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(x³;+y³;)dx-3xy²;dy=0,齊次方程的通解?dy/dx=(x³;+y³;)/3xy²;=(1/3)[(x/y)²;+(y/x)]=(1/3)[1/(y/x)²;+(y/x)]令y/x=u,則y=ux,dy/dx=u+x(du/dx),代入上式得:u+x(du/dx)=(1/3)[(1/u&…
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