Question Number 103411 by abdomsup last updated on 14/Jul/20

solve x^2 y^(′′)  +xy^′  +y =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)

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)