∵dy/dx-2y/(x+1)=0 ==>dy/y=2dx/(x+1)
==>ln|y|=2ln|x+1|+ln|C| (C是积分常数)
==>y=C/(x+1)²
∴设原方程的通解为 y=C(x)/(x+1)² (C(x)表示关于x的函数)
代入原方程整理得 C'(x)=(x+1)^4 ==>C(x)=1/5(x+1)^5+C (C是积分常数)
==>y=(x+1)³/5+C/(x+1)²
故原方程的通解是 y=(x+1)³/5+C/(x+1)² (C是积分常数)