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Question Number 100866 by mathmax by abdo last updated on 29/Jun/20

solve 3x^2 y^(′′) −2xy^′  +4y =0

$$\mathrm{solve}\:\mathrm{3x}^{\mathrm{2}} \mathrm{y}^{''} −\mathrm{2xy}^{'} \:+\mathrm{4y}\:=\mathrm{0} \\ $$

Answered by bramlex last updated on 29/Jun/20

let y = x^r  is solution .  y′ = rx^(r−1)  , y′′=r(r−1)x^(r−2)   ⇔3x^2 .{r(r−1)x^(r−2) }−2x(rx^(r−1) )+4x^r  =0  (3r^2 −3r)x^r −2rx^r +4x^r  =0  ⇔3r^2 −5r+4 = 0  r = ((5 ± (√(25−48)))/6) = ((5 ± i(√(23)))/6)  r = (5/6) ± ((i(√(23)))/6)  ∵ y =  { (x^((5/6)+((i(√(23)))/6)) ),(x^((5/6)−((i(√(23)))/6)) ) :}

$${let}\:{y}\:=\:{x}^{{r}} \:{is}\:{solution}\:. \\ $$$${y}'\:=\:{rx}^{{r}−\mathrm{1}} \:,\:{y}''={r}\left({r}−\mathrm{1}\right){x}^{{r}−\mathrm{2}} \\ $$$$\Leftrightarrow\mathrm{3}{x}^{\mathrm{2}} .\left\{{r}\left({r}−\mathrm{1}\right){x}^{{r}−\mathrm{2}} \right\}−\mathrm{2}{x}\left({rx}^{{r}−\mathrm{1}} \right)+\mathrm{4}{x}^{{r}} \:=\mathrm{0} \\ $$$$\left(\mathrm{3}{r}^{\mathrm{2}} −\mathrm{3}{r}\right){x}^{{r}} −\mathrm{2}{rx}^{{r}} +\mathrm{4}{x}^{{r}} \:=\mathrm{0} \\ $$$$\Leftrightarrow\mathrm{3}{r}^{\mathrm{2}} −\mathrm{5}{r}+\mathrm{4}\:=\:\mathrm{0} \\ $$$${r}\:=\:\frac{\mathrm{5}\:\pm\:\sqrt{\mathrm{25}−\mathrm{48}}}{\mathrm{6}}\:=\:\frac{\mathrm{5}\:\pm\:{i}\sqrt{\mathrm{23}}}{\mathrm{6}} \\ $$$${r}\:=\:\frac{\mathrm{5}}{\mathrm{6}}\:\pm\:\frac{{i}\sqrt{\mathrm{23}}}{\mathrm{6}} \\ $$$$\because\:{y}\:=\:\begin{cases}{{x}^{\frac{\mathrm{5}}{\mathrm{6}}+\frac{{i}\sqrt{\mathrm{23}}}{\mathrm{6}}} }\\{{x}^{\frac{\mathrm{5}}{\mathrm{6}}−\frac{{i}\sqrt{\mathrm{23}}}{\mathrm{6}}} }\end{cases} \\ $$

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