Question

设y=y(x)由方程xe^f(u)=e^y确定,其中f的二阶可导,且f'≠1求d^2(y)/dx^2

Answer 1

xe^f(u)=e^y
x=e^[y-f(u)]
1=e^[y-f(u)][y'-f'(u)u']
y'=e^[f(u)-y]+f'(u)u'
y''={e^[f(u)-y]+f'(u)u'}
=e^[f(u)-y][f'(u)u'-y']+f''(u)（u'）^2+f'(u)u