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Question Number 34311 by prof Abdo imad last updated on 03/May/18

let give the d.e. (1+x^2 )y^(′′) +3xy^′ +y =0find a solution y(x) deveppable at integr serie with∣x∣<1 .

$${let}\:{give}\:{the}\:{d}.{e}.\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}^{''} \:+\mathrm{3}{xy}^{'} \:+{y}\:=\mathrm{0}{find} \\ $$ $${a}\:{solution}\:{y}\left({x}\right)\:{deveppable}\:{at}\:{integr}\:{serie}\: \\ $$ $${with}\mid{x}\mid<\mathrm{1}\:. \\ $$

Answered by candre last updated on 04/May/18

y=Σ_(i=0) ^∞ a_i x^i y′=Σ_(i=0) ^∞ a_(i+1) (i+1)x^i y′′=Σ_(i=0) ^∞ a_(i+2) (i+2)(i+1)x^i (1+x^2 )Σ_(i=0) ^∞ a_(i+2) (i+2)(i+1)x^i +3xΣ_(i=0) ^∞ a_(i+1) (i+1)x^i +Σ_(i=0) ^∞ a_i x^i =0 Σ_(i=0) ^∞ a_(i+2) (i+2)(i+1)x^i +Σ_(i=0) ^∞ a_(i+2) (i+2)(i+1)x^(i+2) +Σ_(i=0) ^∞ 3a_(i+1) (i+1)x^(i+1) +Σ_(i=0) ^∞ a_i x^i =0 Σ_(i=0) ^∞ a_(i+2) (i+2)(i+1)x^i +Σ_(i=2) ^∞ a_i i(i−1)x^i +Σ_(i=1) ^∞ 3a_i ix^i +Σ_(i=0) ^∞ a_i x^i =0 2a_2 +a_0 =0 6a_3 +4a_1 =0 a_(i+2) (i^2 +3i+2)+a_i (i^2 +2i+1)=0 (i≥2)

$${y}=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}} {x}^{{i}} \\ $$ $${y}'=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}+\mathrm{1}} \left({i}+\mathrm{1}\right){x}^{{i}} \\ $$ $${y}''=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}+\mathrm{2}} \left({i}+\mathrm{2}\right)\left({i}+\mathrm{1}\right){x}^{{i}} \\ $$ $$\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}+\mathrm{2}} \left({i}+\mathrm{2}\right)\left({i}+\mathrm{1}\right){x}^{{i}} +\mathrm{3}{x}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}+\mathrm{1}} \left({i}+\mathrm{1}\right){x}^{{i}} +\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}} {x}^{{i}} =\mathrm{0} \\ $$ $$\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}+\mathrm{2}} \left({i}+\mathrm{2}\right)\left({i}+\mathrm{1}\right){x}^{{i}} +\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}+\mathrm{2}} \left({i}+\mathrm{2}\right)\left({i}+\mathrm{1}\right){x}^{{i}+\mathrm{2}} +\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{3}{a}_{{i}+\mathrm{1}} \left({i}+\mathrm{1}\right){x}^{{i}+\mathrm{1}} +\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}} {x}^{{i}} =\mathrm{0} \\ $$ $$\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}+\mathrm{2}} \left({i}+\mathrm{2}\right)\left({i}+\mathrm{1}\right){x}^{{i}} +\underset{{i}=\mathrm{2}} {\overset{\infty} {\sum}}{a}_{{i}} {i}\left({i}−\mathrm{1}\right){x}^{{i}} +\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{3}{a}_{{i}} {ix}^{{i}} +\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}} {x}^{{i}} =\mathrm{0} \\ $$ $$\mathrm{2}{a}_{\mathrm{2}} +{a}_{\mathrm{0}} =\mathrm{0} \\ $$ $$\mathrm{6}{a}_{\mathrm{3}} +\mathrm{4}{a}_{\mathrm{1}} =\mathrm{0} \\ $$ $${a}_{{i}+\mathrm{2}} \left({i}^{\mathrm{2}} +\mathrm{3}{i}+\mathrm{2}\right)+{a}_{{i}} \left({i}^{\mathrm{2}} +\mathrm{2}{i}+\mathrm{1}\right)=\mathrm{0}\:\:\:\left({i}\geqslant\mathrm{2}\right) \\ $$

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