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Question Number 103411 by abdomsup last updated on 14/Jul/20
solve x^2 y^(′′)  +xy^′  +y =0
$${solve}\:{x}^{\mathrm{2}} {y}^{''} \:+{xy}^{'} \:+{y}\:=\mathrm{0} \\ $$
Answered by bemath last updated on 14/Jul/20
set y=x^r   y′=rx^(r−1) ; y′′=r(r−1)x^(r−2)   ⇒x^2 (r^2 −r)x^(r−2) +x(rx^(r−1) )+x^r =0  x^r (r^2 −r+r+1)=0  ⇒r^2 +1=0 , r=±i  y=x^(± i)
$${set}\:{y}={x}^{{r}} \\ $$$${y}'={rx}^{{r}−\mathrm{1}} ;\:{y}''={r}\left({r}−\mathrm{1}\right){x}^{{r}−\mathrm{2}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} \left({r}^{\mathrm{2}} −{r}\right){x}^{{r}−\mathrm{2}} +{x}\left({rx}^{{r}−\mathrm{1}} \right)+{x}^{{r}} =\mathrm{0} \\ $$$${x}^{{r}} \left({r}^{\mathrm{2}} −{r}+{r}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{r}^{\mathrm{2}} +\mathrm{1}=\mathrm{0}\:,\:{r}=\pm{i} \\ $$$${y}={x}^{\pm\:{i}} \\ $$
Answered by OlafThorendsen last updated on 15/Jul/20
y = x^m   y′ = mx^(m−1)   y′′ = m(m−1)x^(m−2)   ⇒ m(m−1)+m+1 = 0  m^2 +1 = 0  m = ±i  ⇒ y = C_1 x^i +C_2 x^(−i)   y = C_1 e^(ilnx) +C_2 e^(−ilnx)   y = (C_1 +C_2 )cos(lnx)+(C_1 −C_2 )isin(lnx)  C_1 −C_2  = 0 ⇒ C_1  = C_2   Finally y = Kcos(lnx)
$${y}\:=\:{x}^{{m}} \\ $$$${y}'\:=\:{mx}^{{m}−\mathrm{1}} \\ $$$${y}''\:=\:{m}\left({m}−\mathrm{1}\right){x}^{{m}−\mathrm{2}} \\ $$$$\Rightarrow\:{m}\left({m}−\mathrm{1}\right)+{m}+\mathrm{1}\:=\:\mathrm{0} \\ $$$${m}^{\mathrm{2}} +\mathrm{1}\:=\:\mathrm{0} \\ $$$${m}\:=\:\pm{i} \\ $$$$\Rightarrow\:{y}\:=\:\mathrm{C}_{\mathrm{1}} {x}^{{i}} +\mathrm{C}_{\mathrm{2}} {x}^{−{i}} \\ $$$${y}\:=\:\mathrm{C}_{\mathrm{1}} {e}^{{i}\mathrm{ln}{x}} +\mathrm{C}_{\mathrm{2}} {e}^{−{i}\mathrm{ln}{x}} \\ $$$${y}\:=\:\left(\mathrm{C}_{\mathrm{1}} +\mathrm{C}_{\mathrm{2}} \right)\mathrm{cos}\left(\mathrm{ln}{x}\right)+\left(\mathrm{C}_{\mathrm{1}} −\mathrm{C}_{\mathrm{2}} \right){i}\mathrm{sin}\left(\mathrm{ln}{x}\right) \\ $$$$\mathrm{C}_{\mathrm{1}} −\mathrm{C}_{\mathrm{2}} \:=\:\mathrm{0}\:\Rightarrow\:\mathrm{C}_{\mathrm{1}} \:=\:\mathrm{C}_{\mathrm{2}} \\ $$$$\mathrm{Finally}\:{y}\:=\:\mathrm{Kcos}\left(\mathrm{ln}{x}\right) \\ $$

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