solve 3x^2 y^(′′) −2xy^′ +4y =0


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

$solve {3 x}^{2} y^{^{″}} - {2 xy}^{^{'}} + 4 y = 0$
	Answered by bramlex last updated on 29/Jun/20
	$let y = x^r is solution . y′ = rx^(r−1) , y′′=r(r−1)x^(r−2) ⇔3x^2 .{r(r−1)x^(r−2) }−2x(rx^(r−1) )+4x^r =0 (3r^2 −3r)x^r −2rx^r +4x^r =0 ⇔3r^2 −5r+4 = 0 r = ((5 ± (√(25−48)))/6) = ((5 ± i(√(23)))/6) r = (5/6) ± ((i(√(23)))/6) ∵ y = { (x^((5/6)+((i(√(23)))/6)) ),(x^((5/6)−((i(√(23)))/6)) ) :}$
	$l e t y = x^{r} i s s o l u t i o n .$ $y^{'} = {r x}^{r - 1}, y^{″} = r (r - 1) x^{r - 2}$ $\Leftrightarrow 3 x^{2} . {r (r - 1) x^{r - 2}} - 2 x ({r x}^{r - 1}) + 4 x^{r} = 0$ $(3 r^{2} - 3 r) x^{r} - 2 {r x}^{r} + 4 x^{r} = 0$ $\Leftrightarrow 3 r^{2} - 5 r + 4 = 0$ $r = \frac{5 \pm \sqrt{25 - 48}}{6} = \frac{5 \pm i \sqrt{23}}{6}$ $r = \frac{5}{6} \pm \frac{i \sqrt{23}}{6}$ $∵ y = {\begin{cases} x^{\frac{5}{6} + \frac{i \sqrt{23}}{6}} \\ x^{\frac{5}{6} - \frac{i \sqrt{23}}{6}} \end{cases}$
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