作业帮求积分,dy/dx=(x-y-1)/(x+y+1)..方程变为dY/dX=(X-Y)/(X+Y),令u=(Y/X),代入,得(1+u)du/(1-2u-u^2)=-dX/X,积分得1-2u-u^2=Cx^2.我不懂：u=(Y/X),代入,得(1+u)du/(1-2u-u^2)=-dX/dX,具体怎么代入得到?
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求积分,dy/dx=(x-y-1)/(x+y+1)..方程变为dY/dX=(X-Y)/(X+Y),令u=(Y/X),代入,得(1+u)du/(1-2u-u^2)=-dX/X,积分得1-2u-u^2=Cx^2.我不懂：u=(Y/X),代入,得(1+u)du/(1-2u-u^2)=-dX/dX,具体怎么代入得到?
求积分,dy/dx=(x-y-1)/(x+y+1)
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方程变为dY/dX=(X-Y)/(X+Y),令u=(Y/X),代入,得(1+u)du/(1-2u-u^2)=-dX/X,积分得1-2u-u^2=Cx^2.
我不懂：u=(Y/X),代入,得(1+u)du/(1-2u-u^2)=-dX/dX,具体怎么代入得到?
求积分,dy/dx=(x-y-1)/(x+y+1)..方程变为dY/dX=(X-Y)/(X+Y),令u=(Y/X),代入,得(1+u)du/(1-2u-u^2)=-dX/X,积分得1-2u-u^2=Cx^2.我不懂：u=(Y/X),代入,得(1+u)du/(1-2u-u^2)=-dX/dX,具体怎么代入得到?
y=ux,dy=udx+xdu,dy/dx=u+xdu/dx=(1-u)/(1+u),xdu/dx=(1-2u-u^2)/(1+u),还有不懂吗
把y=ux两边微分就可以了
设x-y-1=u x+y+1=v
x=(u+v)/2 y=(v-u-2)/2
dy/dx=(dv-du)/(du+dv)=u/v
udu+udv=vdv-vdu
2udu+2udv=2vdv-2vdu
d(u^2-v^2)=d(2uv)
u^2-v^2=2uv+C0
通解(x-y-1)^2-(x+y+1)^2=2(x-y-1)(x+y+1)+C0
u=(Y/X)
Y=U*X
dy/dx=du/dx*x+u;
带入dY/dX=(X-Y)/(X+Y)
du/dx*x+u=(1-u)/(1+u)
化简之后就是结果
∫ P dx Q dy 要证明此种积分与路径无关，只需证 Q/ x= P/ y 令P=x y,Q=x-y，则 Q/ x=1= P/ y ∴曲线积分与路径无关（在整个xoy面