Question Number 57925 by maxmathsup by imad last updated on 14/Apr/19
$${solve}\:{y}^{''} \:−{xy}\:=\mathrm{0}\:\:{by}\:{using}\:{integr}\:{series}. \\ $$
Commented by maxmathsup by imad last updated on 18/Apr/19
$${let}\:{search}\:{develloppable}\:{at}\:{integr}\:{serie}\:\:{let}\:{y}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:{a}_{{n}} {x}^{{n}} \:\Rightarrow \\ $$$${y}^{'} \left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:{na}_{{n}} {x}^{{n}−\mathrm{1}} \:\:{and}\:{y}^{''} \left({x}\right)\:=\sum_{{n}=\mathrm{2}} ^{\infty} \:{n}\left({n}−\mathrm{1}\right){a}_{{n}} {x}^{{n}−\mathrm{2}} \\ $$$$\left({e}\right)\:\Rightarrow\sum_{{n}=\mathrm{2}} ^{\infty} \:{n}\left({n}−\mathrm{1}\right){a}_{{n}} {x}^{{n}−\mathrm{2}} \:−\sum_{{n}=\mathrm{0}} ^{\infty} \:{a}_{{n}} {x}^{{n}+\mathrm{1}} \:=\mathrm{0}\:\Rightarrow \\ $$$$\sum_{{p}=\mathrm{0}} ^{\infty} \:\left({p}+\mathrm{2}\right)\left({p}+\mathrm{1}\right){a}_{{p}+\mathrm{2}} \:{x}^{{p}} \:−\sum_{{p}=\mathrm{1}} ^{\infty} \:{a}_{{p}−\mathrm{1}} {x}^{{p}} \:=\mathrm{0}\:\Rightarrow\mathrm{2}{a}_{\mathrm{2}} \:+\sum_{{p}=\mathrm{1}} ^{\infty} \left\{\:\left({p}+\mathrm{1}\right)\left({p}+\mathrm{2}\right){a}_{{p}+\mathrm{2}} −{a}_{{p}−\mathrm{1}} \right\}{x}^{{p}} =\mathrm{0} \\ $$$$\Rightarrow{a}_{\mathrm{2}} =\mathrm{0}\:\:\:{and}\:\:\left({p}+\mathrm{1}\right)\left({p}+\mathrm{2}\right){a}_{{p}+\mathrm{2}} −{a}_{{p}−\mathrm{1}} =\mathrm{0}\:\:\:\forall\:{p}\geqslant\mathrm{1}\:\Rightarrow{a}_{{p}+\mathrm{2}} =\frac{{a}_{{p}−\mathrm{1}} }{\left({p}+\mathrm{1}\right)\left({p}+\mathrm{2}\right)} \\ $$$$\forall{p}\geqslant\mathrm{1}\:\:{and}\:{a}_{\mathrm{2}} =\mathrm{0}\:\Rightarrow{a}_{\mathrm{2}{n}+\mathrm{2}} =\frac{{a}_{\mathrm{2}{n}−\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{2}\right)}\:\:{and}\:\:{a}_{\mathrm{2}{n}+\mathrm{3}} \:=\frac{{a}_{\mathrm{2}{n}} }{\left(\mathrm{2}{n}+\mathrm{2}\right)\left(\mathrm{2}{n}+\mathrm{3}\right)} \\ $$$${y}\left({x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:{a}_{{n}} {x}^{{n}} \:\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{a}_{\mathrm{2}{n}} {x}^{\mathrm{2}{n}} \:+\sum_{{n}=\mathrm{0}} ^{\infty} \:{a}_{\mathrm{2}{n}+\mathrm{1}} {x}^{\mathrm{2}{n}+\mathrm{1}} \\ $$$$=\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\:\frac{{a}_{\mathrm{2}{n}−\mathrm{3}} }{\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}\:{x}^{\mathrm{2}{n}} \:\:+{a}_{\mathrm{0}} \:\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{{a}_{\mathrm{2}{n}−\mathrm{2}} }{\mathrm{2}{n}\left(\mathrm{2}{n}+\mathrm{1}\right)}\:{x}^{\mathrm{2}{n}+\mathrm{1}} \\ $$$${also}\:{we}\:{can}\:{use}\:{that}\:\:\:\frac{{a}_{\mathrm{2}{n}+\mathrm{2}} }{{a}_{\mathrm{2}{n}−\mathrm{1}} }\:=\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{2}\right)}\:\:…{be}\:{continued}… \\ $$
Answered by 121194 last updated on 14/Apr/19
$${y}=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}} {x}^{{i}} \\ $$$${y}'=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}} {ix}^{{i}−\mathrm{1}} =\underset{{i}=−\mathrm{1}} {\overset{\infty} {\sum}}{a}_{{i}+\mathrm{1}} \left({i}+\mathrm{1}\right){x}^{{i}} =\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}+\mathrm{1}} \left({i}+\mathrm{1}\right){x}^{{i}} \\ $$$${y}''=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}} {i}\left({i}−\mathrm{1}\right){x}^{{i}−\mathrm{2}} =\underset{{i}=−\mathrm{2}} {\overset{\infty} {\sum}}{a}_{{i}+\mathrm{2}} \left({i}+\mathrm{2}\right)\left({i}+\mathrm{1}\right){x}^{{i}} =\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}+\mathrm{2}} \left({i}+\mathrm{2}\right)\left({i}+\mathrm{1}\right){x}^{{i}} \\ $$$${y}''−{xy}=\mathrm{0} \\ $$$$\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}+\mathrm{2}} \left({i}+\mathrm{2}\right)\left({i}+\mathrm{1}\right){x}^{{i}} −{x}\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}} {x}^{{i}} =\mathrm{0} \\ $$$$\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}+\mathrm{2}} \left({i}+\mathrm{2}\right)\left({i}+\mathrm{1}\right){x}^{{i}} −\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}} {x}^{{i}+\mathrm{1}} =\mathrm{0} \\ $$$$\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}+\mathrm{2}} \left({i}+\mathrm{2}\right)\left({i}+\mathrm{1}\right){x}^{{i}} −\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}{a}_{{i}−\mathrm{1}} {x}^{{i}} =\mathrm{0} \\ $$$$\mathrm{2}{a}_{\mathrm{2}} =\mathrm{0}\Rightarrow{a}_{\mathrm{2}} =\mathrm{0} \\ $$$$\left({i}+\mathrm{2}\right)\left({i}+\mathrm{1}\right){a}_{{i}+\mathrm{2}} −{a}_{{i}−\mathrm{1}} =\mathrm{0};{i}>\mathrm{0} \\ $$$${a}_{{i}+\mathrm{2}} =\frac{{a}_{{i}−\mathrm{1}} }{\left({i}+\mathrm{2}\right)\left({i}+\mathrm{1}\right)};{i}>\mathrm{0} \\ $$$${a}_{{i}} =\frac{{a}_{{i}−\mathrm{3}} }{{i}\left({i}−\mathrm{1}\right)};{i}>\mathrm{2} \\ $$$${a}_{\mathrm{0}} ={c}_{\mathrm{1}} \\ $$$${a}_{\mathrm{1}} ={c}_{\mathrm{2}} \\ $$$${a}_{\mathrm{2}} =\mathrm{0} \\ $$$${a}_{\mathrm{3}} =\frac{{a}_{\mathrm{0}} }{\mathrm{6}}=\frac{{c}_{\mathrm{1}} }{\mathrm{6}} \\ $$$${a}_{\mathrm{4}} =\frac{{a}_{\mathrm{1}} }{\mathrm{12}}=\frac{{c}_{\mathrm{2}} }{\mathrm{12}} \\ $$$${a}_{\mathrm{5}} =\frac{{a}_{\mathrm{2}} }{\mathrm{20}}=\mathrm{0} \\ $$$${a}_{\mathrm{6}} =\frac{{a}_{\mathrm{3}} }{\mathrm{30}}=\frac{{c}_{\mathrm{1}} }{\mathrm{180}} \\ $$$$… \\ $$$${y}=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{i}} {x}^{{i}} \\ $$$$=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{\mathrm{3}{i}} {x}^{\mathrm{3}{i}} +\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{\mathrm{3}{i}+\mathrm{1}} {x}^{\mathrm{3}{i}+\mathrm{1}} +\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{\mathrm{3}{i}+\mathrm{2}} {x}^{\mathrm{3}{i}+\mathrm{2}} \\ $$$$={c}_{\mathrm{1}} \underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{3}{i}} }{{f}\left({i}\right)}+{c}_{\mathrm{2}} \underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{\mathrm{3}{i}+\mathrm{1}} }{{g}\left({i}\right)} \\ $$$${f}\left({i}\right)=\begin{cases}{\mathrm{1}}&{{i}=\mathrm{0}}\\{\frac{{f}\left({i}−\mathrm{1}\right)}{\mathrm{3}{i}\left(\mathrm{3}{i}−\mathrm{1}\right)}}&{{i}>\mathrm{0}}\end{cases} \\ $$$${g}\left({i}\right)=\begin{cases}{\mathrm{1}}&{{i}=\mathrm{0}}\\{\frac{{f}\left({i}−\mathrm{1}\right)}{\mathrm{3}{i}\left(\mathrm{3}{i}+\mathrm{1}\right)}}&{{i}>\mathrm{0}}\end{cases} \\ $$