利用曲线积分,求微分表达式的原函数 (x^2+2xy-y^2)dx+(x^2-2xy-y^2)dy

Question

如题所述

Answer 1

设p(x,y)=x²+2xy+y²,q(x,y)=x²-2xy-y²
因为∂p/∂y=2x+2y,∂q/∂x=2x-2y,只有当y=0时才成立∂p/∂y=∂q/∂x,所以所求函数的定义域为整个x轴
设du=(x²+2xy+y²)dx+(x²-2xy-y²)dy,要求出u,因为y=0,dy=0,那么全微分只剩下du=x²dx这一部分,所以u=x³/3+c

Answer 2

(x^2+2xy-y^2)dx+(x^2-2xy-y^2)dy
P=(x^2+2xy-y^2)
Q=(x^2-2xy-y^2)
Py=Qx，积分与路径无关
z(x，y)=∫(x^2+2xy-y^2)dx+(x^2-2xy-y^2)dy
=x^3/3-y^3/3+∫(2xydx+x^2dy)-∫(y^2dx+2xydy
=x^3/3-y^3/3+x^2y-y^2x+C