xy'-y-√(x^2+y^2)=0的通解

来源：学生作业学帮网编辑：学帮网时间：2024/06/04 15:33:21

∵xy'-y-√(x²+y²)=0 ==>xy'-y=√(x²+y²)
==>√(1+(y/x)²)/x=(xy'-y)/x²
==>√(1+(y/x)²)/x=d(y/x)/dx
==>dx/x=d(y/x)/√(1+(y/x)²)
==>ln│x│=arctan(y/x)+ln│C│ (C是积分常数)
==>x=Ce^(arctan(y/x))
∴原方程的通解是x=Ce^(arctan(y/x)) (C是积分常数).