Question

dy/dx=(x+y)^1/2+y/x通解

Answer 1

解：令x+y=t，则扮晌dy/dx=dt/dx-1
代入原方程，得dt/dx-1=√t+(t-x)/x
==>dt/dx=√t+t/x........(1)
令√t=z，则dt/dx=2zdz/dx
代入方程(1)，得2zdz/乎缺晌dx=z+z²/x
==>2dz/dx=1+z/x..........(2)
令z/x=m，则dz/dx=xdm/dx+m
代入方程(2)，得2(xdm/dx+m)=1+m
==>2xdm/dx+m=1
==>2dm/(1-m)=dx/x
==>-2ln│1-m│=ln│x│+ln│C│ (C是积分常数)
==>1/(1-m)²=Cx
==>岁锋1=Cx(1-z/x)² (用z/x=m代换)
==>1=C(x-√t)²/x (用√t=z代换)
==>x=C(x-√(x+y))² (用x+y=t代换)
故原方程的通解是x=C(x-√(x+y))² (C是积分常数)。