一阶微分方程y'=2((y+2)/(x+y-1))^2
问题描述：
一阶微分方程y'=2((y+2)/(x+y-1))^2
答
设m=y+2,n=x-3
∴代入原方程得dm/dn=2[m/(m+n)]²
==>dm/dn=2/(1+n/m)².(1)
再设n/m=t,则dn=tdm+mdt
∴代入方程(1)得dm/(tdm+mdt)=2/(1+t)²
==>t+mdt/dm=(1+t)²/2
==>mdt/dm=(1+t²)/2
==>2dt/(1+t²)=dm/m
==>2arctant=ln│m│-ln│C│ (C是积分常数)
==>e^(2arctant)=m/C
==>m=Ce^(2arctan(n/m))
==>y+2=Ce^(2arctan((x-3)/(y+2))
故原方程的通解是y+2=Ce^(2arctan((x-3)/(y+2)) （C是积分常数）.