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

solve 3x^2 y^(′′) −2xy^′  +4y =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)) ) :}