数学分析考研题1.设z为x,y的可微函数,试将方程(x^2)*(z对x的偏导数)+(y^2)*(z对y的偏导数)=z^2

1个回答

  • du/dy

    =1/(dy/du)

    =(1+uv)^2

    dv/dy

    =1/(dy/dv)

    =-(1+uv)^2/u^2

    dz/dx

    =dz/du*du/dx+dz/dw*dw/du*du/dx

    =[(1+uw)-u(w+udw/du)]/(1+uw)^2*1-u^2/(1+uw)^2*dw/du*1

    =1/(1+uw)^2

    dz/dy

    =dz/du*du/dy+dz/dw*(dw/du*du/dy+dw/dv*dv/dy)

    =[(1+uw)-u(w+udw/du)]/(1+uw)^2*(1+uv)^2

    -u^2/(1+uw)^2*((1+uv)^2*dw/du-(1+uv)^2/u^2*dw/dv)

    x^2*dz/dx+y^2*dz/dy

    =x^2*[1/(1+uw)^2]+y^2*{[(1+uw)-u(w+udw/du)]/(1+uw)^2*(1+uv)^2-u^2/(1+uw)^2*((1+uv)^2*dw/du-(1+uv)^2/u^2*dw/dv)}

    =u^2*[1/(1+uw)^2]+[u/(1+uv)^2]^2*{[(1+uw)-u(w+udw/du)]/(1+uw)^2*(1+uv)^2-u^2/(1+uw)^2*((1+uv)^2*dw/du-(1+uv)^2/u^2*dw/dv)}=...=u^2/(1+uw)^2

    =>(1-u^2*dw/du)=u^2*dw/du-dw/dv

    =>2u^2*dw/du-dw/dv=1