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Question Number 79761 by Najaah last updated on 28/Jan/20

(x^2 +2)y′′ + xy = 0

$$\left({x}^{\mathrm{2}} +\mathrm{2}\right){y}''\:+\:{xy}\:=\:\mathrm{0} \\ $$

Commented by abdomathmax last updated on 28/Jan/20

solution developpable at?integr serie y(x)=Σ_(n=0) ^∞ a_n x^n ⇒y^′ =Σ_(n=1) ^∞ na_n x^(n−1) ⇒ y^((2)) (x)=Σ_(n=2) ^∞ n(n−1)a_n x^(n−2) (e) ⇒(x^2 +2)Σ_(n=2) ^∞ n(n−1)a_n x^(n−2) +Σ_(n=0) ^∞ a_n x^(n+1) =0 ⇒Σ_(n=2) ^∞ n(n−1)a_n x^n +2Σ_(n=2) ^∞ n(n−1)a_n x^(n−2) +Σ_(n=1) ^∞ a_(n−1) x^n =0 ⇒ a_0 +Σ_(n=2) ^∞ {n(n−1)a_n +a_(n−1) }x^n +2a_2 +2Σ_(n=1) ^∞ (n+2)(n+1)a_(n+2) x^n =0 ⇒a_0 +2a_2 +Σ_(n=2) ^∞ {n(n−1)a_n +a_(n−1) +(n+2)(n+1)a_(n+2) }x^n =0 ⇒a_0 +2a_2 =0 and (n+2)(n+1)a_(n+2) +n(n−1)a_n +a_(n−1) =0 ∀n≥2 ...be continued...

$${solution}\:{developpable}\:{at}?{integr}\:{serie} \\ $$$${y}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:{a}_{{n}} {x}^{{n}} \:\Rightarrow{y}^{'} =\sum_{{n}=\mathrm{1}} ^{\infty} {na}_{{n}} {x}^{{n}−\mathrm{1}} \:\Rightarrow \\ $$$${y}^{\left(\mathrm{2}\right)} \left({x}\right)=\sum_{{n}=\mathrm{2}} ^{\infty} {n}\left({n}−\mathrm{1}\right){a}_{{n}} {x}^{{n}−\mathrm{2}} \\ $$$$\left({e}\right)\:\Rightarrow\left({x}^{\mathrm{2}} +\mathrm{2}\right)\sum_{{n}=\mathrm{2}} ^{\infty} {n}\left({n}−\mathrm{1}\right){a}_{{n}} {x}^{{n}−\mathrm{2}} +\sum_{{n}=\mathrm{0}} ^{\infty} {a}_{{n}} {x}^{{n}+\mathrm{1}} =\mathrm{0} \\ $$$$\Rightarrow\sum_{{n}=\mathrm{2}} ^{\infty} {n}\left({n}−\mathrm{1}\right){a}_{{n}} {x}^{{n}} \:+\mathrm{2}\sum_{{n}=\mathrm{2}} ^{\infty} {n}\left({n}−\mathrm{1}\right){a}_{{n}} \:{x}^{{n}−\mathrm{2}} \\ $$$$+\sum_{{n}=\mathrm{1}} ^{\infty} {a}_{{n}−\mathrm{1}} {x}^{{n}} \:=\mathrm{0}\:\Rightarrow \\ $$$${a}_{\mathrm{0}} \:+\sum_{{n}=\mathrm{2}} ^{\infty} \left\{{n}\left({n}−\mathrm{1}\right){a}_{{n}} +{a}_{{n}−\mathrm{1}} \right\}{x}^{{n}} \:\:\:\:\:\:\:\:\:+\mathrm{2}{a}_{\mathrm{2}} \\ $$$$+\mathrm{2}\sum_{{n}=\mathrm{1}} ^{\infty} \:\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right){a}_{{n}+\mathrm{2}} \:{x}^{{n}} =\mathrm{0} \\ $$$$\Rightarrow{a}_{\mathrm{0}} \:+\mathrm{2}{a}_{\mathrm{2}} \:\:+\sum_{{n}=\mathrm{2}} ^{\infty} \left\{{n}\left({n}−\mathrm{1}\right){a}_{{n}} +{a}_{{n}−\mathrm{1}} +\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right){a}_{{n}+\mathrm{2}} \right\}{x}^{{n}} =\mathrm{0} \\ $$$$\Rightarrow{a}_{\mathrm{0}} \:+\mathrm{2}{a}_{\mathrm{2}} =\mathrm{0}\:{and} \\ $$$$\left({n}+\mathrm{2}\right)\left({n}+\mathrm{1}\right){a}_{{n}+\mathrm{2}} +{n}\left({n}−\mathrm{1}\right){a}_{{n}} \:+{a}_{{n}−\mathrm{1}} =\mathrm{0}\:\forall{n}\geqslant\mathrm{2} \\ $$$$...{be}\:{continued}... \\ $$

Answered by Henri Boucatchou last updated on 28/Jan/20

we have ((y′′)/y)=−(x/(x^2 +2)) ⇒ lny′=−(1/2)ln(x^2 +2) ⇒ y′=(x^2 +2)^(−1/2) ⇒y=∫(x^2 +2)^(−1/2) dx=ln(x+(√(x^2 +2)))+Cte

$${we}\:\:{have}\:\:\frac{{y}''}{{y}}=−\frac{{x}}{{x}^{\mathrm{2}} +\mathrm{2}}\:\:\Rightarrow\:{lny}'=−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}} +\mathrm{2}\right) \\ $$$$\Rightarrow\:\:{y}'=\left({x}^{\mathrm{2}} +\mathrm{2}\right)^{−\mathrm{1}/\mathrm{2}} \\ $$$$\Rightarrow{y}=\int\left({x}^{\mathrm{2}} +\mathrm{2}\right)^{−\mathrm{1}/\mathrm{2}} {dx}={ln}\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}\right)+{Cte} \\ $$

Commented by mr W last updated on 28/Jan/20

sir: ((d(ln y′))/dx)=((y′′)/(y′))≠((y′′)/y) ((y′′)/y)=−(x/(x^2 +2)) ⇏ lny′=−(1/2)ln(x^2 +2)+C

$${sir}: \\ $$$$\frac{{d}\left(\mathrm{ln}\:{y}'\right)}{{dx}}=\frac{{y}''}{{y}'}\neq\frac{{y}''}{{y}} \\ $$$$\frac{{y}''}{{y}}=−\frac{{x}}{{x}^{\mathrm{2}} +\mathrm{2}}\:\:\nRightarrow\:{lny}'=−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}} +\mathrm{2}\right)+{C} \\ $$

Commented by abdomathmax last updated on 28/Jan/20

not correct

$${not}\:{correct} \\ $$

Commented by jagoll last updated on 28/Jan/20

how to get it?

$$\mathrm{how}\:\mathrm{to}\:\mathrm{get}\:\mathrm{it}? \\ $$

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