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yDx xDy x 2 y 2 Dx

令y=xu 则y'=u+xu' x(u+xu')-xu=√(x²+x²u²) xu'=√(1+u²) du/√(1+u²)=dx/x 积分:ln(u+√(1+u²))=ln|x|+C1 得u+√(1+u²)=Cx 即y²+√(x²+y²)=Cx²

y dx - x dy = (x² + y²) dx y - x • dy/dx = x² + y² y' = y/x - y²/x - x (令y = - xv,y' = - (xv' + v) = - xv' - v) - xv' - v = - v - xv² - x - v' = - v² - 1 dv/(v² + 1) = 1 arctan(v)...

令y=xu 则y'=u+xu' x(u+xu')-xu=√(x²+x²u²) xu'=√(1+u²) du/√(1+u²)=dx/x 积分:ln(u+√(1+u²))=ln|x|+C1 得u+√(1+u²)=Cx 即y²+√(x²+y²)=Cx²

(ydx-xdy)/(x^2+y^2) = y/(x^2+y^2) dx - x/(x^2+y^2) dy 假设该函数存在, 令该函数 = f(x) = z , 则 dz/dx = y/(x^2+y^2) 1/y dz = 1/(x^2+y^2) dx z/y = 1/y arctan (x/y) + C1(y) z1 =arctan (x/y) + y* C1(y) 同理,dz/dy = -x /(x^2+y^2) z2...

方法为格林公式,但是注意原来的被积函数在L围成的区域中包含奇点(0,0),所以需要补上曲线L1以挖空奇点,参考解法:

解:∵x/ydy-1/ydx=(2+y)/(1-y-y^2)dx ==>(xdy-dx)(1-y-y^2)=y(2+y)dx ==>x(y^2+y-1)dy+(y+1)dx=0 ==>(y^2+y-1)dy/(y+1)+dx/x=0 ==>(y-1/(y+1))dy+dx/x=0 ==>∫(y-1/(y+1))dy+∫dx/x=0 ==>y^2/2-ln│y+1│+ln│x│=ln│C│ (C是非零常数) ==>xe^(y^2/2)/(...

因为l是算得空白部分

xdy-ydx=√(x²+y²)dx xdy=[√(x²+y²)+y]dx dy/dx=√[1+(y/x)²]+y/x 设y/x=u u+xdu/dx=√(1+u²)+u du/√(1+u²)=dx/x arctanu=lnx+C 即arctan(y/x)=lnx+C

(xdy-ydx)/x^2=(1+(y/x)^2)xdx d(y/x)=(1+(y/x)^2)xdx d(y/x)/(1+(y/x)^2)=xdx 两边积分:arctan(y/x)=x^2/2+C y/x=tan(x^2/2+C) y=xtan(x^2/2+C)

有个简单的解法: xdy-ydx=y^2dy变形:(xdy-ydx)/y^2=dy 由于:d(x/y)=(ydx-xdy)/y^2 故:d(x/y)=-dy 通解为:x/y=-y+C 或:x=y(C-y)

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