Question

设函数F(x,y)有连续的偏导数,且F(x,y)(ydx+xdy)是某个函数U(x,y)的全微分,则F(x,y)应满足

Answer 1

F(x,y)(ydx+xdy)=dU(x,y),

yF(x,y)dx+xF(x,y)dy=(δU/δx)dx+(δU/δy)dy,

δU/δx=yF(x,y),δU/δy=xF(x,y),

δ^2U/δxδy=δ[yF(x,y)]/δy=F(x,y)+yδF/δy,

δ^2U/δyδx=δ[xF(x,y)]/δx=F(x,y)+xδF/δx,

δ^2U/δxδy=δ^2U/δyδx,

F(x,y)+yδF/δy=F(x,y)+xδF/δx,

yδF/δy=xδF/δx.