问题描述: 求微分方程的通解.[1+2e^(x/y)]dx+ 2e^(x/y)*[1-x/y]dy=0. 1个回答 分类:数学 2014-12-08 问题解答: 我来补答 令x/y=px=pyx'=p+p'y[1+2e^(x/y)]dx+ 2e^(x/y)*[1-x/y]dy=0[1+2e^(x/y)]dx/dy+ 2e^(x/y)*[1-x/y]=0(1+2e^p)(p+p'y)+2e^p*(1-p)=0p+p'y=-2e^p*(1-p)/(1+2e^p)p'y=-2e^p*(1-p)/(1+2e^p)-p=(-2e^p+2e^p*p-p-2e^p*p)/(1+2e^p)=(-2e^p-p)/(1+2e^p)(1+2e^p)/(2e^p+p)dp=-dy/yd(p+2e^p)/(2e^p+p)=-dy/yln(2e^p+p)=-lny+C12e^p+p=C/y反代即可 展开全文阅读