高数求证设f(x,y,z)是可微函数,且对任意实数t>0 有f(tx,ty,tz)=t^k f(x,y,z),证明：f(x,y,z)满足 xfx’＋yfy’＋zfz’＝kf(x,y,z)
2019-04-14
高数求证
设f(x,y,z)是可微函数,且对任意实数t>0 有f(tx,ty,tz)=t^k f(x,y,z),
证明：f(x,y,z)满足 xfx’＋yfy’＋zfz’＝kf(x,y,z)
优质解答
令 g(t) = f(tx,ty,tz)=t^k f(x,y,z),
dg/dt = xf'tx + yf'ty + zf'tz
= kt^(k-1) f(x,y,z)
取t=1,xfx’＋yfy’＋zfz’＝kf(x,y,z)
令 g(t) = f(tx,ty,tz)=t^k f(x,y,z),
dg/dt = xf'tx + yf'ty + zf'tz
= kt^(k-1) f(x,y,z)
取t=1,xfx’＋yfy’＋zfz’＝kf(x,y,z)