Question

微分方程求导

Answer 1

Answer 2

y=ln(xy)
y'=y/(xy) + xy'/(xy)=1/x + y'/y
y'(1-1/y)=1/x
y'=y/[x(y-1)]=y/(xy-x)
y''=[y'(xy-x)-y(xy-x)']/(xy-x)²
=[y'(xy-x)-y(y+xy'-1)]/(xy-x)²
=(-xy'-y²+y)/(xy-x)²
=[-xy/(xy-x) -y²+y]/(xy-x)²
=[-xy-xy(y-1)²]/(xy-x)³
=[xy(-y²+2y-2)]/(xy-x)³

代入原方程
(xy-x)y''+xy'²+yy'-2y'
=[xy(-y²+2y-2)]/(xy-x)² +xy²/(xy-x)² + y²/(xy-x) -2y/(xy-x)
=[xy(-y²+2y-2)+xy²+y²(xy-x)-2y(xy-x)]/(xy-x)²
=(-xy³+2xy²-2xy+xy²+xy³-xy²-2xy²+2xy)/(xy-x)²
=0