微分方程 x(dx/dy)-y-根号(x^2+y^2)=0的通解

x(dx/dy)-y-√(x^2+y^2)=0,除以y:(x/y)(dx/dy)-1-√((x/y)^2+1)=0令x/y=u ,代入：u(u+yu')=√(u^2+1)+1yu'= (√(u^2+1)+1)/u-u= (√(u^2+1)+1-u^2)/uudu/ (√(u^2+1)+1-u^2)=dy/ydu^2/ (√(u^2+1)+1-u^2)=2dy/y积分...