设p=y'=dy/dx
则y"=dp/dx=dp/dy*dy/dx=pdp/dy
代入原方程：pdp/dy=p^3+p
即：dp/(p^2+1)=dy
积分得：arctanp=y+c
即：p=tan(y+c)
dy/tan(y+c)=dx
dy* cos(y+c)/sin(y+c)=dx
d(sin(y+c))/sin(y+c)=dx
积分：ln|sin(y+c)|=x+c2
得：|sin(y+c)|=c1e^x