方程变形为dy/dx=x/y+y/x.令u=y/x,则y=xu,dy/dx=u++x*du/dx,所以原方程化为
u+x*du/dx=u+1/u.所以udu=dx/x.两边积分1/2*u^2=lnx+lnC.代入u=y/x得通解y^2=2x^2ln(Cx).
另外x≡0也是微分方程的解.
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