xy+sin(πy^2)=0两边同时对x求导,得y+xy'+cos(πy^2)*2πy*y'=0y'=-y/(x+2πycos(πy^2))y'(0,1)=-1/2πcosπ=1/(2π)y''=-(y'(x+2πycos(πy^2))-y(1+2πcos(πy^2)-2πysin(πy^2)*2πyy'))/(x+2πycos(πy^2))^2=...不知道你过程对不对但答案错了答案是1/4π^2我也感觉答案有问题,自己代入验证吧!y''=-(y'(x+2πycos(πy^2))-y(1+2πy'cos(πy^2)-2πysin(πy^2)*2πyy'))/(x+2πycos(πy^2))^2=-(1/(2π)*(-2π)-(1-2π*1/(2π)-0))/(-2π)^2=-(-1)/((2π)^2) =1/4π^2