>>3

定石によれば　y/x=u とおき、
　A + 2Bu + Cuu = 1/xx,
両辺をu で3回微分して
　0 = DDD (1/xx),
ここで、
　D = d/du = (dx/du)(d/dx) = {xx/(xy'-y)}(d/dx)

D (1/xx) = -2/{x(xy'-y)}
DD (1/xx) = 2(xxy''+xy'-y)/(xy'-y)^3
DDD (1/xx) = -2(x^4)(3x(y'')^2+y'''(y-xy'))/(xy'-y)^5

3x(y'')^2 + y'''(y-xy') = 0.

[微分積分スレ.249 260]