题目

求xy-sin(πy^2)=0在（0,-1）的二阶导数.

提问时间：2021-04-01

答案

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