设y'=p(y),则y''=pp'(y),
所以pp'(y)=p^3+p,
分离变量得dp/(p^2+1)=dy,
积分得arctanp=y+c,
所以y'=p=tan(y+c),
所以dy/tan(y+c)=dx,
ln[sin(y+c)]=x+c2,
sin(y+c)=e^(x+c2),为所求。本回答被网友采纳

设 y' = p(y)， 则 y'' = dp/dx = [dp(y)/dy](dy/dx) = p(y)[dp(y)/dy]
微分方程化为 p(y)[dp(y)/dy] = p(y)^3 + p(y)
p(y) = 0, 或 dp(y)/dy = 1+p(y)^2
解 y' = p(y) = 0, 得 y = C
解 dp(y)/dy = 1+p(y)^2， dp(y)/[1+p(y)^2] = dy,
arctanp(y) = y+C1, y' = p(y) = tan(y+C1)
cot(y+C1)dy = dx, ln[sin(y+C1)] = x + lnC2
sin(y+C1) = C2e^x
通解为 sin(y+C1) = C2e^x 或 y = C

令y'=u(x)，解出u(x) = ±1/sqrt(-1+exp(-2*x)*_C1)-->
y(x) =± arctan(sqrt(-1+exp(-2*x)*_C1))+_C2