Question Number 100323 by bobhans last updated on 26/Jun/20

Solve x^2 y′′+2xy′−2y=0

Commented bybemath last updated on 26/Jun/20

since diff a power pushes down the  exponent by one unit, the form of  this eq suggest that we look for  possible solutions of the type y=x^n .  we optain the quadratic eq  n(n−1)+2n−2=0, this has roots  n=1; −2 so are solutions is   y = C_1 x + C_2 x^(−2)  ★

Answered by mathmax by abdo last updated on 27/Jun/20

am^2  +(b−a)m +c =0 with a =1 ,b=2 and c =−2 ⇒  m^2 +m −2 =0 →Δ =1+8 =9 ⇒m_1 =((−1+3)/2) =1 and m_2 =((−1−3)/2) =−2 ⇒  y =ax +b x^(−2)  =ax +(b/x^2 )

Commented bymathmax by abdo last updated on 27/Jun/20

sorry y =αx +β x^(−2)  =αx +(β/x^2 )