多元微积分问题

1个回答

  • (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)