设函数y=y(x)由方程xy+e^y=1所确定,求y"(0)
人气:219 ℃ 时间:2020-03-18 15:53:28
解答
xy+e^y=1
e^y(0) =1
y(0) = 0
xy'+y+e^y y'=0
0+y(0) + y'(0) =0
y'(0) = 0
xy''+y'+ y' + e^y y'' + (y')^2e^y =0
0 +2y'(0)+ y''(0) + (y'(0))^2e^0 =0
y''(0) =0
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