求助一道有关偏导数微分方程的题~求高手解答
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求助一道有关偏导数微分方程的题~求高手解答
![求助一道有关偏导数微分方程的题~求高手解答](/uploads/image/z/5000624-8-4.jpg?t=%E6%B1%82%E5%8A%A9%E4%B8%80%E9%81%93%E6%9C%89%E5%85%B3%E5%81%8F%E5%AF%BC%E6%95%B0%E5%BE%AE%E5%88%86%E6%96%B9%E7%A8%8B%E7%9A%84%E9%A2%98%7E%E6%B1%82%E9%AB%98%E6%89%8B%E8%A7%A3%E7%AD%94)
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
r^2-4=0
r=-2,2
f=Ae^(2u)+Be^(-2u)
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
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
r^2-4=0
r=-2,2
f=Ae^(2u)+Be^(-2u)
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