求xy'+y=y(lnxy+lny)的通解

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令xy=u;
=>
u'=u/x (2lnu-lnx);

v=lnu;
=>
v'=2v/x-lnx/x;

令 v=C(x)x^2;
=>
C'(x)x^2=-lnx/x;
C'(x)=-lnx /x^3;
C(x)=C+1/x^2 *(1/4+1/2 *Ln[x])

==>y=D/x*Exp[1/x^2*(1/4+1/2*Ln[x])];
D is constant
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