设f(u)有一阶连续偏导数,f(0)=2,且z=xf(y/x)+yf(y/x)满足∂z/∂x+∂z/∂y=y/x,求z的表达式设f(u)有一阶连续偏导数,f(0)=2,且z=xf(y/x)+yf(y/x)满足∂z/∂x+∂z/∂y=y/x,求z的表达式

Question

设f(u)有一阶连续偏导数,f(0)=2,且z=xf(y/x)+yf(y/x)满足∂z/∂x+∂z/∂y=y/x,求z的表达式
设f(u)有一阶连续偏导数,f(0)=2,且z=xf(y/x)+yf(y/x)满足∂z/∂x+∂z/∂y=y/x,求z的表达式

Answer 1

z=xf(y/x)+yf(y/x)=(x+y)f(y/x) u=y/x
∂z/∂x=(x+y)f'(u)(-y/x^2)+f(u)
∂z/∂y=(x+y)f'(u)(1/x)+f(u)
(x+y)f'(u)(-y/x^2)+f(u)+(x+y)f'(u)(1/x)+f(u)=u
f'(u)(-u^2-u+1+u)+2f(u)=u
f'(u)+2f(u)/(1-u^2)=u/(1-u^2)
这是一阶线性方程,由通解公式：
f(u)=e^(-∫2du/(1-u^2))(C+∫u/(1-u^2)*e^(∫2du/(1-u^2))du
=(u-1)/(u+1)*(C+∫（u/(1-u^2))*(u+1)/(u-1)*du
=(u-1)/(u+1)*(C-∫u/(u-1)^2)*du)
=(u-1)/(u+1)*(C-ln|u-1|+1/(u-1))
由f(0)=2,代入：C=-1
z=xf(y/x)+yf(x/y)=(x+y)f(y/x)
=（x+y){(y/x-1)/(y/x+1)*(-1-ln(y/x+1)-1/(y/x-1))}
=（y-x)(-1-ln|y/x+1|-x/(y-x))