多元微积分问题
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多元微积分问题
![](http://img.wesiedu.com/upload/6/49/6494ae935ec8596c422eee8432cfd745.jpg)
![](http://img.wesiedu.com/upload/6/49/6494ae935ec8596c422eee8432cfd745.jpg)
![多元微积分问题](/uploads/image/z/19549116-36-6.jpg?t=%E5%A4%9A%E5%85%83%E5%BE%AE%E7%A7%AF%E5%88%86%E9%97%AE%E9%A2%98)
(i) Zxx=d^2z/dx^2,Zyy=d^2y/dy^2
z(x,y)=f(u)=f(e^x*siny)
dz/dx=f'(u)*du/dx=f'(u)*e^x*siny
Zxx=d^2z/dx^2=f''(u)e^xsiny*e^x*siny+f'(u)*e^x*siny
dz/dy=f'(u)*du/dy=f'(u)*e^x*cosy
Zyy=d^2z/dy^2=f''(u)e^xcosy*cosy*e^x+f'(u)(-siny)*e^x
(ii) Zxx+Zyy=e^(2x)*Z
{f''(u)e^xsiny*e^x*siny+f'(u)*e^x*siny}
+{f''(u)e^xcosy*cosy*e^x+f'(u)(-siny)*e^x}
=e^(2x)*Z=e^(2x)*f(u)
=> f''(u)=f(u) => f''(u)-f(u)=0
此为二阶齐次微分方程,特征方程为 r^2-1=0
特征根为 r1,2=±1
∴其通解为 f(u)=C1e^u+C2e^(-u)
f'(u)=C1e^u-C2e^(-u)
带入初始条件f(0)=0,f'(0)=1,可得
f(0)=C1+C2=0,f'(0)=C1-C2=1
可解得 C1=1/2,C2=-1/2
∴f(u)=1/2*e^u-1/2*e^(-u)
z(x,y)=f(u)=f(e^x*siny)
dz/dx=f'(u)*du/dx=f'(u)*e^x*siny
Zxx=d^2z/dx^2=f''(u)e^xsiny*e^x*siny+f'(u)*e^x*siny
dz/dy=f'(u)*du/dy=f'(u)*e^x*cosy
Zyy=d^2z/dy^2=f''(u)e^xcosy*cosy*e^x+f'(u)(-siny)*e^x
(ii) Zxx+Zyy=e^(2x)*Z
{f''(u)e^xsiny*e^x*siny+f'(u)*e^x*siny}
+{f''(u)e^xcosy*cosy*e^x+f'(u)(-siny)*e^x}
=e^(2x)*Z=e^(2x)*f(u)
=> f''(u)=f(u) => f''(u)-f(u)=0
此为二阶齐次微分方程,特征方程为 r^2-1=0
特征根为 r1,2=±1
∴其通解为 f(u)=C1e^u+C2e^(-u)
f'(u)=C1e^u-C2e^(-u)
带入初始条件f(0)=0,f'(0)=1,可得
f(0)=C1+C2=0,f'(0)=C1-C2=1
可解得 C1=1/2,C2=-1/2
∴f(u)=1/2*e^u-1/2*e^(-u)