求解微分方程dy/dx=(2x+y-1)^2/(x-2)^2
问题描述：
求解微分方程dy/dx=(2x+y-1)^2/(x-2)^2
答
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^2Z=v/u,v=zu,dv=udz+zdu,dv/du=udz/du+z=4+4z+z^2udz/du=4+3z+z^2dz/(4+3z+z^2)=du/udz/[25/4+(3/2+z)^2]=du/u两边积分得2/5arctgan(3/5+2/5z)=ln...