Question

xy'-y-√(x^2+y^2)=0的通解

Answer 1

解：∵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)。

Answer 2

P=y/x,即y=px,dy=pdx+xdp,y'=p+xdp/dx
xy'-y-√(x^2+y^2)=0

y'-p-√(x1+P^2)=0

xdp/dx-√(1+p^2)=0
dx/（x）=dp/√(1+p^2)
两边积分lnx=c+arcsinp=c+arcsin(y/x)