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Question Number 57925 by maxmathsup by imad last updated on 14/Apr/19

solve y^(′′) −xy =0 by using integr series.

$${solve}\:{y}^{''} \:−{xy}\:=\mathrm{0}\:\:{by}\:{using}\:{integr}\:{series}. \\ $$

Commented by maxmathsup by imad last updated on 18/Apr/19

let search develloppable at integr serie let y =Σ_(n=0) ^∞ a_n x^n ⇒ y^′ (x) =Σ_(n=1) ^∞ na_n x^(n−1) and y^(′′) (x) =Σ_(n=2) ^∞ n(n−1)a_n x^(n−2) (e) ⇒Σ_(n=2) ^∞ n(n−1)a_n x^(n−2) −Σ_(n=0) ^∞ a_n x^(n+1) =0 ⇒ Σ_(p=0) ^∞ (p+2)(p+1)a_(p+2) x^p −Σ_(p=1) ^∞ a_(p−1) x^p =0 ⇒2a_2 +Σ_(p=1) ^∞ { (p+1)(p+2)a_(p+2) −a_(p−1) }x^p =0 ⇒a_2 =0 and (p+1)(p+2)a_(p+2) −a_(p−1) =0 ∀ p≥1 ⇒a_(p+2) =(a_(p−1) /((p+1)(p+2))) ∀p≥1 and a_2 =0 ⇒a_(2n+2) =(a_(2n−1) /((2n+1)(2n+2))) and a_(2n+3) =(a_(2n) /((2n+2)(2n+3))) y(x) =Σ_(n=0) ^∞ a_n x^n = Σ_(n=0) ^∞ a_(2n) x^(2n) +Σ_(n=0) ^∞ a_(2n+1) x^(2n+1) =Σ_(n=2) ^∞ (a_(2n−3) /((2n−1)(2n+1))) x^(2n) +a_0 +Σ_(n=1) ^∞ (a_(2n−2) /(2n(2n+1))) x^(2n+1) also we can use that (a_(2n+2) /a_(2n−1) ) =(1/((2n+1)(2n+2))) ...be continued...

$${let}\:{search}\:{develloppable}\:{at}\:{integr}\:{serie}\:\:{let}\:{y}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:{a}_{{n}} {x}^{{n}} \:\Rightarrow \\ $$$${y}^{'} \left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:{na}_{{n}} {x}^{{n}−\mathrm{1}} \:\:{and}\:{y}^{''} \left({x}\right)\:=\sum_{{n}=\mathrm{2}} ^{\infty} \:{n}\left({n}−\mathrm{1}\right){a}_{{n}} {x}^{{n}−\mathrm{2}} \\ $$$$\left({e}\right)\:\Rightarrow\sum_{{n}=\mathrm{2}} ^{\infty} \:{n}\left({n}−\mathrm{1}\right){a}_{{n}} {x}^{{n}−\mathrm{2}} \:−\sum_{{n}=\mathrm{0}} ^{\infty} \:{a}_{{n}} {x}^{{n}+\mathrm{1}} \:=\mathrm{0}\:\Rightarrow \\ $$$$\sum_{{p}=\mathrm{0}} ^{\infty} \:\left({p}+\mathrm{2}\right)\left({p}+\mathrm{1}\right){a}_{{p}+\mathrm{2}} \:{x}^{{p}} \:−\sum_{{p}=\mathrm{1}} ^{\infty} \:{a}_{{p}−\mathrm{1}} {x}^{{p}} \:=\mathrm{0}\:\Rightarrow\mathrm{2}{a}_{\mathrm{2}} \:+\sum_{{p}=\mathrm{1}} ^{\infty} \left\{\:\left({p}+\mathrm{1}\right)\left({p}+\mathrm{2}\right){a}_{{p}+\mathrm{2}} −{a}_{{p}−\mathrm{1}} \right\}{x}^{{p}} =\mathrm{0} \\ $$$$\Rightarrow{a}_{\mathrm{2}} =\mathrm{0}\:\:\:{and}\:\:\left({p}+\mathrm{1}\right)\left({p}+\mathrm{2}\right){a}_{{p}+\mathrm{2}} −{a}_{{p}−\mathrm{1}} =\mathrm{0}\:\:\:\forall\:{p}\geqslant\mathrm{1}\:\Rightarrow{a}_{{p}+\mathrm{2}} =\frac{{a}_{{p}−\mathrm{1}} }{\left({p}+\mathrm{1}\right)\left({p}+\mathrm{2}\right)} \\ $$$$\forall{p}\geqslant\mathrm{1}\:\:{and}\:{a}_{\mathrm{2}} =\mathrm{0}\:\Rightarrow{a}_{\mathrm{2}{n}+\mathrm{2}} =\frac{{a}_{\mathrm{2}{n}−\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{2}\right)}\:\:{and}\:\:{a}_{\mathrm{2}{n}+\mathrm{3}} \:=\frac{{a}_{\mathrm{2}{n}} }{\left(\mathrm{2}{n}+\mathrm{2}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)} \\ $$$${y}\left({x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:{a}_{{n}} {x}^{{n}} \:\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{a}_{\mathrm{2}{n}} {x}^{\mathrm{2}{n}} \:+\sum_{{n}=\mathrm{0}} ^{\infty} \:{a}_{\mathrm{2}{n}+\mathrm{1}} {x}^{\mathrm{2}{n}+\mathrm{1}} \\ $$$$=\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\:\frac{{a}_{\mathrm{2}{n}−\mathrm{3}} }{\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}\:{x}^{\mathrm{2}{n}} \:\:+{a}_{\mathrm{0}} \:\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{{a}_{\mathrm{2}{n}−\mathrm{2}} }{\mathrm{2}{n}\left(\mathrm{2}{n}+\mathrm{1}\right)}\:{x}^{\mathrm{2}{n}+\mathrm{1}} \\ $$$${also}\:{we}\:{can}\:{use}\:{that}\:\:\:\frac{{a}_{\mathrm{2}{n}+\mathrm{2}} }{{a}_{\mathrm{2}{n}−\mathrm{1}} }\:=\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{2}\right)}\:\:...{be}\:{continued}... \\ $$

Answered by 121194 last updated on 14/Apr/19

y=Σ_(i=0) ^∞ a_i x^i y′=Σ_(i=0) ^∞ a_i ix^(i−1) =Σ_(i=−1) ^∞ a_(i+1) (i+1)x^i =Σ_(i=0) ^∞ a_(i+1) (i+1)x^i y′′=Σ_(i=0) ^∞ a_i i(i−1)x^(i−2) =Σ_(i=−2) ^∞ a_(i+2) (i+2)(i+1)x^i =Σ_(i=0) ^∞ a_(i+2) (i+2)(i+1)x^i y′′−xy=0 Σ_(i=0) ^∞ a_(i+2) (i+2)(i+1)x^i −xΣ_(i=0) ^∞ a_i x^i =0 Σ_(i=0) ^∞ a_(i+2) (i+2)(i+1)x^i −Σ_(i=0) ^∞ a_i x^(i+1) =0 Σ_(i=0) ^∞ a_(i+2) (i+2)(i+1)x^i −Σ_(i=1) ^∞ a_(i−1) x^i =0 2a_2 =0⇒a_2 =0 (i+2)(i+1)a_(i+2) −a_(i−1) =0;i>0 a_(i+2) =(a_(i−1) /((i+2)(i+1)));i>0 a_i =(a_(i−3) /(i(i−1)));i>2 a_0 =c_1 a_1 =c_2 a_2 =0 a_3 =(a_0 /6)=(c_1 /6) a_4 =(a_1 /(12))=(c_2 /(12)) a_5 =(a_2 /(20))=0 a_6 =(a_3 /(30))=(c_1 /(180)) ... y=Σ_(i=0) ^∞ a_i x^i =Σ_(i=0) ^∞ a_(3i) x^(3i) +Σ_(i=0) ^∞ a_(3i+1) x^(3i+1) +Σ_(i=0) ^∞ a_(3i+2) x^(3i+2) =c_1 Σ_(i=0) ^∞ (x^(3i) /(f(i)))+c_2 Σ_(i=0) ^∞ (x^(3i+1) /(g(i))) f(i)= { (1,(i=0)),(((f(i−1))/(3i(3i−1))),(i>0)) :} g(i)= { (1,(i=0)),(((f(i−1))/(3i(3i+1))),(i>0)) :}

$${y}=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}} {x}^{{i}} \\ $$$${y}'=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}} {ix}^{{i}−\mathrm{1}} =\underset{{i}=−\mathrm{1}} {\overset{\infty} {\sum}}{a}_{{i}+\mathrm{1}} \left({i}+\mathrm{1}\right){x}^{{i}} =\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}+\mathrm{1}} \left({i}+\mathrm{1}\right){x}^{{i}} \\ $$$${y}''=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}} {i}\left({i}−\mathrm{1}\right){x}^{{i}−\mathrm{2}} =\underset{{i}=−\mathrm{2}} {\overset{\infty} {\sum}}{a}_{{i}+\mathrm{2}} \left({i}+\mathrm{2}\right)\left({i}+\mathrm{1}\right){x}^{{i}} =\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}+\mathrm{2}} \left({i}+\mathrm{2}\right)\left({i}+\mathrm{1}\right){x}^{{i}} \\ $$$${y}''−{xy}=\mathrm{0} \\ $$$$\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}+\mathrm{2}} \left({i}+\mathrm{2}\right)\left({i}+\mathrm{1}\right){x}^{{i}} −{x}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}} {x}^{{i}} =\mathrm{0} \\ $$$$\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}+\mathrm{2}} \left({i}+\mathrm{2}\right)\left({i}+\mathrm{1}\right){x}^{{i}} −\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}} {x}^{{i}+\mathrm{1}} =\mathrm{0} \\ $$$$\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}+\mathrm{2}} \left({i}+\mathrm{2}\right)\left({i}+\mathrm{1}\right){x}^{{i}} −\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}{a}_{{i}−\mathrm{1}} {x}^{{i}} =\mathrm{0} \\ $$$$\mathrm{2}{a}_{\mathrm{2}} =\mathrm{0}\Rightarrow{a}_{\mathrm{2}} =\mathrm{0} \\ $$$$\left({i}+\mathrm{2}\right)\left({i}+\mathrm{1}\right){a}_{{i}+\mathrm{2}} −{a}_{{i}−\mathrm{1}} =\mathrm{0};{i}>\mathrm{0} \\ $$$${a}_{{i}+\mathrm{2}} =\frac{{a}_{{i}−\mathrm{1}} }{\left({i}+\mathrm{2}\right)\left({i}+\mathrm{1}\right)};{i}>\mathrm{0} \\ $$$${a}_{{i}} =\frac{{a}_{{i}−\mathrm{3}} }{{i}\left({i}−\mathrm{1}\right)};{i}>\mathrm{2} \\ $$$${a}_{\mathrm{0}} ={c}_{\mathrm{1}} \\ $$$${a}_{\mathrm{1}} ={c}_{\mathrm{2}} \\ $$$${a}_{\mathrm{2}} =\mathrm{0} \\ $$$${a}_{\mathrm{3}} =\frac{{a}_{\mathrm{0}} }{\mathrm{6}}=\frac{{c}_{\mathrm{1}} }{\mathrm{6}} \\ $$$${a}_{\mathrm{4}} =\frac{{a}_{\mathrm{1}} }{\mathrm{12}}=\frac{{c}_{\mathrm{2}} }{\mathrm{12}} \\ $$$${a}_{\mathrm{5}} =\frac{{a}_{\mathrm{2}} }{\mathrm{20}}=\mathrm{0} \\ $$$${a}_{\mathrm{6}} =\frac{{a}_{\mathrm{3}} }{\mathrm{30}}=\frac{{c}_{\mathrm{1}} }{\mathrm{180}} \\ $$$$... \\ $$$${y}=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}} {x}^{{i}} \\ $$$$=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{\mathrm{3}{i}} {x}^{\mathrm{3}{i}} +\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{\mathrm{3}{i}+\mathrm{1}} {x}^{\mathrm{3}{i}+\mathrm{1}} +\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{\mathrm{3}{i}+\mathrm{2}} {x}^{\mathrm{3}{i}+\mathrm{2}} \\ $$$$={c}_{\mathrm{1}} \underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{3}{i}} }{{f}\left({i}\right)}+{c}_{\mathrm{2}} \underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{3}{i}+\mathrm{1}} }{{g}\left({i}\right)} \\ $$$${f}\left({i}\right)=\begin{cases}{\mathrm{1}}&{{i}=\mathrm{0}}\\{\frac{{f}\left({i}−\mathrm{1}\right)}{\mathrm{3}{i}\left(\mathrm{3}{i}−\mathrm{1}\right)}}&{{i}>\mathrm{0}}\end{cases} \\ $$$${g}\left({i}\right)=\begin{cases}{\mathrm{1}}&{{i}=\mathrm{0}}\\{\frac{{f}\left({i}−\mathrm{1}\right)}{\mathrm{3}{i}\left(\mathrm{3}{i}+\mathrm{1}\right)}}&{{i}>\mathrm{0}}\end{cases} \\ $$

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