高等数学偏导数

2024-11-17 06:34:36
推荐回答(1个)
回答1:

记 p=x/y, q=y/x, 则 u=yf(x/y)+xf(y/x)=yf(p)+xf(q),
u'=f'

+f(q)-(y/x)f', u'=f(p)-(x/y)f'

+f',
u''=(-y/x^2)f'(q)+(1/y)f''+(y/x^2)f'+(y^2/x^3)f''
=(1/y)f''+(y^2/x^3)f''
u''=(-x/y^2)f'(p)+(1/x)f''+(x/y^2)f'

+(x^2/y^3)f''
=(1/x)f''+(x^2/y^3)f''
x^2*u-y^2*u
=(x^2/y)f''+(y^2/x)f'' - (y^2/x)f''- (x^2/y)f''=0.
选D。