Question

设z=xφ(x+y)+yψ(x+y),其中φ,ψ具有二阶连续导数,验证x^2*x的二阶偏导-2*混合偏导(先x后y)+y

Answer 1

z=xφ(x+y)+yψ(x+y)

z‘x=φ+xφ'+yψ'

z'y=xφ'+ψ+yψ'

z‘'xx=2φ'+xφ''+yψ''

z'yy=xφ''+2ψ'+yψ''

z‘'xy=φ'+xφ''+ψ'+yψ''

所以：z‘'xx-2z‘'xy+z'yy

=(2φ'+xφ''+yψ'')-2(φ'+xφ''+ψ'+yψ'')+(xφ''+2ψ'+yψ'')

=(2-2)φ'+(x-2x+x)φ''+(2-2)ψ'+(y-2y+y)ψ''=0

看来,你那验证x^2.中x^2是多的啦