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y'-y/x-√[1-(y/x)^2]=1 , y(1)=1令u=y/x则y=xu, y'=u+xu'代入原方程得：u+xu'-u-√(1-u^2)=1xu'=1+√(1-u^2)du/[1+√(1-u^2)]=dx/x再令u=sint, 则du=cost dt,代入上式得：costdt/[1+cost]=dx/xdt[1-1/(1+cost)]=dx/x积分：t-∫dt/(1+cost)=lnx+c1t-∫sec^2(t/2)d(t/2)=lnx+c1t-tan(t/2)=lnx+c1arcsinu-tan[(arcsinu)/2]=lnx+c1将x=1,y=1, u=1代入上式得：c1=π/2-1... |
