令p=y'
则y"=pdp/dy
代入原式：pdp/dy+p=py
dp/dy+1=y
dp=(y-1)dy
积分：p=(y-1)²/2+c1
即dy/dx=(y-1)²/2+c1
2dy/[(y-1)²+2c1]=dx
积分：
若c1=0,有-2/(y-1)=x+C2
若c1>0,有√(2/c1)arctan[(y-1)/√(2c1)]=x+c2
若c1
令p=y'
则y"=pdp/dy
代入原式：pdp/dy+p=py
dp/dy+1=y
dp=(y-1)dy
积分：p=(y-1)²/2+c1
即dy/dx=(y-1)²/2+c1
2dy/[(y-1)²+2c1]=dx
积分：
若c1=0,有-2/(y-1)=x+C2
若c1>0,有√(2/c1)arctan[(y-1)/√(2c1)]=x+c2
若c1