Question

微分方程 x(dx/dy)-y-根号(x^2+y^2)=0的通解

Answer 1

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)+1
yu'=
(√(u^2+1)+1)/u-u=
(√(u^2+1)+1-u^2)/u
udu/
(√(u^2+1)+1-u^2)=dy/y
du^2/
(√(u^2+1)+1-u^2)=2dy/y
积分∫dt/(√(t+1)+1-t)可令√(t+1)=z化成有理分式函数求解得:
∫dt/(√(t+1)+1-t)=
(-2/3)ln(√(t+1)+1)+(-4/3)ln(√(t+1)-2),代入得通解:
(-2/3)ln(√(u^2+1)+1)+(-4/3)ln(√(u^2+1)-2)=2lny+(-2/3)lnC
或通解:(√(u^2+1)(√(u^2+1)-2)^2=C/y^3
其中:u=x/y