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(2ysinx/y 3xCosx/y)Dy%(3yCosx/y)Dx=0

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...

你好,满意请采纳哦! P(x,y)=2xy^3-y^2cosx,Q(x,y)=1-2ysinx+3x^2y^2 αP/αy=αQ/αx=6xy^2-2ycosx 因此本题积分与路径无关,可自选积分路线

解:∵y(3+ycosx)dx+(3x+2ysinx)dy=0 ==>3(ydx+xdy)+(y^2cosxdx+2ysinxdy)=0 ==>3d(xy)+d(y^2sinx)=0 ==>∫3d(xy)+∫d(y^2sinx)=0 ==>3xy+y^2sinx=C (C是常数) ∴此方程的通解是3xy+y^2sinx=C。

P=2xy³-y²cosx,Q=1-2ysinx+3x²y² 易验证:∂Q/∂x=∂P/∂y=6xy²-2ycosx 因此本题积分与路径无关,可自选积分路线 选从(0,0)到(π/2,1)的折线, L1:y=0,x:0--->π/2 L2:x=π/2,y:0--->1 则原...

P(x,y)=2xy^3-y^2cosx,Q(x,y)=1-2ysinx+3x^2y^2αP/αy=αQ/...π/2) (0-0)dx +∫(0~1) (1-2ysin(π/2)+3(π/2)^2y^2)dy...

若原函数为f(x,y)=x+x2y3-y2sinx+y+c,则分别对x,y求偏导时,df(x,y)=(1+2xy3-y2cosx)dx+(1-2ysinx+3x2y2)dy,所以A选项正确.而f(x,y)=y-y2sinx+x2y3+c,df(x,y)=(-y2cosx+2xy3)dx+(1-2ysinx+3x2y2)dy,所以选项B错误.而...

证明:由题意,P=2xcosy-y2sinx,Q=2ycosx-x2siny,在整个平面上具有一阶连续偏导数,且?P?y=?2xsiny?2ysinx=?Q?x∴曲线积分I与积分路径无关.取路径从(0,0)到(2,0)再到(0,3),则I=∫202xdx+∫30(2ycos2?4siny)dy=4+9cos2+4cos3-4=9cos...

e^(x+y)-3x=2y^2 -5=0 (1+dy/dx) e^(x+y)-3=4ydy/dx dy/dx = (e^(x+y)-3)/(4y- e^(x+y))

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