Question

求解微分方程dy/dx=(2x+y-1)^2/(x-2)^2

Answer 1

令u=x-2,v=y+3，du=dx,dv=dy,
dy/dx=dv/du=((2u+v)/u)^2=4+4v/u+v^2/u^2
Z=v/u,v=zu,dv=udz+zdu,
dv/du=udz/du+z=4+4z+z^2
udz/du=4+3z+z^2
dz/(4+3z+z^2)=du/u
dz/[25/4+(3/2+z)^2]=du/u
两边积分得
2/5arctgan(3/5+2/5z)=lnu+C
2/5arctgan(3/5+2v/5u)=lnu+C
2/5arctgan{3/5+[2(y+3)/5(x-2)]}=lnu+C

Answer 2

dy/dx=[(2x+y-1)/(x-2)]^2
=[2+(y+3)/(x-2)]^2
(y+3)/(x-2)=u
dy=d(y+3)=ud(x-2)+(x-2)du,dx=d(x-2)
u+(x-2)du/d(x-2)=(2+u)^2
(x-2)du/d(x-2)=u^2+u+1
du/(u^2+u+1)=d(x-2)/(x-2)
du/[(u+1/2)^2+3/4]=dln(x-2)
(2/√3)darctan(2u/√3+1/√3)=dln(x-2)
(2/√3)arctan(2u/√3+1/√3)=ln(x-2)+c0
(2/√3)arctan(2(y+3)/(√3x-2√3)+1/√3)=ln(x-2)+C0