+\mathrm{2}\right)\left({i}+\mathrm{1}\right){x}^{{i}+\mathrm{2}} +\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{3}{a}_{{i}+\mathrm{1}} \left({i}+\mathrm{1}\right){x}^{{i}+\mathrm{1}} +\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}} {x}^{{i}} =\mathrm{0} \\ $$$$\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}+\mathrm{2}} \left({i}+\mathrm{2}\right)\left({i}+\mathrm{1}\right){x}^{{i}} +\underset{{i}=\mathrm{2}} {\overset{\infty} {\sum}}{a}_{{i}} {i}\left({i}−\mathrm{1}\right){x}^{{i}} +\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{3}{a}_{{i}} {ix}^{{i}} +\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}} {x}^{{i}} =\mathrm{0} \\ $$$$\mathrm{2}{a}_{\mathrm{2}} +{a}_{\mathrm{0}} =\mathrm{0} \\ $$$$\mathrm{6}{a}_{\mathrm{3}} +\mathrm{4}{a}_{\mathrm{1}} =\mathrm{0} \\ $$$${a}_{{i}+\mathrm{2}} \left({i}^{\mathrm{2}} +\mathrm{3}{i}+\mathrm{2}\right)+{a}_{{i}} \left({i}^{\mathrm{2}} +\mathrm{2}{i}+\mathrm{1}\right)=\mathrm{0}\:\:\:\left({i}\geqslant\mathrm{2}\right) \\ $$