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yDx 2y

解:∵e^ydx+(xe^y-2y)dy=0 ==>(e^ydx+xe^ydy)-2ydy=0 ==>d(xe^y)-d(y^2)=0 ==>∫d(xe^y)-∫d(y^2)=0 (积分) ==>xe^y-y^2=C (C是任意常数) ∴此方程的通解是xe^y-y^2=C。

用格林公式:奇点(0,0)不在积分域内. I = ∮L (ydx - xdy)/(x^2 + y^2) = ∫∫D [(x^2 - y^2)/(x^2 + y^2)^2 - (x^2 - y^2)/(x^2 + y^2)^2] dxdy = 0 用参数方程. { x = 1 + cost、dx = - sint dt { y = 1 + sint、dy = cost dt 0 ≤ t ≤ 2π ∮L (ydx...

猜[2ysin(x/y)+ 3xcos(x/y)]dy-3ycos(x/y)dx=0, 两边都除以y,得[2sin(x/y)+(3x/y)cos(x/y)]dy-3cos(x/y)dx=0, 设x=uy,则dx=ydu+udy,上式变为 [2sinu+3ucosu]dy-2cosu(ydu+udy)=0, 整理得(2sinu+ucosu)dy=2ycosudu, 分离变量得dy/y=2cosudu/(2s...

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