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解:∵微分方程为(x²+2xy-y²)dx+(y²+2xy-x² )dy=0 ∴方程化为
dy/dx=(x²+2xy-y²)/(x²-2xy-y²)
∴设y=ux,有d(ux)/dx=
(1+2u-u²)/(1-2u-u²),
u+xdu/dx=(1+2u-u²)/(1-2u-u²),
xdu/dx=(1+u+u²+u³)/(1-2u-u²),
(u²+2u-1)du/(1+u+u²+u³)=-dx/x,
[-1/(u+1)+2u/(u²+1)]du=-dx/x,
ln|u²+1|-ln|u+1|=-ln|x|-ln|c|
(c为任意非零常数)
∴(u+1)/(u²+1)=cx,u+1=cx(u²+1),
x(u+1)=cx²(u²+1),得:
方程的通解为y+x=cy²+cx²
dy/dx=(x²+2xy-y²)/(x²-2xy-y²)
∴设y=ux,有d(ux)/dx=
(1+2u-u²)/(1-2u-u²),
u+xdu/dx=(1+2u-u²)/(1-2u-u²),
xdu/dx=(1+u+u²+u³)/(1-2u-u²),
(u²+2u-1)du/(1+u+u²+u³)=-dx/x,
[-1/(u+1)+2u/(u²+1)]du=-dx/x,
ln|u²+1|-ln|u+1|=-ln|x|-ln|c|
(c为任意非零常数)
∴(u+1)/(u²+1)=cx,u+1=cx(u²+1),
x(u+1)=cx²(u²+1),得:
方程的通解为y+x=cy²+cx²
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