Question

求助一道有关偏导数微分方程的题~求高手解答

Answer 1

zx=f'(e^xcosy)*e^xcosy
zxx=f''(e^xcosy)(e^xcosy)^2+f'(e^xcosy)*e^xcosy
zy=f'(e^xcosy)*e^x(-siny)
zyy=f''(e^xcosy)*[e^x(-siny)]^2+f'(e^xcosy)*e^x(-cosy)
zxx+zyy=f''(e^xcosy)(e^xcosy)^2+f'(e^xcosy)*e^xcosy+f''(e^xcosy)*[e^x(siny)]^2-f'(e^xcosy)*e^xcosy
=f''(e^xcosy)e^(2x)
=4[f(e^xcosy)+e^xcosy]e^(2x)
f''(u)=4f(u)+4u
f''-4f=4u
齐次解：r^2-4=0
r=-2,2
f=Ae^(2u)+Be^(-2u)
非齐次解：f=-u 因为右端是一次函数
f=Ae^(2u)+Be^(-2u)-u
u=0,f=A+B=0
f'=2Ae^(2u)-2Be^(-2u)-1
u=0,f'=2A-2B-1=0
A=1/4,B=-1/4
所以
f(u)=(1/4)e^(2u)-(1/4)e^(-2u)-u