证明积分f^(23))(x+2y)dx+(2x-y)dy 在整个f^(23))(x+2y)dx+(2x-y)dy平面上与路线无关，并计算积分值

+2y)dx+(2x+3(y)^2)dy在整个xoy平面内是某一函数+2y)dx+(2x+3(y)^2)dy的全微分，则+2y)dx+(2x+3(y)^2)dy

已知曲线积分(int )_(t)^yf(x)dx+((x)^2+y)dy与路径无关，则(int )_(t)^yf(x)dx+((x)^2+y)dy_____.已

计算 =(int )_(L)dfrac ((x-y)dx+(x+y)dy)({x)^2+(y)^2}-|||-f[(x-y)(x+(x+y)dy)/(x^2+x

(2)(e^x+y-e^x)dx+(e^x+y+e^y)dy=0;(2)$(e^{x+y}-e^{x})dx+(e^{x+y}+e^{y})dy=0;$

[单选题]方程（x2+y）dx+（x-2y）dy=0的通解是（）．A . B . C . D .

(3)(x^2+2xy-y^2)dx+(y^2+2xy-x^2)dy=0,y|_(x=1)=1;(3)$(x^{2}+2xy-y^{2})dx+(y^{2}+2

微分方程((a)^x+y-(a)^x)dx+((a)^x+y+(a)^y)dy=0的通解为( )((a)^x+y-(a)^x)dx+((a)^x+y+(a)^y

(2)(x^2+2xy-y^2)dx+(y^2+2xy-x^2)dy=0, y|_(x=1)=1.(2)$(x^{2}+2xy-y^{2})dx+(y^{2}+

6.求下列全微分的原函数:-|||-(1) ((x)^2+2xy-(y)^2)dx+((x)^2-2xy-(y)^2)dy-|||-(2) ^x[ e(x-y+

dfrac (dy)(dx)=(x)^2+(y)^2 B . dfrac (dy)(dx)=(x)^2+(y)^2 C .dfrac (dy)(dx)=(x)^