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请问 设y=y(x)有方程2x-tan(x-y)=∫上限x-y下限0 [sec(t)]^2d所确定,求d^2y/dx^2

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请问 设y=y(x)有方程2x-tan(x-y)=∫上限x-y下限0 [sec(t)]^2d所确定,求d^2y/dx^2
请问 设y=y(x)有方程2x-tan(x-y)=∫上限x-y下限0 [sec(t)]^2d所确定,求d^2y/dx^2
2x-tan(x-y)=∫(0,x-y) [sec(t)]^2dt
两边对x求导得:
2-sec²(x-y)(1-y')=sec²(x-y)(1-y')
sec²(x-y)(1-y')=1
y'=1-cos²(x-y)=sin²(x-y)
y''=2sin(x-y)cos(x-y)(1-y')
=sin(2x-2y)*cos²(x-y)