Question

一道高数题求助在线等在线采纳

求解这个怎么解出来的

Answer 1

let
u=y'
xy''-y'+2 =0
xu' -u +2 =0
xu' = u-2
∫du/(u-2) = ∫dx/x
ln|u-2| =ln|x| + C'
u-2 = C1'.x
dy/dx = C1'.x +2
y =∫ [C1'.x +2] dx
= (1/2)C1'x^2 +2x + C2
=C1.x^2 +2x + C2