xy'+y=y(lny+lnx)
xu'e^u+e^u=e^u(u+lnx)
u'-u/x=(lnx-1)/x
为一阶线性微分方程，根据公式
u=Ce^∫dx/x)*[1+∫(lnx-1)/x*e^(-∫dx/x)*dx]
=Cx[1+∫(lnx-1)/x*1/x)dx]
∫(lnx-1)/x*1/x)dx=∫lnx/x^2*dx-∫dx/x^2=-∫lnx/*d(1/x)+1/x
=-lnx/x+∫dx/x^2+1/x=-lnx/x
所以u=Cx(1-lnx/x)=Cx-Clnx
即lny=Cx-Clnx
y=e^(Cx)/x^C
xy'/y+1=lny+lnx
令t=lny
方程化为xt'+1=t+lnx
即(xt'-t)/(x^2)=(lnx-1)/(x^2)
积分，有t/x=-lnx/x+C
那么，y=(Ce^x)/x本回答被提问者采纳
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