Q1-find-tow-power-series-solutions-of-the-given-D-E-about-x-0-y-2xy-y-0-Q2-use-the-power-series-method-to-solve-the-given-intial-value-problem-y-2xy-8y-0-y-0-3y-0-0-

Question Number 89624 by M±th+et£s last updated on 18/Apr/20

$$\left.{Q}\mathrm{1}\right){find}\:{tow}\:{power}\:{series}\:{solutions}\:{of}\:{the}\: \\ $$$${given}\:{D}.{E}\:{about}\:{x}=\mathrm{0} \\ $$$${y}^{''} −\mathrm{2}{xy}^{'} +{y}=\mathrm{0} \\ $$$$ \\ $$$$\left.{Q}\mathrm{2}\right){use}\:{the}\:{power}\:{series}\:{method}\:\:{to}\:{solve}\:{the} \\ $$$${given}\:{intial}\:{value}\:{problem} \\ $$$${y}^{''} −\mathrm{2}{xy}^{'} +\mathrm{8}{y}=\mathrm{0} \\ $$$${y}\left(\mathrm{0}\right)=\mathrm{3}{y}^{'} \left(\mathrm{0}\right)=\mathrm{0} \\ $$

Commented by mathmax by abdo last updated on 18/Apr/20

y^(′′) −2xy^′ +y =0 let y =Σ_(n=0) ^∞ a_n x^n ⇒y^′ =Σ_(n=1) ^∞ na_n x^(n−1) =Σ_(n=0) ^∞ (n+1)a_(n+1) x^n and y^((2)) =Σ_(n=2) ^∞ n(n−1)a_n x^(n−2) =Σ_(n=0) ^∞ (n+2)(n+1)a_(n+2) x^n (e)⇒Σ_(n=0) ^∞ (n+2)(n+1)a_(n+2) x^n −2Σ_(n=0) ^∞ (n+1)a_(n+1) x^(n+1) +Σ_(n=0) ^∞ a_n x^n =0 ⇒ Σ_(n=0) ^∞ (n+1)(n+2)a_(n+2) x^n −2Σ_(n=1) ^∞ na_n x^n +Σ_(n=0) ^∞ a_n x^n =0 ⇒2a_2 +Σ_(n=1) ^∞ {(n+1)(n+2)a_(n+2) +(1−2n)a_n }x^n =0 ⇒ { ((a_2 =0)),(((n+1)(n+2)a_(n+2) =(2n−1)a_n ∀n≥1 ⇒)) :} { ((a_2 =0)),((a_(n+2) =((2n−1)/((n+1)(n+2)))a_n ∀n≥1)) :} a_(2n+2) =((4n−1)/((2n+1)(2n+2)))a_(2n) ⇒Π_(k=2) ^n (a_(2k+2) /a_(2k) ) =Π_(k=2) ^n ((4k−1)/(2(2k+1)(k+1))) (a_6 /a_4 )......(a_(2n) /a_(2n−2) ).(a_(2n+2) /a_(2n) ) =(1/2^n )Π_(k=1) ^n ((4k−1)/((k+1)(2k+1))) ⇒ a_(2n+2) =(a_4 /2^n )((3/(2.3))×(7/(3.5))×((11)/(4×7))×.....((4n−1)/((n+1)(2n+1)))) we follow the same way to find a_(2n+1) ...

$${y}^{''} −\mathrm{2}{xy}^{'} \:+{y}\:=\mathrm{0}\:\:{let}\:{y}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:{a}_{{n}} {x}^{{n}} \:\:\Rightarrow{y}^{'} =\sum_{{n}=\mathrm{1}} ^{\infty} {na}_{{n}} {x}^{{n}−\mathrm{1}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \left({n}+\mathrm{1}\right){a}_{{n}+\mathrm{1}} {x}^{{n}} \\ $$$${and}\:{y}^{\left(\mathrm{2}\right)} =\sum_{{n}=\mathrm{2}} ^{\infty} {n}\left({n}−\mathrm{1}\right){a}_{{n}} {x}^{{n}−\mathrm{2}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right){a}_{{n}+\mathrm{2}} {x}^{{n}} \\ $$$$\left({e}\right)\Rightarrow\sum_{{n}=\mathrm{0}} ^{\infty} \left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right){a}_{{n}+\mathrm{2}} {x}^{{n}} \:−\mathrm{2}\sum_{{n}=\mathrm{0}} ^{\infty} \left({n}+\mathrm{1}\right){a}_{{n}+\mathrm{1}} {x}^{{n}+\mathrm{1}} \\ $$$$+\sum_{{n}=\mathrm{0}} ^{\infty} {a}_{{n}} {x}^{{n}} =\mathrm{0}\:\Rightarrow \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right){a}_{{n}+\mathrm{2}} {x}^{{n}} \:−\mathrm{2}\sum_{{n}=\mathrm{1}} ^{\infty} {na}_{{n}} {x}^{{n}} \:+\sum_{{n}=\mathrm{0}} ^{\infty} {a}_{{n}} {x}^{{n}} =\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}{a}_{\mathrm{2}} \:+\sum_{{n}=\mathrm{1}} ^{\infty} \left\{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right){a}_{{n}+\mathrm{2}} +\left(\mathrm{1}−\mathrm{2}{n}\right){a}_{{n}} \right\}{x}^{{n}} =\mathrm{0}\:\Rightarrow \\ $$$$\begin{cases}{{a}_{\mathrm{2}} \:=\mathrm{0}}\\{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right){a}_{{n}+\mathrm{2}} =\left(\mathrm{2}{n}−\mathrm{1}\right){a}_{{n}} \:\:\forall{n}\geqslant\mathrm{1}\:\Rightarrow}\end{cases} \\ $$$$\begin{cases}{{a}_{\mathrm{2}} =\mathrm{0}}\\{{a}_{{n}+\mathrm{2}} =\frac{\mathrm{2}{n}−\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}{a}_{{n}} \:\:\forall{n}\geqslant\mathrm{1}}\end{cases} \\ $$$${a}_{\mathrm{2}{n}+\mathrm{2}} =\frac{\mathrm{4}{n}−\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{2}\right)}{a}_{\mathrm{2}{n}} \:\Rightarrow\prod_{{k}=\mathrm{2}} ^{{n}} \:\frac{{a}_{\mathrm{2}{k}+\mathrm{2}} }{{a}_{\mathrm{2}{k}} }\:=\prod_{{k}=\mathrm{2}} ^{{n}} \:\frac{\mathrm{4}{k}−\mathrm{1}}{\mathrm{2}\left(\mathrm{2}{k}+\mathrm{1}\right)\left({k}+\mathrm{1}\right)} \\ $$$$\frac{{a}_{\mathrm{6}} }{{a}_{\mathrm{4}} }……\frac{{a}_{\mathrm{2}{n}} }{{a}_{\mathrm{2}{n}−\mathrm{2}} }.\frac{{a}_{\mathrm{2}{n}+\mathrm{2}} }{{a}_{\mathrm{2}{n}} }\:=\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\prod_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{4}{k}−\mathrm{1}}{\left({k}+\mathrm{1}\right)\left(\mathrm{2}{k}+\mathrm{1}\right)}\:\Rightarrow \\ $$$${a}_{\mathrm{2}{n}+\mathrm{2}} =\frac{{a}_{\mathrm{4}} }{\mathrm{2}^{{n}} }\left(\frac{\mathrm{3}}{\mathrm{2}.\mathrm{3}}×\frac{\mathrm{7}}{\mathrm{3}.\mathrm{5}}×\frac{\mathrm{11}}{\mathrm{4}×\mathrm{7}}×…..\frac{\mathrm{4}{n}−\mathrm{1}}{\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}\right) \\ $$$${we}\:{follow}\:{the}\:{same}\:{way}\:{to}\:{find}\:\:{a}_{\mathrm{2}{n}+\mathrm{1}} … \\ $$

Commented by M±th+et£s last updated on 18/Apr/20

$${thanx}\:{sir}\:{and}\:{what}\:{about}\:{Q}\mathrm{2} \\ $$

Commented by mathmax by abdo last updated on 18/Apr/20

$${add}\:{the}\:{inital}\:{condition}… \\ $$

Answered by ajfour last updated on 18/Apr/20

y=c_0 +c_1 x+c_2 x^2 +c_3 x^3 +...c_n x^n y′=c_1 +2c_2 x+3c_3 x^2 +....+nc_n x^(n−1) y′′=2c_2 +2.3c_3 x+....+n(n−1)c_n x^(n−2) n(n−1)c_n −2(n−2)c_(n−2) +c_(n−2) =0 (n+1)(n+2)c_(n+2) =(2n−1)c_n c_(n+2) =((2n−1)/((n+1)(n+2)))c_n 2c_2 =c_0 , c_3 =(c_1 /6)

$${y}={c}_{\mathrm{0}} +{c}_{\mathrm{1}} {x}+{c}_{\mathrm{2}} {x}^{\mathrm{2}} +{c}_{\mathrm{3}} {x}^{\mathrm{3}} +…{c}_{{n}} {x}^{{n}} \\ $$$${y}'={c}_{\mathrm{1}} +\mathrm{2}{c}_{\mathrm{2}} {x}+\mathrm{3}{c}_{\mathrm{3}} {x}^{\mathrm{2}} +….+{nc}_{{n}} {x}^{{n}−\mathrm{1}} \\ $$$${y}''=\mathrm{2}{c}_{\mathrm{2}} +\mathrm{2}.\mathrm{3}{c}_{\mathrm{3}} {x}+….+{n}\left({n}−\mathrm{1}\right){c}_{{n}} {x}^{{n}−\mathrm{2}} \\ $$$${n}\left({n}−\mathrm{1}\right){c}_{{n}} −\mathrm{2}\left({n}−\mathrm{2}\right){c}_{{n}−\mathrm{2}} +{c}_{{n}−\mathrm{2}} =\mathrm{0} \\ $$$$\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right){c}_{{n}+\mathrm{2}} =\left(\mathrm{2}{n}−\mathrm{1}\right){c}_{{n}} \\ $$$${c}_{{n}+\mathrm{2}} =\frac{\mathrm{2}{n}−\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}{c}_{{n}} \\ $$$$\mathrm{2}{c}_{\mathrm{2}} ={c}_{\mathrm{0}} \:\:\:,\:\:\:{c}_{\mathrm{3}} =\frac{{c}_{\mathrm{1}} }{\mathrm{6}} \\ $$$$ \\ $$

Commented by M±th+et£s last updated on 18/Apr/20