参考
y[x]= -(C[1]/(2 x^2)) + C[2]
let
u= x^3.y'
du/dx = x^3.y'' +3x^2.y'
y''= [du/dx - (3/x)u] /x^3
//
xy''+3y'=0
x{ [du/dx - (3/x)u] /x^3 } + 3u/x^3 =0
x.du/dx=0
u= ∫ dx/x
= lnx + C1
x^3. dy/dx = lnx + C1
dy/dx = (lnx + C1)/x^3
y= ∫ (lnx + C1)/x^3 dx
= -(C1/2)(1/x^2) + ∫ lnx /x^3 dx
= -(C1/2)(1/x^2) -(1/2)∫ lnx d(1/x^2)
= -(C1/2)(1/x^2) -(1/2)( lnx/x^2 ) +(1/2)∫ dx/x^3
= -(C1/2)(1/x^2) -(1/2)( lnx/x^2 ) -(1/4)(1/x^2) +C2
= k1.(1/x^2) -(1/2)( lnx/x^2 ) +C2
淘宝:精油基地