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已知xy-e^y=0,求d^2y/dx^2
人气:166 ℃ 时间:2020-04-07 09:41:20
解答
xy-e^y=0y+xdy/dx-e^y·dy/dx=0dy/dx=y/(e^y-x)d²y/dx²=[dy/dx·(e^y-x)-y(e^y·dy/dx-1)]/(e^y-x)²把dy/dx=y/(e^y-x)代入上式得d²y/dx²=[y-y²e^y/(e^y-x)]/(e^y-x)²
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