当然是对y求导了,这是隐函数,关键是把y当作x的复合函数看待来解
原式=y+xy'-cos(πy^2)*2πy*y'=0
y'(x-2πycos(πy^2))=-y
y'=-y /x-2πycos(πy^2)
y''=(-y'(x-2πycos(πy^2))+y(x-2πycos(πy^2))')/(x-2πycos(πy^2))^2=...........以下打出来太繁,但算起来简单的,算到最后一步再把
y'=-y /x-2πycos(πy^2)
代进去就行了
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