設(shè)y/x=t,則y=xt,dy=xdt+tdx
∵(y²+x²)dy-xydx=0 ==>(y/x+x/y)dy-dx=0
==>(t+1/t)(xdt+tdx)=dx
==>x(t²+1)dt/t+(t²+1)dx=dx
==>x(t²+1)dt/t+t²dx=0
==>(1/t+1/t³)dt+dx/x=0
==>ln│t│-1/(2t²)+ln│x│=ln│C│ (C是積分常數(shù))
==>xt=Ce^(1/(2t²))
==>y=Ce^(x²/(2y²)) (用t=y/x代換)
∴原微分方程的通解是y=Ce^(x²/(2y²)) (C是積分常數(shù)).